Maker Pro
Maker Pro

MOSFETs, Avalanche Energy and Solenoids

A

Alan Samet

Jan 1, 1970
0
I have a simple circuit where I have a solenoid connected to the drain
of an N-channel MOSFET. Right now, I'm using a clamp diode for
protection, which I would like to eliminate. There are a lot of
MOSFETs with Avalanche Energy ratings. I don't know the
characteristics, other than voltage and power consumption, of the
solenoid. I've seen a couple white papers with vague mathematical
formulas for dealing with Avalanche Energy. My question is this:

Is there a simple way to determine whether a particular MOSFET can
safely support my solenoid without a clamp diode? For instance, if I
cycle the circuit at 100 Hz for a few seconds or minutes and it still
works, is this acceptable?

Is there a way to *measure* the actual Avalanche Energy the MOSFET
should expect to encounter from the solenoid? All I have access to is
an oscilloscope, a multifunction meter, and components that I can use
to build test drivers.

What should I look for?

-Alan
http://www.htmlwindows.net/
 
G

Genome

Jan 1, 1970
0
| I have a simple circuit where I have a solenoid connected to the drain
| of an N-channel MOSFET. Right now, I'm using a clamp diode for
| protection, which I would like to eliminate. There are a lot of
| MOSFETs with Avalanche Energy ratings. I don't know the
| characteristics, other than voltage and power consumption, of the
| solenoid. I've seen a couple white papers with vague mathematical
| formulas for dealing with Avalanche Energy. My question is this:
|
| Is there a simple way to determine whether a particular MOSFET can
| safely support my solenoid without a clamp diode? For instance, if I
| cycle the circuit at 100 Hz for a few seconds or minutes and it still
| works, is this acceptable?
|
| Is there a way to *measure* the actual Avalanche Energy the MOSFET
| should expect to encounter from the solenoid? All I have access to is
| an oscilloscope, a multifunction meter, and components that I can use
| to build test drivers.
|
| What should I look for?
|
| -Alan
| http://www.htmlwindows.net/

You need to know the energy stored by the solenoid. E=0.5LI^2.

I will be limited by the resistance of the coil. I mean I not Me.

You need to look at.....

1) The absolute maximum ratings.... could be meaningless.

Both the single and continuous ratings will be based on the assumption
that you hold the junction temperature at 25C.

Better still.

2) The transient thermal impedance curves.

That's when the fun begins.... They are a statement of the various
thermal impedances and thermal capacities of the bits and pieces that go
to make up your mosfet. That's why they are bendy.

If I were doing the job then I'd look a bit harder at them. Since I'm
not I won't.

Sorry

DNA
 
M

mike

Jan 1, 1970
0
Genome said:
| I have a simple circuit where I have a solenoid connected to the drain
| of an N-channel MOSFET. Right now, I'm using a clamp diode for
| protection, which I would like to eliminate. There are a lot of
| MOSFETs with Avalanche Energy ratings. I don't know the
| characteristics, other than voltage and power consumption, of the
| solenoid. I've seen a couple white papers with vague mathematical
| formulas for dealing with Avalanche Energy. My question is this:
|
| Is there a simple way to determine whether a particular MOSFET can
| safely support my solenoid without a clamp diode? For instance, if I
| cycle the circuit at 100 Hz for a few seconds or minutes and it still
| works, is this acceptable?
|
| Is there a way to *measure* the actual Avalanche Energy the MOSFET
| should expect to encounter from the solenoid? All I have access to is
| an oscilloscope, a multifunction meter, and components that I can use
| to build test drivers.
|
| What should I look for?
|
| -Alan
| http://www.htmlwindows.net/

You need to know the energy stored by the solenoid. E=0.5LI^2.

I will be limited by the resistance of the coil. I mean I not Me.

You need to look at.....

1) The absolute maximum ratings.... could be meaningless.

Both the single and continuous ratings will be based on the assumption
that you hold the junction temperature at 25C.

Better still.

2) The transient thermal impedance curves.

That's when the fun begins.... They are a statement of the various
thermal impedances and thermal capacities of the bits and pieces that go
to make up your mosfet. That's why they are bendy.

If I were doing the job then I'd look a bit harder at them. Since I'm
not I won't.

Sorry

DNA

Is there a tutorial on this effect?
I downloaded countless application notes from the irf.com website and
never found info on how to interpret/design for inductive loads.
Yes, there's a graph in the spec sheet, but what the heck does it mean
in the context of a stepper driver when there are rpeptitive ventures
into avalanche? The hexfets claim to be avalanche rated, but I can't
figure out how to validate that claim for my application???
Thanks, mike

--
Return address is VALID.
Bunch of stuff For Sale and Wanted at the link below.
Toshiba & Compaq LiIon Batteries, Test Equipment
Honda CB-125S $800 in PDX
Yaesu FTV901R Transverter, 30pS pulser
Tektronix Concept Books, spot welding head...
http://www.geocities.com/SiliconValley/Monitor/4710/
 
W

Winfield Hill

Jan 1, 1970
0
mike wrote...
No, because you'll have no idea what the safety margin is.

One caution is that L isn't constant for a solenoid. For example,
I have a fairly-large Guardian 4HD-INT-24D solenoid here, a 24V
part with a DC resistance of 15 ohms. This means 1.6A of current
at 24V, or nearly 40W of dissipation, hence the intermittent rating.
Its inductance measures 70mH (with the plunger out by about 0.25")
to 84mH (plunger fully in).

[FYI, the solenoid's coil alone is 24.4mH, or 32.9mH assembled in
the frame. The coil & plunger, outside the frame, measures 95mH.]

My solenoid is storing E = 0.5 * 1.6A squared * 84mH = 108mJ of
inductive energy just before turnoff.

We'll consider switching my 1.6A solenoid with an IRF540, a classic
workhorse TO-220 power MOSFET. The original version's datasheet
spec'd a single-pulse avalanche rating of 230mJ, so it appears with
this FET I could eliminate a shutoff clamp diode with my solenoid.

But a bit closer examination may be in order. For example, a fine-
print note on the datasheet says, Vdd = 25V, starting at Tj = 25C,
L = 440uH, Rg = 25 ohms, Ias = 28A (see fig 12). What's that mean?

Another issue is that IR's original IRF540 has been updated with a
newer IRF540N rated at 13mJ (with L=1.5mH Ias=15A), and an IRF540Z
rated at 83mJ. OOPS! 230mJ to 13mJ to 83mJ, WHAT'S THE DEAL?

Upon further examination, we see that all the "2nd-source" '540-type
FETs have wildly different ratings. For example, Fairchild's nice
IRF540A is rated at 10.7mJ, Motorola's IRF540 is 365mJ (but it's
discontinued), Philips' IRF540 retains IR's original 230mJ (under
different test conditions than IR's original spec, ST rates theirs
220mJ, Siliconix (Vishay) has a rating of 36mJ, and Samsung rates
their IRF540A much higher, at 523mJ (L=1mH, Ias=28A). So there!

Sheesh! We have specs ranging from 13 - 523mJ, a huge 40:1 range.

One issue is that most newer FET designs have a smaller die area,
which means they'll have less thermal mass, and therefore a poorer
single-pulse avalanche rating. That could be the explanation.

Yet, despite apparent serious avalanche-rating spec differences,
I'll assert that all these FETs are more alike than they appear.
That's because their avalanche ratings all specify different
inductance values, which means different avalanche time durations.
To understand this we'll have to do the math to determine the FET
avalanche time, t, and explore the concept of transient thermal
impedance, etc.
Is there a tutorial on this effect? [ snip ] The hexfets claim
to be avalanche rated, but I can't figure out how to validate that
claim for my application???

IRF refers us to their AN-1005, but I haven't been able to find a
copy. I'll keep looking, or else later I'll post a short tutorial
here, with the formulas and FET avalanche evaluation methods I use
and an explanation of the different FETs specs outlined above.

Thanks,
- Win

whill_at_picovolt-dot-com
 
G

Genome

Jan 1, 1970
0
| mike wrote...
| >
| > Genome wrote:
| >> Alan Samet wrote ...
| >>|
| >>| I have a simple circuit where I have a solenoid connected to the
| >>| drain of an N-channel MOSFET. Right now, I'm using a clamp diode
| >>| for protection, which I would like to eliminate. There are a lot
| >>| of MOSFETs with Avalanche Energy ratings. I don't know the
| >>| characteristics, other than voltage and power consumption, of the
| >>| solenoid. I've seen a couple white papers with vague mathematical
| >>| formulas for dealing with Avalanche Energy. My question is this:
| >>|
| >>| Is there a simple way to determine whether a particular MOSFET can
| >>| safely support my solenoid without a clamp diode? For instance, if
| >>| I cycle the circuit at 100 Hz for a few seconds or minutes and it
| >>| still works, is this acceptable?
|
| No, because you'll have no idea what the safety margin is.
|
| >>| Is there a way to *measure* the actual Avalanche Energy the MOSFET
| >>| should expect to encounter from the solenoid? All I have access to
| >>| is an oscilloscope, a multifunction meter, and components that I
| >>| can use to build test drivers.
| >>
| >> You need to know the energy stored by the solenoid. E = 0.5 LI^2.
| >> I will be limited by the resistance of the coil. I mean I not Me.
| >> You need to look at.....
| >>
| >> 1) The absolute maximum ratings.... could be meaningless.
| >>
| >> Both the single and continuous ratings will be based on the
| >> assumption that you hold the junction temperature at 25C.
| >>
| >> Better still.
| >>
| >> 2) The transient thermal impedance curves.
|
| One caution is that L isn't constant for a solenoid. For example,
| I have a fairly-large Guardian 4HD-INT-24D solenoid here, a 24V
| part with a DC resistance of 15 ohms. This means 1.6A of current
| at 24V, or nearly 40W of dissipation, hence the intermittent rating.
| Its inductance measures 70mH (with the plunger out by about 0.25")
| to 84mH (plunger fully in).
|
| [FYI, the solenoid's coil alone is 24.4mH, or 32.9mH assembled in
| the frame. The coil & plunger, outside the frame, measures 95mH.]
|
| My solenoid is storing E = 0.5 * 1.6A squared * 84mH = 108mJ of
| inductive energy just before turnoff.
|
| We'll consider switching my 1.6A solenoid with an IRF540, a classic
| workhorse TO-220 power MOSFET. The original version's datasheet
| spec'd a single-pulse avalanche rating of 230mJ, so it appears with
| this FET I could eliminate a shutoff clamp diode with my solenoid.
|
| But a bit closer examination may be in order. For example, a fine-
| print note on the datasheet says, Vdd = 25V, starting at Tj = 25C,
| L = 440uH, Rg = 25 ohms, Ias = 28A (see fig 12). What's that mean?
|
| Another issue is that IR's original IRF540 has been updated with a
| newer IRF540N rated at 13mJ (with L=1.5mH Ias=15A), and an IRF540Z
| rated at 83mJ. OOPS! 230mJ to 13mJ to 83mJ, WHAT'S THE DEAL?
|
| Upon further examination, we see that all the "2nd-source" '540-type
| FETs have wildly different ratings. For example, Fairchild's nice
| IRF540A is rated at 10.7mJ, Motorola's IRF540 is 365mJ (but it's
| discontinued), Philips' IRF540 retains IR's original 230mJ (under
| different test conditions than IR's original spec, ST rates theirs
| 220mJ, Siliconix (Vishay) has a rating of 36mJ, and Samsung rates
| their IRF540A much higher, at 523mJ (L=1mH, Ias=28A). So there!
|
| Sheesh! We have specs ranging from 13 - 523mJ, a huge 40:1 range.
|
| One issue is that most newer FET designs have a smaller die area,
| which means they'll have less thermal mass, and therefore a poorer
| single-pulse avalanche rating. That could be the explanation.
|
| Yet, despite apparent serious avalanche-rating spec differences,
| I'll assert that all these FETs are more alike than they appear.
| That's because their avalanche ratings all specify different
| inductance values, which means different avalanche time durations.
| To understand this we'll have to do the math to determine the FET
| avalanche time, t, and explore the concept of transient thermal
| impedance, etc.
|
| > Is there a tutorial on this effect? [ snip ] The hexfets claim
| > to be avalanche rated, but I can't figure out how to validate that
| > claim for my application???
|
| IRF refers us to their AN-1005, but I haven't been able to find a
| copy. I'll keep looking, or else later I'll post a short tutorial
| here, with the formulas and FET avalanche evaluation methods I use
| and an explanation of the different FETs specs outlined above.
|
| Thanks,
| - Win
|
| whill_at_picovolt-dot-com
|

Being pissed as a bucket which hasn't been pissed in I'll suggest that
yuo also look at the difference between the pulse avalanche current and
the absolute maximum pulsed draain current. One is bonding wire limited
and the other is something else to do with the avarage.

I think that avalanche breakdown is to do with the parasitic body source
diode which is a transistor but uses up less of the die area perhaps
thas why there is a discrepency. Perhaps it's somthing to do with
secondary breakdown.

If you calculate the something or other to do with the test circuit then
yopu end up with the energy quoted somewhere else. All I can thingk is
that since the test circuit includes and RG then turn of durinmg
avalanche is not restricted to cionduction by the body source
diode/transisitor alone but involves channel conduction as well with the
gate being enhanced.

blah blah blaht.

DNA
 
T

Tony Williams

Jan 1, 1970
0
I have a simple circuit where I have a solenoid connected to the
drain of an N-channel MOSFET. Right now, I'm using a clamp diode
for protection, which I would like to eliminate.
[snip]

Why? With a clamp diode most of the solenoid's
stored energy is dissipated within the solenoid
itself. Using an external clamp (such as the
MOSFET avalanching) means that the external clamp
now has to handle to dissipation.

[snip]
Is there a way to *measure* the actual Avalanche Energy the
MOSFET should expect to encounter from the solenoid? All I have
access to is an oscilloscope, a multifunction meter, and
components that I can use to build test drivers.

You can get a rough idea with the circuit below.

---+---------+-- Vsupply
| |
) |
Ls ) |
) |
| |
/ /
Rs \ Rext \
/ /
| |
+---------+--->Scope
|
|--+
----|
|--+
|
---+-------------- 0v

Ls/Rs are the solenoid's inductance and resistance.
Rext is an external resistance, of value equal to Rs.
The scope looks at the voltage across Rext, and all
measurements are taken with respect to Vsupply.

At each release of the solenoid, the current in Ls
(Vs/Rs) will drive Rs+Rext up to a peak voltage and
there will then be an initial exponential decay with
a time constant of T= Ls/(Rs+Rext), where T is the
time taken for the peak current (Vs/Rs) to fall to
1/3 of Vs/Rs.

So measure the time taken for the voltage across
Rext to fall from Vpk to 1/3 Vpk and then calculate
Ls = T*(Rs+Rext).

The energy in Ls for each flyback is then simply
'1/2.Ls.I-squared', where I is again that Vs/Rs.

Note the careful phrase "initial exponential decay"
used above. The tidy exponential decay lasts until
the current has dropped to about a quarter of the
peak, at which time the solenoid will start to drop
out. When it does it will generate a motoring emf
and the inductance will start to change. This will
change the flyback waveshape...... but we don't care
because we've already done the measurement.
 
W

Winfield Hill

Jan 1, 1970
0
Tony Williams wrote...
You can get a rough idea with the circuit below.

---+---------+-- Vsupply
| |
) /
Ls ) Rext \
) /
| |
/ +--->Scope
Rs \ _|_
/ /_\ <-- added diode
| |
+---------'
|
|--+
----|
|--+
|
---+-------------- 0v

Ls/Rs are the solenoid's inductance and resistance.
Rext is an external resistance, of value equal to Rs.
The scope looks at the voltage across Rext, and all
measurements are taken with respect to Vsupply.

At each release of the solenoid, the current in Ls
(Vs/Rs) will drive Rs+Rext up to a peak voltage and
there will then be an initial exponential decay with
a time constant of T = Ls/(Rs+Rext), where T is the
time taken for the peak current (Vs/Rs) to fall to
1/3 of Vs/Rs.

So measure the time taken for the voltage across
Rext to fall from Vpk to 1/3 Vpk and then calculate
Ls = T*(Rs+Rext).

The energy in Ls for each flyback is then simply
'1/2.Ls.I-squared', where I is again that Vs/Rs.

Note the careful phrase "initial exponential decay"
used above. The tidy exponential decay lasts until
the current has dropped to about a quarter of the
peak, at which time the solenoid will start to drop
out. When it does it will generate a motoring emf
and the inductance will start to change. This will
change the flyback waveshape... but we don't care
because we've already done the measurement.

Tony, that's a nice simple way to measure in-circuit
inductance and possibly even catch its value changing
in-the-act (curve fitting) as the solenoid opens!

I added a diode to minimize the "on-current" impact
on the power supply and FET. Also, small point, Rext
can be any convenient value smaller than Rs, since
you include Rext in all the formulas.

Thanks,
- Win

whill_at_picovolt-dot-com
 
M

mike

Jan 1, 1970
0
Genome wrote:
snip
Being pissed as a bucket which hasn't been pissed in I'll suggest that
yuo also look at the difference between the pulse avalanche current and
the absolute maximum pulsed draain current. One is bonding wire limited
and the other is something else to do with the avarage.

I think that avalanche breakdown is to do with the parasitic body source
diode which is a transistor but uses up less of the die area perhaps
thas why there is a discrepency. Perhaps it's somthing to do with
secondary breakdown.

If you calculate the something or other to do with the test circuit then
yopu end up with the energy quoted somewhere else. All I can thingk is
that since the test circuit includes and RG then turn of durinmg
avalanche is not restricted to cionduction by the body source
diode/transisitor alone but involves channel conduction as well with the
gate being enhanced.

blah blah blaht.

DNA
I think you have a computer virus that deleted all the sense from your post.
You should repost after fixing the virus.
mike
 
T

Tony Williams

Jan 1, 1970
0
Winfield Hill said:
Tony, that's a nice simple way to measure in-circuit
inductance......

It's quick and dirty, but probably gives a more
meaningful result for the effective 'storage
inductance' than a bridge measurement around 0,0.
.......and possibly even catch its value changing
in-the-act (curve fitting) as the solenoid opens!

I looked at several contactor decay waveforms and
have never been able to interpret what is happening
as the armature starts to move.
I added a diode to minimize the "on-current" impact
on the power supply and FET. Also, small point, Rext
can be any convenient value smaller than Rs, since
you include Rext in all the formulas.

Ok.
 
A

Alan Samet

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

Tony, that's a nice simple way to measure in-circuit
inductance and possibly even catch its value changing
in-the-act (curve fitting) as the solenoid opens!

I added a diode to minimize the "on-current" impact
on the power supply and FET. Also, small point, Rext
can be any convenient value smaller than Rs, since
you include Rext in all the formulas.

Thanks,
- Win

whill_at_picovolt-dot-com

I really appreciate all the help y'all have given me on this. I
wouldn't have gotten far enough to ask this question if it weren't for
"The Art of Electronics," which is what got me started a few months
back. Thank you.

It looks as if I should keep the clamp (I imagine the diode needs to
be able to dissipate the same amount of current as the inductor
consumes). I just saw this rating on the datasheets and a foot-note
related to eliminating the need for a protection diode; I thought I
may have been putting an unneeded component in my design and couldn't
find the answer.

-Alan
http://www.htmlwindows.net/
 
W

Winfield Hill

Jan 1, 1970
0
Genome wrote...
Being pissed as a bucket which hasn't been pissed in I'll suggest that
yuo also look at the difference between the pulse avalanche current and
the absolute maximum pulsed draain current. One is bonding wire limited
and the other is something else to do with the avarage.

Bonding-wire limits may be an issue for some large-area FETs that are
fully on with los Rds voltage drops, but they are not for any high Vds
conditions, such as avalanche.
I think that avalanche breakdown is to do with the parasitic body
source diode which is a transistor but uses up less of the die area
perhaps thas why there is a discrepency. Perhaps it's somthing to do
with secondary breakdown.

No, in most power MOSFETs the avalanche breakdown happily takes place
over the entire channel area (avoiding sensitive gate regions, thanks
to careful design) and has no dangerous localizing hot spots, thanks
to the positive Vdss/T coefficient for avalanche. It's a well-behaved
self-stabilizing process. Because the FET's entire body is involved,
one can use ordinary data-sheet thermal parameters in the calculations.
If you calculate the something or other to do with the test circuit
then you end up with the energy quoted somewhere else. All I can
think is that since the test circuit includes and RG then turn of
during avalanche is not restricted to conduction by the body source
diode/transisitor alone but involves channel conduction as well with
the gate being enhanced.

blah blah blaht.

As far as I've been able to tell, the test-set Rg value isn't very
critical, except perhaps to give an upper limit, to be sure the
transition from Rds(on) to avalanche is quick.

I'm glad you're thinking about the issues, DNA. Over the years I've
researched this to my satisfaction, and intend to post interesting
details, but I've not had time to finish the post I started writing
Sunday morning. Hint: It's all a matter of Vsupply, Vdss, Id and L,
and thermal mass. Everything works out analytically and numerically.

Stay tuned.

Thanks,
- Win

whill_at_picovolt-dot-com
 
T

Tony Williams

Jan 1, 1970
0
Alan Samet said:
I really appreciate all the help y'all have given me on this. I
wouldn't have gotten far enough to ask this question if it
weren't for "The Art of Electronics," which is what got me
started a few months back. Thank you.
It looks as if I should keep the clamp......
[snip]

If it's still all there on the bench in front of you
it would be worth doing the suggested experiment,
both as a learning exercise, and to gain more data
about that solenoid....... You get some idea of the
stored energy, and how long it takes for the solenoid
to drop out with just a diode clamp.
 
W

Winfield Hill

Jan 1, 1970
0
Winfield Hill wrote...
mike wrote...

No, because you'll have no idea what the safety margin is.

(I apologize for being too short on time to complete my remarks, but
I'll make one quick but useful point and leave the rest for later.)

Although the inductor's stored energy is E = 0.5 L I^2, the energy
dissipated by the avalanching FET is higher. The correct formula is

.. Vdss
.. E = 0.5 L I^2 ---------
.. Vdss - Vdd

where Vdd is the supply voltage. The term Vdss/(Vdss-Vdd) can also
be expressed as 1/(1 - Vdd/Vdss), revealing that at high FET breakdown
voltages the energy equation approaches the usual E = 1/2 L I^2.

With a diode the solenoid's inductance discharge is an exponential
decay with the time constant L/R. But with FET avalanche discharge
the decay is faster and appears to have a nearly constant di/dt down
to zero current. The exact decay time to zero is

.. L Vdss
.. t = --- ln ( --------- )
.. R Vdss - Vdd

This can of course also be written, t = (L/R) ln (1 - Vdd/Vdss)^-1.

For example, the current in a 12V solenoid, switched off with a FET
with breakdown Vdss = 100V, would slew to zero in just 0.128 time
constants, more than 15x faster than a simple reverse-clamp diode.

[Note, another expression, t = (L/R)/((Vdss/Vdd)-0.5)}, avoids using
the natural log and is pretty close, < 4% error for Vdss > 2 Vdd.
A commonly-used form, t = (L/R)/((Vdss/Vdd)-1)}, ignores the voltage
across the relay or solenoid's internal resistance during the decay,
and is wrong; it has a much larger error.]

It's my intention to return later to this thread and expand on the
issue of how much avalanch energy a MOSFET can safely handle, and
explain the wildly-different specs on FET manufacturer's data sheets.

Thanks,
- Win

whill_at_picovolt-dot-com
 
W

Winfield Hill

Jan 1, 1970
0
Winfield Hill wrote...
Although the inductor's stored energy is E = 0.5 L I^2, the energy
dissipated by the avalanching FET is higher. The correct formula is

. Vdss
. E = 0.5 L I^2 ---------
. Vdss - Vdd

where Vdd is the supply voltage. The term Vdss/(Vdss-Vdd) can also
be expressed as 1/(1 - Vdd/Vdss), revealing that at high FET breakdown
voltages the energy equation approaches the usual E = 1/2 L I^2.

I got the "correct" formula above from manufacturer's notes years ago,
but a careful derivation now apparently shows it's off a bit. Assuming
the avalanche current drops linearly from I to 0, an "exact" formula is,

.. Vdss
.. E = 0.5 L I^2 ln ( --------- )
.. Vdss - Vdd

The linear current-drop assumption fails below Vdss = 2 Vdd or so.
As with the avalanche-time formula, there's a simple non-log form,

.. Vdss
.. E = 0.5 L I^2 --------------
.. Vdss - 0.5 Vdd

that's accurate for Vdss >> Vdd, and has a 4% error for Vdss = 2 Vdd.

These are smaller corrections to the common 0.5 L I^2 energy formula
than the original formula, so failing to use the correction leads to
< 14% error for breakdown voltages more than four times the supply.

I plan to discuss the strange MOSFET Eas data-sheet scene later.

Thanks,
- Win

whill_at_picovolt-dot-com (use hill_at_rowland-dot-org for now)
 
R

R.Legg

Jan 1, 1970
0
It looks as if I should keep the clamp (I imagine the diode needs to
be able to dissipate the same amount of current as the inductor
consumes). I just saw this rating on the datasheets and a foot-note
related to eliminating the need for a protection diode; I thought I
may have been putting an unneeded component in my design and couldn't
find the answer.

An avalanching mosfet also produces excessive EMI, compared to that of
a diode or even a zener clamp.

RL
 
W

Winfield Hill

Jan 1, 1970
0
R.Legg wrote...
An avalanching mosfet also produces excessive EMI, compared
to that of a diode or even a zener clamp.

What's that based on, the sharp current cutoff after the
avalanche is finished, and the sharp 100V drain-step edge
and the RF ringing that follows? A small parallel cap
would get it back into the zener EMI category.

Thanks,
- Win

whill_at_picovolt-dot-com (use hill_at_rowland-dot-org for now)
 
L

legg

Jan 1, 1970
0
R.Legg wrote...

What's that based on, the sharp current cutoff after the
avalanche is finished, and the sharp 100V drain-step edge
and the RF ringing that follows? A small parallel cap
would get it back into the zener EMI category.

It's based on the inevitable non-optimization of the application.

- The voltage is not controlled and can be much larger than needs be,
were a zener present.
- The body size of the radiator is larger, it's shape is loopy and it
is often in intimate contact with even larger metal paths (heatsinks)
for assisted coupling at the frequencies of interest. Obviously
smaller parts performing lower-powered tasks are less prone.
- The part may still even be in active operation, depending on gate
drive configuration (assumed as simple and low-frequency to control a
simple relay) and is free to oscillate through it's active terminal,
at a nasty frequency of it's own choice. These oscillations eat
drivers and PICs for lunch.

Even with added (we were originally concerned with part minimization,
I believe) external capacitance, it is extremely difficult to get
external parts to compensate for internal and body strays, due to the
simple physical limitations of layout above 10MHz. The added capacity
does not damp the circuit, but it may reduce the rate of rise, assist
in turn-off, and lower the frequency to one less happily broadcast.

The poster is asking a fairly simple question; we are obliged to give
fairly simple answers, when possible.

RL
 
W

Winfield Hill

Jan 1, 1970
0
legg wrote...
The poster is asking a fairly simple question; we are obliged
to give fairly simple answers, when possible.

That's right, and Alan promptly got his answer: use a diode.

But here on sci.electronics.design, there's a place to explore
the more exotic aspects of design possibilities, which to me
means to examine the details of alternate approaches. Only
then can we truly do quick back-of-the-envelope evaluations of
different approaches, to discover a good non-obvious avenue to
further optimize a design. Indeed, I have found understanding
avalanche issues very useful several times once over the years.

Still, the "art" of electronics means we need to be aware of
"off-topic" issues that could countermand an off-the-beaten-path
solution. Your point about EMI problems is an example of this.
Knowing these issues could mandate some added element, like an
end-of-avalanche snubber (which I used in my last application).

With the extra parts in place on paper, one can evaluate whether
an exotic solution still has the appeal it first appeared to have.
In the case of using avalanche to save one part (a diode) clearly
the answer would be no; more parts would be required.

Increasingly FET and power IC manufacturers are promoting the use
of their component's avalanche capability. The question will come
up more often, and the ruggedness of the component to withstand an
avalanche will be established. So it'll be increasingly important
to evaluate any other relevant issues.

Thanks,
- Win

whill_at_picovolt-dot-com (use hill_at_rowland-dot-org for now)
 
T

Tony Williams

Jan 1, 1970
0
[snip]
The linear current-drop assumption fails below Vdss = 2 Vdd or
so. As with the avalanche-time formula, there's a simple non-log
form,
. Vdss
. E = 0.5 L I^2 --------------
. Vdss - 0.5 Vdd
that's accurate for Vdss >> Vdd, and has a 4% error for Vdss = 2
Vdd.


Interesting. The formula above highlights the fact that
a voltage clamp connected between Drain and Source will
always have to dissipate more energy than was originally
stored in the solenoid. In the case of (say) Vdss = 2x
Vdd it would be 1.3*(0.5 L I^2).

The extra energy comes from the power supply of course.

High values of Vdss reduce the multiplying factor but
result in high values of dI/dT, which could then cause
Vdd (or it's 0v) to bounce (and/or require explicit
attention to the supply impedance between the Source
and the low end of the solenoid).

It starts to make a clamp directly across the solenoid,
(ofany voltage), look much more attractive..........
 
M

mike

Jan 1, 1970
0
Tony said:
[snip]
The linear current-drop assumption fails below Vdss = 2 Vdd or
so. As with the avalanche-time formula, there's a simple non-log
form,

. Vdss
. E = 0.5 L I^2 --------------
. Vdss - 0.5 Vdd

that's accurate for Vdss >> Vdd, and has a 4% error for Vdss = 2
Vdd.



Interesting. The formula above highlights the fact that
a voltage clamp connected between Drain and Source will
always have to dissipate more energy than was originally
stored in the solenoid. In the case of (say) Vdss = 2x
Vdd it would be 1.3*(0.5 L I^2).

The extra energy comes from the power supply of course.

High values of Vdss reduce the multiplying factor but
result in high values of dI/dT, which could then cause
Vdd (or it's 0v) to bounce (and/or require explicit
attention to the supply impedance between the Source
and the low end of the solenoid).

It starts to make a clamp directly across the solenoid,
(ofany voltage), look much more attractive..........

All that energy has to go somewhere. If you want the solenoid
to drop out quickly, or the stepper motor to go fast, you have
to let the voltage go high. You've already got a fet with a
heat sink. If we could reliably interpret the FET spec
to determine whether the FET could handle this reliably,
that should be the way to go. I believe in staying WELL
within the spec, just can't determine what that spec is????
mike

--
Return address is VALID.
Bunch of stuff For Sale and Wanted at the link below.
Toshiba & Compaq LiIon Batteries, Test Equipment
Honda CB-125S $800 in PDX
Yaesu FTV901R Transverter, 30pS pulser
Tektronix Concept Books, spot welding head...
http://www.geocities.com/SiliconValley/Monitor/4710/
 
Top