# Simple pulsing supply that acts like a switching supply.

Discussion in 'Electronic Basics' started by Scott Wiper, May 23, 2004.

1. ### Scott WiperGuest

I would like to know if this curcuit
will work to the spec I have calculated.

This is a simple pulsing power supply useing
transistors set in darlington pairs with ZENER diodes as a referance voltage
so I can shunt regulate voltage and sustain high currents when in
use with minimal heating of the power transistors. Yes They will be fully
mounted on a 4"* 8" piece of matel for heat sinking.

A Breif expanation:

(http://www.travel-net.com/~swiper/switch.gif)

I will be useng a large step down transformer to reduce mains voltage
to 16.0VAC at 80VA (Watts/5A) But I just got a 25.2VAC at 100VA (Watts/4A).
This marked T1 on my schmatic.

The filter will comprise of a large high current bridge rectifer and a
10,000 uF cap C1 that will charge to 25.2*1.414=35.63VDC or
16.0*1.414=22.62. This will be fed into Q1, Q2 and Q3 (2N3055). At 16.1
referance it should have a emitter voltage of 14.0 volts with no load. D1 is
th part of the curcuit protect Q3 from back voltage of the battery or
batterys. If you use more AH of storage you should get a longer run times on
the 5 volt section. B1's charge is controlled by R4 that useing the spec
float of 13.5 -13.8VDC that is well with the specs of most lead acid
batterys. D8 is to reduce B1 by .6volts so when it becomes fully gharged
it's voltage will always be .6 volts less of the Vout from Q3. D8 will stop
reverse voltage on Q3 as well.

The perpose of C2 and C7,C8 is to smooth out voltage and reduce noise from
the pulsing outputs from Q3 and Q5. The battery or batterys will the 5 volt
part of the supply functioning with the constany frequency that is applied
through U1(NE556) and U2(CD4013). U2 is a garrenteed 50% duty cycle at 1/2
Frequency. The whole idea is to make a simple 5VDC UPS but with the features
of a switching power supply. This one is going to power the swiper7000 light

For better viewing and printing. http://www.expresspcb.com
http://www.travel-net.com/~swiper/switch.sch
Run you spyware utility because this one is loaded with it.

Question 1.

Will I have to place an inductor on both the 12 and 5 volt outputs as
current tanks? (C1 C7 C8)

Question 2.

If inductors have to be used. Will they have to be placed in series with the
output caps or in parrallel?

2. ### John PopelishGuest

First of all, you should review this tutorial on switching regulator
http://www.national.com/appinfo/power/files/f5.pdf

I see lots of other problems.

Q1, with 35 volts on its emitter will never turn off with either the 0
or +12 volt output of U2A. In fact, you are likely to turn U2 into an
SCR that shorts its 12 volt supply by the current R7 will dump into
its output.

Any time you connect two capacitors together with a switch, there must
be a waste of energy (in some combination of the switch's resistance
and the capacitors' resistance) that equals the energy you transfer
from one capacitor to the other. This is why inductors are put in
series with such circuits.

If Q3 really represents a good, low resistance switch, then when it
turns on, the voltage on C2 will jump up to 9/10ths of the voltage on
C1 when the switch closes. How do you imagine that the voltage will
rise only to 14 volts?

As to your note at the top about Q3, Q3 (or Q3 and C1 and C2) must
dissipate about (35-14)*average current through Q3. Switching Q3 on
and off doesn't help in this configuration (connecting two capacitors
together with a switch).

These same comments apply to the 12 volt to 5 volt reduction circuit,
sdince it also connects two capacitors together with a switch.

3. ### Jan PanteltjeGuest

A shunt regulator is a regulator tha tis in PARALLEL with the output,
yours is not.

Pulsing will NOT cause less heat, as the heat is proportional
to teh power dissipated, and the power = i x u x t, where, I is the current
through the transistor, U the voltage across it, and t teh time it is on.
For teh SAME amount of output current (average) after C2, you can for
example have 3 A for 100% of the time, and 6 A for 50% of the time,
So in the fist case power is 3 x U and in the secon example the same:
3 x Iu * .5
etc.
So you 'switching' idea without the use of an inductor sucks, and brings
JP

4. ### Jan PanteltjeGuest

errata:
6 x I * .5 of cause

5. ### Scott WiperGuest

First of all, you should review this tutorial on switching regulator
http://www.national.com/appinfo/power/files/f5.pdf

I see lots of other problems.
I see the mistake on Q1 I will have to add a reverse block diode before R7.
This will be corrected.

I have built a non pulsing proto type here
http://www.travel-net.com/~swiper/5volt.gif. What I am attempting to do is
turn the transitors on and on rapidly This enegy is dumped D4 that charges
the battery or batterys. Transistor do funny things on their emitters. they
have at tendency to drop .7 volts to what ever Base/Emitter drive voltage is
applied. So if you Base/Emitter follow through two transistors it
Vref-.7-.7=Vout. EG: Q2 and Q3. http://www.travel-net.com/~swiper/switch.gif

This why audio amplifiers have their npn and pnp transistors set on common
emitter with a no signal voltage of .65 volts to hold them off. There two
..5ohm resistors tying the emitters together to the speaker output from both
+ rail and the - rail.
A reverse block will also will have to be placed on Q6 as well you will see
the changes after this post. As for traansistor theory. I am dealing with
mosfet types I am useing old fashioned plain bipolar types. These have a
good charatoristic of limit emitter voltage that will allways be .7volts
less then Vbe. What I am doing is useing a low frequency square wave at 50%
duty cycle for short periods of time resulting in C2 will only charge to Vbe
set Vref on Q2's base.

So lets review bipolar transistors. Vc (Collector) this is the input is 35.6
from the heavy stepdown transformer. Q2 and Q3 are in a base/emitter
follower so the emitter voltage on Q3 will 16.1()Vref-1.4=14.7V this with Q3
held on all the time. R2 to v+. What will happen when Q3 sees several amps
on it's emittier it's going to heat up quite abit. By turning on and off Q3
at 60 hutrz This will allow for up to 10Amps without the transistor getting
so hot you can cook a "McDonalds Pork burger" (1.5A constant has been
tested) on it like the other one put going "POOP" and with rotten small and
white smoke.

6. ### Scott WiperGuest

The whole Idea Of my website is to keep it simple If I can achieve this
simplistity I will reached the holely grail. I will be useing the spice
simulator to test this theory out. If I can switch one voltage from a
stepdown coil and to a logic voltage without inductors so be it. Bipolar
transistor can have there tranfer of resistance set by Vbe even when they
are set in darlington pairs. This when you have two Bipolar transistors with
the base/emitter of QA following through QB. So Vref=16.1V set by zener
diode. So Vc = 35.6VDC Where Q2 and Q3 are commoned. So here is the math.
(Vref)16.1 - 1.4(VbeQ2/Q3)=14.1VDC(Ve Q3). This project is suppsed be a
wanna be that acts like a switching supply, but it fact it's not. This why
I will recommend that you put a large aluminum plate where you intend to
mount the transistors oruse your case as a heat sink. They will get warm but
not buring hot to cook Mcdonalds pork burgers on the caseing of the
transistors. If they get to warm thats what 12VDC box fans are for.

7. ### Jan PanteltjeGuest

I had rather you follwed Mr John Popelish recommendation, and read a good
book on switchmodes, or the link he gave you.
In your circuit you can leave out the switching, and it would work better.
At 3A your serial transistor dissipates 3 x 25 = 75 W.
If you allow for 150 C degrees juncion temp in the transistor, and 40 C
outside temp, you can rize 110 degrees.
So your total thermal resistance is 110 / 75 or about 1.5 C / Watt.
Substract the Rth for a 3055, the Rth for any isolation whasher,
you are out of luck, the Rth for a 2n30055 is 1.5 and the Tmax = 200 (just
looked it up), and the transistor will go kaput, because you ALSO need
to add the Rth for the heatsink.
In fact I think you CAN fry egs on a 75 W heater, think about it,
how much is your soldering iron ;-)
So pleae, yes it IS a good idea to use the LTspice, and feel free to ask more
once you get indo switchmodes.
JP
y

8. ### R.LeggGuest

I see no e4vidence of calculations.
You are attempting to reduce circuit losses by passing all energy
through a 'linear' regulator circuit only during phase angles that
ensure a low voltage drop across the series elements. Capacitive
storage is relied upon to control output droop during all other phase
conditions.

A large peak-to-average current waveform is required from the source
and in all conducting elements of such a regulator. This produces
agravated rms current values.

Series elements that heat under the influence of average current will
not be as severely effected by this method as those that heat up under
the influence of rms current. Feel free to make your own estimation of
the effect on simple square waves of equal 'average' value, but of
reducing duty cycle.

The last time I saw this miraculous technique demonstrated, you could
smell the transformer varnish outgassing, due to straight copper
losses and the resulting internal transformer rises. The effect has
the same relative severity at all power levels.

The only current proponents of this type of regulation (that I am
aware of) is the 'Research and Development Centre for Hybrid
Microelectronics and Resistors' - primarily for use in battery
chargers, where ripple voltage and storage capacitance is not an
issue.

http://www.prz.rzeszow.pl/imaps2001/artykul/T_24.ZIP

Note the number of taps and switches that are inevitably employed -
effectively forming a multi-pulse system - all in an attempt to reduce
the peak-to-average current and increase the useful phase angle of a
power system. Note that no claim is made concerning input harmonic
compliance, after the first order effects of series element

Power conversion expertise is not a noted strength of this
institution.

RL

9. ### Jan PanteltjeGuest

I have actually been thinking a bit, and in the old times (sixtioes)
thyristors became very popular, and many a battery charger with these
was designed.

In such a case, you can probably leave out 90% of the comnponents,
I remember Motorol had a nice application book for their thyristors,
lets see how the web is:
http://pemclab.cn.nctu.edu.tw/peclub/w3cnotes/cn06.電力電子簡介/html/cn06.htm
Here you see the diagram, AND all the formula you asked for ;-)

ac bridge thryristor resistor battery
full wave in
______________________|\|_______________________\/\/\/\_________ +12
| | | |/| | R2
| | | \ |
/ | | | |
\ <-- | ^^^^ pulse transformer 1:1
/ Uout | ^^^^
\ 220k | | |
| |-- | |
|------| UJT | |
| |-- | |
| |____________| |
=== |
| 220nF |
| C1 |
------------------------------------------------------------------

See, in this circuit the thrystor is turned on late in the AC cycle,
so you can control the output voltage.
You can replace the potmeter by a PNP current source, and control its
input voltage.
This way if you use an error amplifier and voltage reference, and filter,
you have a switching in the sense that perhaps you wanted...
R2 limits the peak current into the battery...
UJT is an unijunction transistor, fro example 2n2646
You will have to make sure the UJT does not get to much voltage, else use
some divider, I think the 2n2646 is about 30V.
http://www.americanmicrosemi.com/tutorials/unijunction.htm

Anyways, the idea is that you trigger the thyrister when the half
sinewave drops below the required output voltage (so after the zero
crossing), always in the last 90 degrees...
The potmeter with C1 forms the timeconstant.
The UJT will conduct when the voltage on C1 reaches half the input here,
and discharge C1 via the pulse transformer into the thyristor gate.
A rather crude circuit, but it works up to 30 amps or so.
JP

10. ### Scott WiperGuest

I used the formal that was supplied with the datasheet on the transistor
selected I may have gotten it wrong. I am just begining to grasp the
function and math.
Your point is taken... I DID COOK a tranformer by over currenting the
secondary winding it gave good white smoke and burnt varnish after the
thermal fuse whent in the prinary. This is why I have placed a large ballast
resistor on the positive terminal of the battery. My Prototype is working
just fine even after I outgassed the varnish of a small 25.2 1A stepdown
coil. I am surprised that the first transistor survived the smoking. The 2A
transformer do even get warm with the ballast resistor on the battery
positive terminal

The only current proponents of this type of regulation (that I am
aware of) is the 'Research and Development Centre for Hybrid
Microelectronics and Resistors' - primarily for use in battery
chargers, where ripple voltage and storage capacitance is not an
issue.

http://www.prz.rzeszow.pl/imaps2001/artykul/T_24.ZIP

Note the number of taps and switches that are inevitably employed -
effectively forming a multi-pulse system - all in an attempt to reduce
the peak-to-average current and increase the useful phase angle of a
power system. Note that no claim is made concerning input harmonic
compliance, after the first order effects of series element