Maker Pro
Maker Pro

Low Volts High Current!

W

Winfield Hill

Jan 1, 1970
0
Winfield Hill wrote...
I'm going to sign off now, get breakfast, and shovel the snow. :>)
Then I'll poke around and find an old post of mine to edit and put
up as my entry for Numan's 150A linear regulator.

I came across this old beast, wihch uses 75 pass transistors, for
amusement. It's a 1.8V 250A quad Pentium linear regulator. :>)

Winfield Hill wrote,
 
J

James Meyer

Jan 1, 1970
0
The car battery you have there consists of a number of cells, which are
connected in series. The voltage between the two outermost placed
contacts is 12 Volt.

In between them there are other connections which carry different
voltages. If you can remove the plastiv cover on the battery, or drill
through it, you can connect to any number of cells you like, in series.

I think you have the most satisfactory solution.

The actual voltage isn't extremely important as long as it doesn't cause
the motor to explode immediately upon application. Since a computer is
involved, the power, amps times voltage, is available as one of the parameters
and that should be good enough even if the voltage is a little bit off from the
desired 7.2 volt figure or if it sags a little during the test.

Why throw a lot of electronics at the problem when a little software can
do the job?

Jim
 
W

Winfield Hill

Jan 1, 1970
0
Winfield Hill wrote...
.. 7.2V 150A Linear Regulator
.. BATT 250A fuse by Winfield Hill
.. POS __or breaker
.. =====|__|==(O)===+===+===+===+===+===+===+===+===+===+===+===+===+===+===+
.. +12.5-14.5V | | | | | | | | | | | | | | |
.. ,---+----+------+ | | | | | | | | | | | | | |
.. | | | | | | | | | | | | | | | | | |
.. | | 2R7 | | | | | | | | | | | | | | |
.. | 470 5W | | | | | |
.. | | | | | | Q3 to Q17 = BDW83C, TIP42 or 2n6284 | | |
.. | | |/ Q2 | | | bank of 15 Darlington transistors | | |
.. | +--| ZTX851 | | | | | |
.. | | |\ Q3 | | | | | | | | | | | | | Q17
.. | | V | | | | | | | | | | | | | | |
.. | | | |/ |/ |/ |/ |/ |/ |/ |/ |/ |/ |/ |/ |/ |/ |/
.. | | +====|===|===|===|===|===|===|===|===|===|===|===|===|===|===|
.. | | | |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\
.. | | | V V V V V V V V V V V V V V V
.. | | 220 | | | | | | | | | | | | | | |
.. | | 1W 20m 20m 20m 20m 20m 20m 20m 20m 20m 20m 20m 20m 20m 20m 20m
.. | | | 3W 3W 3W 3W 3W 3W 3W 3W 3W 3W 3W 3W 3W 3W 3W
.. 2k7 | | | | | | | | | | | | | | | | |
.. | | | '===+===+===+===+===+===++==+++=+===+===+===+===+===+==='
.. | | '---------------------, | ||
.. | | | | || +7.20V 0 - 150A
.. | +----100--, | | ++====(O)=============++
.. | | | 0.01 | | ||
.. | '---, +--||--, ,----|---------|---------(O)--------, ||
.. | | | | | | | + sense \ ||
.. | Q1 | __|__ 2k7 4k75 | ,-----+ '--++
.. | |/ | | | | | | | 1000uF ||
.. +-1k--| | |---+----+ | 22 |+ 16V MOTOR
.. | |\ |_____| 2.50V | | 5W === ||
.. | V | IC1 2k49 | | | ,--++
.. | | | TL431 | | | | - sense / ||
.. o +-----+-----------+----+---+-----+---------(o)--------' ||
.. o---33--' ||
.. close = ON ||
.. optional 50mV ||
.. ===========================================+= meter shunt =+=======++
.. BATT NEG

Fifteen places NPN power Darlington, mounted on large aluminum heat sink
with thin heat-sink grease, and no insulating pads. Insulate heat sink from
the chassis and add plastic to protect it from exposure to metal tools, etc.
---

To analyze the circuit we'll start with the 10-milli-ohm emitter resistors,
which are required to insure equal current sharing among the transistors.
These are Ohmite 630HR020 metal-element types, at DigiKey for $0.42 each,
and they'll drop 0.2V at the nominal 10A current each transistor carries at
full load. The Darlington power transistors will have a Vbe drop of up to
2V at 10A, severely eating into the voltage difference available between the
battery and the motor at 150A, so we'll have to be careful with what's left.
(The high Vbe drop is one reason we're limiting each pass transistor to 10A,
which then requires us to use 15 parallel transistors in the pass bank.)

Although we have used high-gain Darlington transistors in an attempt to get
the base drive current down to reasonable levels, assuming a minimum gain of
500 at 10A means Q2 will have to provide over 300mA at full output. Q2 must
have high beta at 300mA, and dissipate up to 1W for high 14.5V battery (i.e.
under charge). A Zetex ZTX851, etc. (available at DigiKey), should work OK.

We cannot use a Darlington transistor at this spot because too much voltage
has been used up by the 15 power Darlington transistor's base-emitter drop
at full current. We have to save some overhead to allow the battery to sag.

In the event of an output short, Q2's 2.7-ohm collector resistor limits its
current, but the power transistors must fend for themselves until the fuse
blows (you can get 250A fuses at auto parts stores). Alternately a ~ 200A
current limit function could be added by amplifying the 50mV shunt voltage,
comparing it to a fixed voltage, and pulling down Q2's base.

Another valuable feature would be a comparator to tell whenever battery
sag means the circuit is running out of headroom, which can be seen by
looking at the control-loop voltage on the TL431's output terminal.

Some folks will say this linear power regulator design illustrates why a
buck switching converter should be used instead. I will not contest their
point, except to point out that I've learned from making SMPS in the 100
to 1000A region that there are many non-trivial issues one will encounter,
dictating knowledge, experience, and good instruments on your workbench.
On the other hand, this circuit can be made and tested by a hobbyist.
 
R

Rich Grise

Jan 1, 1970
0
I have taken the liberty of cross-posting this to s.e.d. because it
can be nicely considered as a nontrivial electronics design problem.
I suggested a commercial 1 or 2kW power supply, such as a Xantrex
XHR 7.5-130, spec'd at 7.5V 130A (but can be externally programmed
to operate to 150A for a short time, and I gave a $350 eBay link,
http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&item=3850054721
That one finished, but the seller still has it, plus a few more...

I was thinking, "Battery charger/boost starter" (among other ideas,
most of which involve huge banks of very pricey batteries), but
then realized:

Welder!

I've seen welders that can hold a very stiff voltage, at arbitrarily
high currents up to hundreds of amps. Search on constant voltage
MIG or TIG or FCAW, and you could lurk in .

Just for the sake of contrariness, I do want to mention that it
sounds very much like they're spec'ing starter motors. :)

<half-anecdote>
Once, at a science fair, some kid asked how he'd get 50 amps at
about 12 volts, and being into ham radio at the time, and since
I practically carried the "Allied" catalog around with me, the
first thing I thought of was stud diodes. Nobody at the time
thought to use a car battery, like was suggested a couple of
months later in Scientific American "Amateur Scientist". It
was for a plasma jet. Nowadays, you just go buy or rent a plasma
cutter. :)
I don't know what the kid finally settled on. But I won a blue
ribbon that year for my two-transistor AM radio. :)
I think it was a winner because the kit came with one transistor,
and I bought another and added it as an audio amp, and kinda
made up the circuit, even though I really didn't have much
Clue at the time. :) The Radio Amateur's Handbook hadn't
heard of transistors yet. :)

Now that I think about it, I think I might have accidentally
invented negative feedback - I had the base bias resistor
going right to that transistor's own collector. Hey, it was
a Lucky WAG, and sound came out!
</half-anecdote>

Good Luck!
Rich
 
R

Rich Grise

Jan 1, 1970
0
Winfield Hill wrote...

I came across this old beast, wihch uses 75 pass transistors, for
amusement. It's a 1.8V 250A quad Pentium linear regulator. :>)

Winfield Hill wrote,

[Absolutely Gorgeous, albeit Mind-Boggling, Circuit Snipped - Thanks!]

"Seat Warmer???!?!?!?" Try "Whole-Carcass Toaster Oven!" ;-)

Cheers!
Rich
 
F

Fred Bloggs

Jan 1, 1970
0
Winfield said:
Winfield Hill wrote...


. 7.2V 150A Linear Regulator
. BATT 250A fuse by Winfield Hill
. POS __or breaker
. =====|__|==(O)===+===+===+===+===+===+===+===+===+===+===+===+===+===+===+
. +12.5-14.5V | | | | | | | | | | | | | | |
. ,---+----+------+ | | | | | | | | | | | | | |
. | | | | | | | | | | | | | | | | | |
. | | 2R7 | | | | | | | | | | | | | | |
. | 470 5W | | | | | |
. | | | | | | Q3 to Q17 = BDW83C, TIP42 or 2n6284 | | |
. | | |/ Q2 | | | bank of 15 Darlington transistors | | |
. | +--| ZTX851 | | | | | |
. | | |\ Q3 | | | | | | | | | | | | | Q17
. | | V | | | | | | | | | | | | | | |
. | | | |/ |/ |/ |/ |/ |/ |/ |/ |/ |/ |/ |/ |/ |/ |/
. | | +====|===|===|===|===|===|===|===|===|===|===|===|===|===|===|
. | | | |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\
. | | | V V V V V V V V V V V V V V V
. | | 220 | | | | | | | | | | | | | | |
. | | 1W 20m 20m 20m 20m 20m 20m 20m 20m 20m 20m 20m 20m 20m 20m 20m
. | | | 3W 3W 3W 3W 3W 3W 3W 3W 3W 3W 3W 3W 3W 3W 3W
. 2k7 | | | | | | | | | | | | | | | | |
. | | | '===+===+===+===+===+===++==+++=+===+===+===+===+===+==='
. | | '---------------------, | ||
. | | | | || +7.20V 0 - 150A
. | +----100--, | | ++====(O)=============++
. | | | 0.01 | | ||
. | '---, +--||--, ,----|---------|---------(O)--------, ||
. | | | | | | | + sense \ ||
. | Q1 | __|__ 2k7 4k75 | ,-----+ '--++
. | |/ | | | | | | | 1000uF ||
. +-1k--| | |---+----+ | 22 |+ 16V MOTOR
. | |\ |_____| 2.50V | | 5W === ||
. | V | IC1 2k49 | | | ,--++
. | | | TL431 | | | | - sense / ||
. o +-----+-----------+----+---+-----+---------(o)--------' ||
. o---33--' ||
. close = ON ||
. optional 50mV ||
. ===========================================+= meter shunt =+=======++
. BATT NEG

Fifteen places NPN power Darlington, mounted on large aluminum heat sink
with thin heat-sink grease, and no insulating pads. Insulate heat sink from
the chassis and add plastic to protect it from exposure to metal tools, etc.
---

To analyze the circuit we'll start with the 10-milli-ohm emitter resistors,
which are required to insure equal current sharing among the transistors.
These are Ohmite 630HR020 metal-element types, at DigiKey for $0.42 each,
and they'll drop 0.2V at the nominal 10A current each transistor carries at
full load. The Darlington power transistors will have a Vbe drop of up to
2V at 10A, severely eating into the voltage difference available between the
battery and the motor at 150A, so we'll have to be careful with what's left.
(The high Vbe drop is one reason we're limiting each pass transistor to 10A,
which then requires us to use 15 parallel transistors in the pass bank.)

Although we have used high-gain Darlington transistors in an attempt to get
the base drive current down to reasonable levels, assuming a minimum gain of
500 at 10A means Q2 will have to provide over 300mA at full output. Q2 must
have high beta at 300mA, and dissipate up to 1W for high 14.5V battery (i.e.
under charge). A Zetex ZTX851, etc. (available at DigiKey), should work OK.

We cannot use a Darlington transistor at this spot because too much voltage
has been used up by the 15 power Darlington transistor's base-emitter drop
at full current. We have to save some overhead to allow the battery to sag.

In the event of an output short, Q2's 2.7-ohm collector resistor limits its
current, but the power transistors must fend for themselves until the fuse
blows (you can get 250A fuses at auto parts stores). Alternately a ~ 200A
current limit function could be added by amplifying the 50mV shunt voltage,
comparing it to a fixed voltage, and pulling down Q2's base.

Another valuable feature would be a comparator to tell whenever battery
sag means the circuit is running out of headroom, which can be seen by
looking at the control-loop voltage on the TL431's output terminal.

Some folks will say this linear power regulator design illustrates why a
buck switching converter should be used instead. I will not contest their
point, except to point out that I've learned from making SMPS in the 100
to 1000A region that there are many non-trivial issues one will encounter,
dictating knowledge, experience, and good instruments on your workbench.
On the other hand, this circuit can be made and tested by a hobbyist.

Eh- you have approximately doubled the required AH of the battery.
 
F

Fred Bloggs

Jan 1, 1970
0
Winfield said:
Winfield Hill wrote...


. 7.2V 150A Linear Regulator
. BATT 250A fuse by Winfield Hill
. POS __or breaker
. =====|__|==(O)===+===+===+===+===+===+===+===+===+===+===+===+===+===+===+
. +12.5-14.5V | | | | | | | | | | | | | | |
. ,---+----+------+ | | | | | | | | | | | | | |
. | | | | | | | | | | | | | | | | | |
. | | 2R7 | | | | | | | | | | | | | | |
. | 470 5W | | | | | |
. | | | | | | Q3 to Q17 = BDW83C, TIP42 or 2n6284 | | |
. | | |/ Q2 | | | bank of 15 Darlington transistors | | |
. | +--| ZTX851 | | | | | |
. | | |\ Q3 | | | | | | | | | | | | | Q17
. | | V | | | | | | | | | | | | | | |
. | | | |/ |/ |/ |/ |/ |/ |/ |/ |/ |/ |/ |/ |/ |/ |/
. | | +====|===|===|===|===|===|===|===|===|===|===|===|===|===|===|
. | | | |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\
. | | | V V V V V V V V V V V V V V V
. | | 220 | | | | | | | | | | | | | | |
. | | 1W 20m 20m 20m 20m 20m 20m 20m 20m 20m 20m 20m 20m 20m 20m 20m
. | | | 3W 3W 3W 3W 3W 3W 3W 3W 3W 3W 3W 3W 3W 3W 3W
. 2k7 | | | | | | | | | | | | | | | | |
. | | | '===+===+===+===+===+===++==+++=+===+===+===+===+===+==='
. | | '---------------------, | ||
. | | | | || +7.20V 0 - 150A
. | +----100--, | | ++====(O)=============++
. | | | 0.01 | | ||
. | '---, +--||--, ,----|---------|---------(O)--------, ||
. | | | | | | | + sense \ ||
. | Q1 | __|__ 2k7 4k75 | ,-----+ '--++
. | |/ | | | | | | | 1000uF ||
. +-1k--| | |---+----+ | 22 |+ 16V MOTOR
. | |\ |_____| 2.50V | | 5W === ||
. | V | IC1 2k49 | | | ,--++
. | | | TL431 | | | | - sense / ||
. o +-----+-----------+----+---+-----+---------(o)--------' ||
. o---33--' ||
. close = ON ||
. optional 50mV ||
. ===========================================+= meter shunt =+=======++
. BATT NEG

Fifteen places NPN power Darlington, mounted on large aluminum heat sink
with thin heat-sink grease, and no insulating pads. Insulate heat sink from
the chassis and add plastic to protect it from exposure to metal tools, etc.
---

To analyze the circuit we'll start with the 10-milli-ohm emitter resistors,
which are required to insure equal current sharing among the transistors.
These are Ohmite 630HR020 metal-element types, at DigiKey for $0.42 each,
and they'll drop 0.2V at the nominal 10A current each transistor carries at
full load. The Darlington power transistors will have a Vbe drop of up to
2V at 10A, severely eating into the voltage difference available between the
battery and the motor at 150A, so we'll have to be careful with what's left.
(The high Vbe drop is one reason we're limiting each pass transistor to 10A,
which then requires us to use 15 parallel transistors in the pass bank.)

Although we have used high-gain Darlington transistors in an attempt to get
the base drive current down to reasonable levels, assuming a minimum gain of
500 at 10A means Q2 will have to provide over 300mA at full output. Q2 must
have high beta at 300mA, and dissipate up to 1W for high 14.5V battery (i.e.
under charge). A Zetex ZTX851, etc. (available at DigiKey), should work OK.

We cannot use a Darlington transistor at this spot because too much voltage
has been used up by the 15 power Darlington transistor's base-emitter drop
at full current. We have to save some overhead to allow the battery to sag.

In the event of an output short, Q2's 2.7-ohm collector resistor limits its
current, but the power transistors must fend for themselves until the fuse
blows (you can get 250A fuses at auto parts stores). Alternately a ~ 200A
current limit function could be added by amplifying the 50mV shunt voltage,
comparing it to a fixed voltage, and pulling down Q2's base.

Another valuable feature would be a comparator to tell whenever battery
sag means the circuit is running out of headroom, which can be seen by
looking at the control-loop voltage on the TL431's output terminal.

Some folks will say this linear power regulator design illustrates why a
buck switching converter should be used instead. I will not contest their
point, except to point out that I've learned from making SMPS in the 100
to 1000A region that there are many non-trivial issues one will encounter,
dictating knowledge, experience, and good instruments on your workbench.
On the other hand, this circuit can be made and tested by a hobbyist.

The answer to this may be a hobbyized version of a multiphase buck
converter- maybe summing currents at the load destination - to break the
problem into manageable pieces:
http://powerelectronics.com/mag/power_capacitor_ripple_current/
 
W

Winfield Hill

Jan 1, 1970
0
Fred Bloggs wrote...
Eh- you have approximately doubled the required AH of the battery.

Indeed, but you'll have to agree it's a big efficiency improvement
over my 42V to 1.8V linear regulator / seat warmer.
 
K

Ken Smith

Jan 1, 1970
0
150A regulator. He delivers his kW power level for only a few
seconds, and during that time dissipates 150A * 5V at most (likely
less, due to battery and cabling voltage drops), which amounts to
about 450 * 3 watt-sec = 1350J during that time. Unlike ordinary
linear kW supplies, this won't present a serious heat sink problem.

The TIP35 and TIP36 transistors are fairly cheap. I think I'd start like
this:

10 each TIP35
------------------- ---/\/\---
! \ /e !
! ----- !
! TIP36 ! !
+-/\/\/-- -----------/\/\------+
! e \ / !
! ------ !
! ! --------- !
---+--/\/\/\/--+-----! LM317 !---------+----[/QUOTE]
 
W

Winfield Hill

Jan 1, 1970
0
Ken Smith wrote...
Winfield Hill wrote:
[.....]
150A regulator. He delivers his kW power level for only a few
seconds, and during that time dissipates 150A * 5V at most (likely
less, due to battery and cabling voltage drops), which amounts to
about 450 * 3 watt-sec = 1350J during that time. Unlike ordinary
linear kW supplies, this won't present a serious heat sink problem.

The TIP35 and TIP36 transistors are fairly cheap. I think I'd start
like this:

10 each TIP35
------------------- ---/\/\---
! \ /e !
! ----- !
! TIP36 ! !
+-/\/\/-- -----------/\/\------+
! e \ / !
! ------ !
! ! --------- !
---+--/\/\/\/--+-----! LM317 !---------+----

Hmm, TIP35 transistors are not Darlington types, which means low
beta, but also lower Vbe. If each transistor conducts 15A with a
beta of 15, then the TIP36 PNP has to handle 10A, which it can do
just fine. If its gain is 25, the LM317 has to handle 0.5A, which
is also fine. At 500mA the LM317's dropout voltage is 1.7V, which
with a Vbe of 2V for a TIP36 at 10A gives us an overall regulator
dropout voltage of 3.7 volts Adding this to the 7.2V output tells
us the battery must stay above 10.9 volts for regulation at 150A.
If the transistor betas are a bit higher, and if the Vbe voltages
are a bit lower, the battery will be able to sag more, and there
may be room left for a cable drop. Now working the TIP35 pathway,
say the emitter resistors drop 0.2V and Vbe = 2V, this leaves 1.5
volts for the TIP36 Vce(sat), assuming a 10.9V battery. Everything
is working out, because 1.5 volts is what the TIP36 datasheet tells
us to expect for Vce(sat) at 10A.

So this is a viable approach, perhaps even better than my drawing.

Here's a small NPN power transistor spec comparison table. All
are big TO-218 style plastic packages, except the 2n6284 is TO-3
(rated Tj = 200C). The TIP35 and TIP36 are NOT Darlington types.

part thermal Hfe @ Ic Vbe @ Ic
number C/W Pd 10A 15A 10A 15A
------ --- --- ---- ---- ---- ----
TIP35 1.0 125W 30 >15 < 2V 2V (also PNP TIP36)
2n6284 1.09 160 750 500 < 2 2.5
BDW83C 0.96 130 750 >100 < 2.5 ?
TIP142 1.0 125 500 ? 2 ~ 2.5

All these transistors are inexpensive and widely available, but
they're comparatively wimpy. For example, a 2n5686 (see below) is
a 50A npn TO-3, rated at 300W, and it's much better suited for the
task, with higher beta and lower Vbe. But it's $11.76 at DigiKey.
The 60A MJ14002 is a bit less at $10.36 qty 10, stocked by DigiKey.

part thermal Hfe @ Ic Vbe @ Ic
number C/W Pd 10A 15A 10A 15A
------ ----- ---- ---- ---- ---- ----
2n5686 0.584 300W 70 50 0.85V 0.9V TO-3 non-Darlington
MJ14002 0.584 300W 100 70 0.8V 0.85V " "

Using one of these transistors would open the possibility of battery
voltages lower than 10.9V, although using a low-dropout LT1085, etc.,
instead of a LM317 (0.9V compared to 1.7V at 0.5A) would be required.
Then a 10V sagging battery + cable voltage could be allowed at 150A.
 
K

Ken Smith

Jan 1, 1970
0
Ken Smith wrote...

[...]
For the OP I'm adding a bit of extra working things out.

I've added a capacitor to the drawing called "Cbig" and labled some
resistors.

The TIP35 and TIP36 are slow. They are extra slow when their collector
voltage drops below about 2V. Along with the usual bypass an LM317 needs,
you also need some added largish capacitor to make up for the fact that it
will take the TIP35,36 stuff time to throttle up and down. The
minimum value of this capacitor I would extimate like this:

Assume that the 150A draw starts suddenly.
Assume that the voltage can droop or spike up by 0.1V.

The TIP35 and 26 have a rise and fall time of about 0.25uS (Check data
sheet to be sure)

The TIP36 has to turn on and then the TIP35 so the total time is 0.5uS.

Rather than having to integrate, we just say the TIP35 and TIP36 stuff
doesn't do anything for 0.5uS and get an over estimate of the minimum.

In 0.5uS, 150A will transfer a charge of 75uC.

The voltage change on the Cbig will be:

V = Q / C

where
Q is the charge
C is the capacitance

We solve for C


C = Q / V = 75uC / 0.1 = 750uF

Now here's the messy bit. We have assumed tha Cbig has no resistance.
All real parts have some resistance. We now need to find the data sheet
on some lets say, 1000uF capacitor with a nice low Equivelent Series
Resistance (ESR) to see how much drop will appear in the ESR. If that
plus the droop in the 1000uF is more than the specification, you need a
larger capacitor.


Finding R1:

If the circuit is lightly loaded, we want the LM317 to pass all of the
current. This means that there should be less than about 0.6V on R1 when
a "small current" is drawn. We will define a small current as 50mA os we
can say

R1 = 0.6V / 0.05A = 12 Ohms

So we make R1 about 12 Ohms.


About R2:

R2 could in most cases be zero. It lowers the gain of the TIP36 part of
things and works to reduce the risk of oscillation. I'd make this a low
inductance resistor of about 0.01 Ohms or a short length of wire.


About R3:

We want about an Amp flowing in the TIP36 before the TIP35s are brought
into things. R3 is something like 0.1 to 1.0 Ohms.


About R4:

This is actually 10 resistors, one resistor in the emitter leg of each
TIP35. When a bipolar transistor heats up, its Vbe drop decreases. If R4
wasn't there, the hottest TIP35 would end up taking on more than its share
of the load, get hotter, draw more, get hotter and etc. This could
destroy one transistor.

Thermal resistance is a lot like electrical resistance. Heat sources are
like current sources. All of the heat must flow out of one transistor
through all of the stuff it goes through to get out to the environment.
In process, one transistor will tend to warm up its brothers. To find the
minimum R4 we start with assuming a current flows in one transistor and
work out how much its temperature rises above the coolest of its brothers.
To make the math easy, just assume 1A flows.

One you have the temperature rise, assume that the individual R4 must add
at least 2.5mV of drop for every amp flowing. It is likely that this will
be such a small value that using 10 wires running from the TIP35s to the
load instead of one big oun will be enough to do it.



A note about layout:

You want Cbig to be as close to the load as practial. Ideally, there
should be 12 wires to the (+) side of the load. The LM317, the TIP36 and
all the TIP35s each have a wire. These wires should be tightly bundled
together.


A note about fuses:

They make special fuses for things like car starter motors. They look
like a thick cable with a wide section in it. Put one of these between
the car battery and the circuit. A car battery can produce nearly
infinite current for a short time.
 
T

Tom Seim

Jan 1, 1970
0
Hello everyone, I need some help,

I have to convert voltage from a 12 volt car battery to 7.2 volts, BUT
be able to draw 150 amps for a short time (approximately 2 or 3
seconds). The voltage output must be accurate, can anyone help
regarding design and recommended components please? As I understand it
a potentiometer would be no good as it couldnt cope with the high
current or am I wrong? Assistance please.

Mark.

I recently built an electronic load that could handle 100 A using a
power hexfet from Intl. Rectifier. This one can handle 150 A with
proper heatsinking:
http://www.irf.com/product-info/datasheets/data/fb180sa10.pdf
You will need a feedback control circuit to regulate the voltage at
your load. An op amp that has your setpoint voltage to the + input,
the load voltage to the - input and the output to the hexfet gate
should do the job.
 
R

Rich Grise

Jan 1, 1970
0
Win Hill, You "Slut"! ;-) You're not supposed to be _giving_ this stuff
away! This kind of stuff is why you get paid to write books!

Thanks Again!
Rich

[quoted awe-inspiring(as usual) exposition retained below]
Ken Smith wrote...
Winfield Hill wrote:
[.....]
150A regulator. He delivers his kW power level for only a few
seconds, and during that time dissipates 150A * 5V at most (likely
less, due to battery and cabling voltage drops), which amounts to
about 450 * 3 watt-sec = 1350J during that time. Unlike ordinary
linear kW supplies, this won't present a serious heat sink problem.

The TIP35 and TIP36 transistors are fairly cheap. I think I'd start
like this:

10 each TIP35
------------------- ---/\/\---
! \ /e !
! ----- !
! TIP36 ! !
+-/\/\/-- -----------/\/\------+
! e \ / !
! ------ !
! ! --------- !
---+--/\/\/\/--+-----! LM317 !---------+----

Hmm, TIP35 transistors are not Darlington types, which means low
beta, but also lower Vbe. If each transistor conducts 15A with a
beta of 15, then the TIP36 PNP has to handle 10A, which it can do
just fine. If its gain is 25, the LM317 has to handle 0.5A, which
is also fine. At 500mA the LM317's dropout voltage is 1.7V, which
with a Vbe of 2V for a TIP36 at 10A gives us an overall regulator
dropout voltage of 3.7 volts Adding this to the 7.2V output tells
us the battery must stay above 10.9 volts for regulation at 150A.
If the transistor betas are a bit higher, and if the Vbe voltages
are a bit lower, the battery will be able to sag more, and there
may be room left for a cable drop. Now working the TIP35 pathway,
say the emitter resistors drop 0.2V and Vbe = 2V, this leaves 1.5
volts for the TIP36 Vce(sat), assuming a 10.9V battery. Everything
is working out, because 1.5 volts is what the TIP36 datasheet tells
us to expect for Vce(sat) at 10A.

So this is a viable approach, perhaps even better than my drawing.

Here's a small NPN power transistor spec comparison table. All
are big TO-218 style plastic packages, except the 2n6284 is TO-3
(rated Tj = 200C). The TIP35 and TIP36 are NOT Darlington types.

part thermal Hfe @ Ic Vbe @ Ic
number C/W Pd 10A 15A 10A 15A
------ --- --- ---- ---- ---- ----
TIP35 1.0 125W 30 >15 < 2V 2V (also PNP TIP36)
2n6284 1.09 160 750 500 < 2 2.5
BDW83C 0.96 130 750 >100 < 2.5 ?
TIP142 1.0 125 500 ? 2 ~ 2.5

All these transistors are inexpensive and widely available, but
they're comparatively wimpy. For example, a 2n5686 (see below) is
a 50A npn TO-3, rated at 300W, and it's much better suited for the
task, with higher beta and lower Vbe. But it's $11.76 at DigiKey.
The 60A MJ14002 is a bit less at $10.36 qty 10, stocked by DigiKey.

part thermal Hfe @ Ic Vbe @ Ic
number C/W Pd 10A 15A 10A 15A
------ ----- ---- ---- ---- ---- ----
2n5686 0.584 300W 70 50 0.85V 0.9V TO-3 non-Darlington
MJ14002 0.584 300W 100 70 0.8V 0.85V " "

Using one of these transistors would open the possibility of battery
voltages lower than 10.9V, although using a low-dropout LT1085, etc.,
instead of a LM317 (0.9V compared to 1.7V at 0.5A) would be required.
Then a 10V sagging battery + cable voltage could be allowed at 150A.
 
W

Winfield Hill

Jan 1, 1970
0
Ken Smith wrote...
Ken Smith wrote...

[...]
For the OP I'm adding a bit of extra working things out.

I've added a capacitor to the drawing called "Cbig" and labled some
resistors.

The TIP35 and TIP36 are slow. They are extra slow when their collector
voltage drops below about 2V. Along with the usual bypass an LM317 needs,
you also need some added largish capacitor to make up for the fact that it
will take the TIP35,36 stuff time to throttle up and down. The
minimum value of this capacitor I would extimate like this:

Assume that the 150A draw starts suddenly.
Assume that the voltage can droop or spike up by 0.1V.

The TIP35 and 26 have a rise and fall time of about 0.25uS (Check data
sheet to be sure)

The TIP36 has to turn on and then the TIP35 so the total time is 0.5uS.

Rather than having to integrate, we just say the TIP35 and TIP36 stuff
doesn't do anything for 0.5uS and get an over estimate of the minimum.

In 0.5uS, 150A will transfer a charge of 75uC.

The voltage change on the Cbig will be:

V = Q / C

where
Q is the charge
C is the capacitance

We solve for C


C = Q / V = 75uC / 0.1 = 750uF

Now here's the messy bit. We have assumed tha Cbig has no resistance.
All real parts have some resistance. We now need to find the data sheet
on some lets say, 1000uF capacitor with a nice low Equivelent Series
Resistance (ESR) to see how much drop will appear in the ESR. If that
plus the droop in the 1000uF is more than the specification, you need a
larger capacitor.


Finding R1:

If the circuit is lightly loaded, we want the LM317 to pass all of the
current. This means that there should be less than about 0.6V on R1 when
a "small current" is drawn. We will define a small current as 50mA os we
can say

R1 = 0.6V / 0.05A = 12 Ohms

So we make R1 about 12 Ohms.


About R2:

R2 could in most cases be zero. It lowers the gain of the TIP36 part of
things and works to reduce the risk of oscillation. I'd make this a low
inductance resistor of about 0.01 Ohms or a short length of wire.


About R3:

We want about an Amp flowing in the TIP36 before the TIP35s are brought
into things. R3 is something like 0.1 to 1.0 Ohms.


About R4:

This is actually 10 resistors, one resistor in the emitter leg of each
TIP35. When a bipolar transistor heats up, its Vbe drop decreases. If R4
wasn't there, the hottest TIP35 would end up taking on more than its share
of the load, get hotter, draw more, get hotter and etc. This could
destroy one transistor.

Thermal resistance is a lot like electrical resistance. Heat sources are
like current sources. All of the heat must flow out of one transistor
through all of the stuff it goes through to get out to the environment.
In process, one transistor will tend to warm up its brothers. To find the
minimum R4 we start with assuming a current flows in one transistor and
work out how much its temperature rises above the coolest of its brothers.
To make the math easy, just assume 1A flows.

One you have the temperature rise, assume that the individual R4 must add
at least 2.5mV of drop for every amp flowing. It is likely that this will
be such a small value that using 10 wires running from the TIP35s to the
load instead of one big oun will be enough to do it.



A note about layout:

You want Cbig to be as close to the load as practial. Ideally, there
should be 12 wires to the (+) side of the load. The LM317, the TIP36 and
all the TIP35s each have a wire. These wires should be tightly bundled
together.


A note about fuses:

They make special fuses for things like car starter motors. They look
like a thick cable with a wide section in it. Put one of these between
the car battery and the circuit. A car battery can produce nearly
infinite current for a short time.

Winfield Hill wrote...
. 7.2V 150A Linear Regulator
. BATT 250A fuse by Winfield Hill
. POS __or breaker
. =====|__|==(O)===+===+===+===+===+===+===+===+===+===+===+===+===+===+===+
. +12.5-14.5V | | | | | | | | | | | | | | |

Yes that's a good point. I learned about this by examining my 1985 Cougar
with the help of Chilton and Haynes manuals. The engine compartment is
filled with 20, 30, 40 and 60A distribution fuses for various things, plus
single 175A "Mega fuse" to the generator and "distribution block" (called
a fuseable link on some models). Mine is a removable plastic-encased fuse.

There is also a 15A and an 80A fuse bypassing the Mega fuse. In the case
of these Ford models the starter motor is _not_ fused. With some effort
these big replaceable fuses can be procured as automotive spare parts.
 
M

Mark Jones

Jan 1, 1970
0
Winfield said:
I have taken the liberty of cross-posting this to s.e.d. because it
can be nicely considered as a nontrivial electronics design problem.







NumanR also wrote ...

Hmm... once I ripped 6 huge electrolytic caps from some old machine
- it might have been a photocopying machine, dunno. It had some kind
of primitive fixed disk in it (and a bubble memory of the magnetic
bead variety, imagine that.) Anyways the caps were at least 4" in
diameter by 10" long and were rated for 16V/250,000uF each. So yep, I
wired them all in parallel and charged them to 12V through a 5kW
autotransformer (used as a potentiometer.)

WOW, instant spot-welder. Jumping wires, permanent magnetization of
nearby metal objects, neat. Today, you could probably do even better
with the 2.0 Farad "supercaps" sold to car audio enthusiasts. (Their
explosion-resistance should be checked though...)

But the point is this, an idea might be to use a linear supply to
charge an array of supercaps to about 8V, then (using an automotive
solenoid or two) short these to your motor. This will provide a large
inrush current for a short time, enough to get your motor spun up.
Also switch on a big 7.2V linear supply to power the motor as the caps
discharge.

Or better yet, use a PWM method and straight 12V to drive the motor,
as car batteries are cheap compared to kW-rated supplies. If RPM is
the goal, PWM is the best method to obtain accurate speed of a DC
motor. A microcontroller could sample the shaft RPM and base the PWM
duty cycle off that easily. If the design goal is measuring how much
power it takes to spin the motor at that speed (how resistive the load
is) then use PWM for accurate speed but measure the motor's
steady-state current and calculate the power delta from that.

IGBT's make a good PWM driver stage with their built-in fast
commutating diodes, high voltage and current ratings, and simplified
drive characteristics. If a very low Rds is achieved, (max Vgs),
(relatively) small heatsinks could be used with (relatively) little
loss. (Might not make a full seat-warmer, but that motor is going to
get hot no matter what.) And hey, the motor placed under your seat
would give a massage along with heat!

-M
 
G

Guy Macon

Jan 1, 1970
0
Mark said:
Hmm... once I ripped 6 huge electrolytic caps from some old machine
- it might have been a photocopying machine, dunno. It had some kind
of primitive fixed disk in it (and a bubble memory of the magnetic
bead variety, imagine that.) Anyways the caps were at least 4" in
diameter by 10" long and were rated for 16V/250,000uF each. So yep, I
wired them all in parallel and charged them to 12V through a 5kW
autotransformer (used as a potentiometer.)

WOW, instant spot-welder. Jumping wires, permanent magnetization of
nearby metal objects, neat. Today, you could probably do even better
with the 2.0 Farad "supercaps" sold to car audio enthusiasts. (Their
explosion-resistance should be checked though...)

The Supercaps have such a high ESR that you can't get much current
out of them. You also can't charge them fast or discharge them fast,
which can be a real pain when designing high-speed testers for
products that include them.
 
A

Asa Cannell

Jan 1, 1970
0
You could charge up some ultracapacitors and then place them across
your load with a big switch. You can buy some 2700 FARAD 2.5V
ultracapacitors at http://www.exoticelectron.com. You would need 3 in
series, so your capacitance would be 900F. Charge them up to 7.2V with
any power supply (might take awhile), then place them across your load
with a huge switch or relay (dont leave it on or the caps will
overheat). 150amps will lower the voltage by 166mV per second. The
caps are rated at 100amps max discharge for 5 seconds, they might be
able to peak at 150amps for a shorter time.

Asa
 
F

Frithiof Andreas Jensen

Jan 1, 1970
0
Winfield Hill said:
Winfield Hill wrote...
Some folks will say this linear power regulator design illustrates why a
buck switching converter should be used instead.

I'd say instead that switchers are too complicated for this application in
the first place and they have the very real disadvantage that it is harder
to take advantage of an intermittent load - because switchers are smaller
and lighter, there is less thermal inertia and they must be rated closer to
the peak load - meaning we have to buy power handling capabilities that we
don't need.

I *would* consider modifying the circuit to use (an) IGBT's instead of
bipolars for the pass transistors; IGBT's are available in bigger packages
than bipolar with screw connectors more suitable to the bus-wiring one needs
for the 150A. The IGBT would have to go 'below' the motor in order to drive
it.
 
K

Ken Smith

Jan 1, 1970
0
Frithiof Andreas Jensen said:
I'd say instead that switchers are too complicated for this application in
the first place

I disagree with this. A 150A switcher is not all that hard to do and
would make a lot less heat.
and they have the very real disadvantage that it is harder
to take advantage of an intermittent load - because switchers are smaller
and lighter, there is less thermal inertia and they must be rated closer to
the peak load - meaning we have to buy power handling capabilities that we
don't need.

No, if you have to make the devices rated at X, you have to make them
rated at X. This is not a capability you don't need. By definition, you
need it.

I *would* consider modifying the circuit to use (an) IGBT's instead of
bipolars for the pass transistors; IGBT's are available in bigger packages
than bipolar with screw connectors more suitable to the bus-wiring one needs
for the 150A. The IGBT would have to go 'below' the motor in order to drive
it.

I think the one IGBT would cost as much as the 11 transistors in my design
and not be as good of a design. IGBTs that are rated for linear operation
are less common than those rated for switching. The electronics to
control the IGBT would be more complex than my proposed circuit. All
round, I think this application is better with bipolars.
 
G

Genome

Jan 1, 1970
0
Winfield Hill said:
Winfield Hill wrote...

. 7.2V 150A Linear Regulator
. BATT 250A fuse by Winfield Hill
. POS __or breaker
. =====|__|==(O)===+===+===+===+===+===+===+===+===+===+===+===+===+===+===+
. +12.5-14.5V | | | | | | | | | | | | | | |
. ,---+----+------+ | | | | | | | | | | | | | |
. | | | | | | | | | | | | | | | | | |
. | | 2R7 | | | | | | | | | | | | | | |
. | 470 5W | | | | | |
. | | | | | | Q3 to Q17 = BDW83C, TIP42 or 2n6284 | | |
. | | |/ Q2 | | | bank of 15 Darlington transistors | | |
. | +--| ZTX851 | | | | | |
. | | |\ Q3 | | | | | | | | | | | | | Q17
. | | V | | | | | | | | | | | | | | |
. | | | |/ |/ |/ |/ |/ |/ |/ |/ |/ |/ |/ |/ |/ |/ |/
. | | +====|===|===|===|===|===|===|===|===|===|===|===|===|===|===|
. | | | |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\
. | | | V V V V V V V V V V V V V V V
. | | 220 | | | | | | | | | | | | | | |
. | | 1W 20m 20m 20m 20m 20m 20m 20m 20m 20m 20m 20m 20m 20m 20m 20m
. | | | 3W 3W 3W 3W 3W 3W 3W 3W 3W 3W 3W 3W 3W 3W 3W
. 2k7 | | | | | | | | | | | | | | | | |
. | | | '===+===+===+===+===+===++==+++=+===+===+===+===+===+==='
. | | '---------------------, | ||
. | | | | || +7.20V 0 - 150A
. | +----100--, | | ++====(O)=============++
. | | | 0.01 | | ||
. | '---, +--||--, ,----|---------|---------(O)--------, ||
. | | | | | | | + sense \ ||
. | Q1 | __|__ 2k7 4k75 | ,-----+ '--++
. | |/ | | | | | | | 1000uF ||
. +-1k--| | |---+----+ | 22 |+ 16V MOTOR
. | |\ |_____| 2.50V | | 5W === ||
. | V | IC1 2k49 | | | ,--++
. | | | TL431 | | | | - sense / ||
. o +-----+-----------+----+---+-----+---------(o)--------' ||
. o---33--' ||
. close = ON ||
. optional 50mV ||
. ===========================================+= meter shunt =+=======++
. BATT NEG

Fifteen places NPN power Darlington, mounted on large aluminum heat sink
with thin heat-sink grease, and no insulating pads. Insulate heat sink from
the chassis and add plastic to protect it from exposure to metal tools, etc.
---

To analyze the circuit we'll start with the 10-milli-ohm emitter resistors,
which are required to insure equal current sharing among the transistors.
These are Ohmite 630HR020 metal-element types, at DigiKey for $0.42 each,
and they'll drop 0.2V at the nominal 10A current each transistor carries at
full load. The Darlington power transistors will have a Vbe drop of up to
2V at 10A, severely eating into the voltage difference available between the
battery and the motor at 150A, so we'll have to be careful with what's left.
(The high Vbe drop is one reason we're limiting each pass transistor to 10A,
which then requires us to use 15 parallel transistors in the pass bank.)

Although we have used high-gain Darlington transistors in an attempt to get
the base drive current down to reasonable levels, assuming a minimum gain of
500 at 10A means Q2 will have to provide over 300mA at full output. Q2 must
have high beta at 300mA, and dissipate up to 1W for high 14.5V battery (i.e.
under charge). A Zetex ZTX851, etc. (available at DigiKey), should work OK.

We cannot use a Darlington transistor at this spot because too much voltage
has been used up by the 15 power Darlington transistor's base-emitter drop
at full current. We have to save some overhead to allow the battery to sag.

In the event of an output short, Q2's 2.7-ohm collector resistor limits its
current, but the power transistors must fend for themselves until the fuse
blows (you can get 250A fuses at auto parts stores). Alternately a ~ 200A
current limit function could be added by amplifying the 50mV shunt voltage,
comparing it to a fixed voltage, and pulling down Q2's base.

Another valuable feature would be a comparator to tell whenever battery
sag means the circuit is running out of headroom, which can be seen by
looking at the control-loop voltage on the TL431's output terminal.

Some folks will say this linear power regulator design illustrates why a
buck switching converter should be used instead. I will not contest their
point, except to point out that I've learned from making SMPS in the 100
to 1000A region that there are many non-trivial issues one will encounter,
dictating knowledge, experience, and good instruments on your workbench.
On the other hand, this circuit can be made and tested by a hobbyist.

What happens to your circuit after you have spun the motor up to 150A and it
(your circuit) decides to reduce the current?

DNA
 
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