12 volt 5 amp dc power

Thread Starter

newtopsupply

Joined Jul 1, 2015
24
Please see attached

I have an electronic 240V to 12 V 60 Watt electronic transformer to a bridge rectifier as pictured.
Across the terminals I have caps as pictured and a resister. I put in a 12V 5W zener to bring the voltage down. It burnt out in 10sec.
Whats the quickest simplest way for a newby to sort it out so I get my 5 amps OK. I run it to an electronic timer and relay to an electronic motor speed controller for a 12V wiper motor. Amperage test gave me 1.8A but under load this increases.
 

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JohnInTX

Joined Jun 26, 2012
4,787
Your zener burned out because you didn't have a current limit / ballast resistor between the bridge and the cathode of the zener. You probably can't use a zener directly but need something like this. Click on 'images' in the search results. The zener provides a reference voltage for the pass transistor to drive high currents.

Welcome to AAC.
 

Thread Starter

newtopsupply

Joined Jul 1, 2015
24
Your zener burned out because you didn't have a current limit / ballast resistor between the bridge and the cathode of the zener. You probably can't use a zener directly but need something like this. Click on 'images' in the search results. The zener provides a reference voltage for the pass transistor to drive high currents.

Welcome to AAC.
Thanks John
Your zener burned out because you didn't have a current limit / ballast resistor between the bridge and the cathode of the zener. You probably can't use a zener directly but need something like this. Click on 'images' in the search results. The zener provides a reference voltage for the pass transistor to drive high currents.

Welcome to AAC.
Thanks John

Its the first site that's given me an idea how to work it out. Fantastic reply to my query. If I can ask your thoughts on the revised circuit attached?

Regards
 

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JohnInTX

Joined Jun 26, 2012
4,787
Sure. First, your caps are drawn funny. These are electrolytic caps and are polarity sensitive. Using the proper symbol (straight line for positive, curved line for negative) - avoids confusion. Avoid jumparounds like in the base of the transistor. The more clear your skiz is, the easier to get help. Other than that, its OK.

Your circuit is basically correct. A few things to know:
There is no current limit provided - maybe not necessary but good to know.
The output voltage is the 12V zener voltage PLUS Vbe of the transistor (.5-.6V) so spec the zener voltage accordingly. If you want 12V out, use a 11.5V zener or so. These two voltages are in parallel with the load so they must add up to the load voltage.
The transistor regulates by controlling its collector-emitter current to satisfy the load requirements, excess voltage is dropped across the transistor and the resulting power will be dissipated as heat. You may need a heat sink on the transistor.

The resistor / zener values should work OK with most NPN power transistors, 2N3055/TIP48 etc.
 

JohnInTX

Joined Jun 26, 2012
4,787
EDIT: I see 5 amps as your output current. The TIP48 I threw out won't do 5A. The 2N3055 will - prepare to heat-sink it.
 

Thread Starter

newtopsupply

Joined Jul 1, 2015
24
Sure. First, your caps are drawn funny. These are electrolytic caps and are polarity sensitive. Using the proper symbol (straight line for positive, curved line for negative) - avoids confusion. Avoid jumparounds like in the base of the transistor. The more clear your skiz is, the easier to get help. Other than that, its OK.

Your circuit is basically correct. A few things to know:
There is no current limit provided - maybe not necessary but good to know.
The output voltage is the 12V zener voltage PLUS Vbe of the transistor (.5-.6V) so spec the zener voltage accordingly. If you want 12V out, use a 11.5V zener or so. These two voltages are in parallel with the load so they must add up to the load voltage.
The transistor regulates by controlling its collector-emitter current to satisfy the load requirements, excess voltage is dropped across the transistor and the resulting power will be dissipated as heat. You may need a heat sink on the transistor.

The resistor / zener values should work OK with most NPN power transistors, 2N3055/TIP48 etc.

Brilliant. Thanks again John. Sorry about the dwg. Have to be a little more careful in future. You're absolutely right when you say it could lead to confusion.
At the moment I'm running off a car battery at about 12.5V so should be all good.
Good of you to take the time.
Till next time.
 

JohnInTX

Joined Jun 26, 2012
4,787
At the moment I'm running off a car battery at about 12.5V so should be all good.
You probably won't get the full effect from the 12.5V battery since the circuit (first order) outputs 12.5V which doesn't allow for any drop in the NPN pass device - but you're on the right track with the 16V input.
 

Thread Starter

newtopsupply

Joined Jul 1, 2015
24
You probably won't get the full effect from the 12.5V battery since the circuit (first order) outputs 12.5V which doesn't allow for any drop in the NPN pass device - but you're on the right track with the 16V input.

Right. Ok, I'll test the circuit for voltage before I hook up to my system. If my voltage is high I'll drop my ZD.
Thanks for the tip. I'll get there eventually.
 

AnalogKid

Joined Aug 1, 2013
10,987
At 5 A there is almost 1 V of Vf drop in each diode in the bridge, so the peak voltage across the bulk filter cap is around 15 V. Thus the max zener current is 13 mA. This means that the max. base current is 13 mA. To have the pass transistor not pull out of regulation, it needs to have a gain of 5/.013 = 367. That's about 10 times what a 3055 can do at 5 A collector current.

Your basic plan is ok, but for a 5 A output you need to recalc the zener power and the series resistor. The peak voltage across the resistor is 3 V. Start with the base current you need for the output transistor to get the resistance and the power, then use that current to calc the power rating for the zener. At low output currents, most of the bias current runs through the zener. At high output currents it runs through the transistor. If the required base current is too high for the resistor and zener you want to use, you can use a Darlington or Sziklai pair for the output transistor.

Consider a 3-terminal regulator as a more simple approach. The LM338 is rated for 5 A continuous output current and has both thermal and current limiting to protect itself.

ak
 

Thread Starter

newtopsupply

Joined Jul 1, 2015
24
At 5 A there is almost 1 V of Vf drop in each diode in the bridge, so the peak voltage across the bulk filter cap is around 15 V. Thus the max zener current is 13 mA. This means that the max. base current is 13 mA. To have the pass transistor not pull out of regulation, it needs to have a gain of 5/.013 = 367. That's about 10 times what a 3055 can do at 5 A collector current.

Your basic plan is ok, but for a 5 A output you need to recalc the zener power and the series resistor. The peak voltage across the resistor is 3 V. Start with the base current you need for the output transistor to get the resistance and the power, then use that current to calc the power rating for the zener. At low output currents, most of the bias current runs through the zener. At high output currents it runs through the transistor. If the required base current is too high for the resistor and zener you want to use, you can use a Darlington or Sziklai pair for the output transistor.

Consider a 3-terminal regulator as a more simple approach. The LM338 is rated for 5 A continuous output current and has both thermal and current limiting to protect itself.

ak

Good afternoon AK

Thanks very much for your input.
I recalculated based on what I believe you may have been saying.
My voltage is 16.8 at 1.4 times the 12V output of the transformer.
This gels with what my MM said on the bench.
So, if I use a 5W resistor at 100 ohm and a 12V 5W ZD, I believe if I have the right of it that the gain becomes 29?
Is this then acceptable?

Regards
New
 

AnalogKid

Joined Aug 1, 2013
10,987
Thanks very much for your input.
I recalculated based on what I believe you may have been saying.
My voltage is 16.8 at 1.4 times the 12V output of the transformer.
This gels with what my MM said on the bench.
So, if I use a 5W resistor at 100 ohm and a 12V 5W ZD, I believe if I have the right of it that the gain becomes 29?
Is this then acceptable?
Don't think so. 12 x 1.414 = 16.97 = 17 V.
minus two diode Vf drops in the bridge = 15 V (Note, at 5A, the diode forward drop is larger than the textbook 0.6 V)
15 - 12 = 3 V, voltage drop across the resistor
3 V / 100 ohms = 30 mA, resistor current
5 A collector current / 30 mA base current = 167. This is the required gain of a single output transistor like a 2N3055

A power darlington can easily do this, but the output voltage will be about 10.5 V instead of 12. You can compensate for this by putting two 1n4004 diodes in series with the 12 V zener diode.

To stay with a single output transistor like a 2N3055, with a gain of 30, then:
5 A / 30 = 167 mA base current.
3 V (resistor voltage drop) / 0.167 = 18 ohms, the series resistor
9 (resistor voltage squared) / 18 = 0.5 W, the worst case resistor power dissipation. A 2 W resistor is sufficient.
0.028 (worst case zener current squared) x 12 = 0.33 W. A 1 W zener will work, but it will get warm.

ak
 

ian field

Joined Oct 27, 2012
6,536
Please see attached

I have an electronic 240V to 12 V 60 Watt electronic transformer to a bridge rectifier as pictured.
Across the terminals I have caps as pictured and a resister. I put in a 12V 5W zener to bring the voltage down. It burnt out in 10sec.
Whats the quickest simplest way for a newby to sort it out so I get my 5 amps OK. I run it to an electronic timer and relay to an electronic motor speed controller for a 12V wiper motor. Amperage test gave me 1.8A but under load this increases.
Your "electronic transformer" probably outputs the AC at considerably higher frequency than mains (probably about 20kHz or so) so you may find the bridge rectifier is getting hot too. Its a lot easier to make you're own bridge rectifier out of individual fast rectifier diodes - especially at that sort of current.

As others have pointed out, you can't just dump a zener across the output without a ballast resistor to limit the current - but that will also limit the current your load can draw.

Some sort of regulator is probably needed - possibly a zener stabilised emitter-follower power transistor.
 

ian field

Joined Oct 27, 2012
6,536
At 5 A there is almost 1 V of Vf drop in each diode in the bridge, so the peak voltage across the bulk filter cap is around 15 V. Thus the max zener current is 13 mA. This means that the max. base current is 13 mA. To have the pass transistor not pull out of regulation, it needs to have a gain of 5/.013 = 367. That's about 10 times what a 3055 can do at 5 A collector current.

Your basic plan is ok, but for a 5 A output you need to recalc the zener power and the series resistor. The peak voltage across the resistor is 3 V. Start with the base current you need for the output transistor to get the resistance and the power, then use that current to calc the power rating for the zener. At low output currents, most of the bias current runs through the zener. At high output currents it runs through the transistor. If the required base current is too high for the resistor and zener you want to use, you can use a Darlington or Sziklai pair for the output transistor.

Consider a 3-terminal regulator as a more simple approach. The LM338 is rated for 5 A continuous output current and has both thermal and current limiting to protect itself.

ak
The Lucas electrics on the old British motor cycles used a dirty great zener as the sole means of regulating the alternator, the actual voltage was 15, but "back then" I looked at similar devices from Philips that were available in the usual voltage increments. AFAICR: the power rating was somewhere in the general direction of 100W on a heatsink mounted in the air flow.

But I'd hesitate to suggest zener clamping the output of an electronic transformer!
 

Thread Starter

newtopsupply

Joined Jul 1, 2015
24
Don't think so. 12 x 1.414 = 16.97 = 17 V.
minus two diode Vf drops in the bridge = 15 V (Note, at 5A, the diode forward drop is larger than the textbook 0.6 V)
15 - 12 = 3 V, voltage drop across the resistor
3 V / 100 ohms = 30 mA, resistor current
5 A collector current / 30 mA base current = 167. This is the required gain of a single output transistor like a 2N3055

A power darlington can easily do this, but the output voltage will be about 10.5 V instead of 12. You can compensate for this by putting two 1n4004 diodes in series with the 12 V zener diode.

To stay with a single output transistor like a 2N3055, with a gain of 30, then:
5 A / 30 = 167 mA base current.
3 V (resistor voltage drop) / 0.167 = 18 ohms, the series resistor
9 (resistor voltage squared) / 18 = 0.5 W, the worst case resistor power dissipation. A 2 W resistor is sufficient.
0.028 (worst case zener current squared) x 12 = 0.33 W. A 1 W zener will work, but it will get warm.

ak
Sorry mate but I really am a bit thick on this stuff

2 W stuff not available through our local. Its 5 or 1W.

Do you mean like the attached but with a 18 ohm resistor where my 82 is?
And diodes in line in this manner?

I am getting voltage fluctuations with my current configuration, that's for sure! (About 1 to 1.5 V)

Also should say here that test bears you out about voltage. Got 9V at output.

Regards
New
 

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Last edited:

Thread Starter

newtopsupply

Joined Jul 1, 2015
24
Sorry mate but I really am a bit thick on this stuff

2 W stuff not available through our local. Its 5 or 1W.

Do you mean like the attached but with a 18 ohm resistor where my 82 is?
And diodes in line in this manner?

I am getting voltage fluctuations with my current configuration, that's for sure! (About 1 to 1.5 V)

Regards
New
Don't think so. 12 x 1.414 = 16.97 = 17 V.
minus two diode Vf drops in the bridge = 15 V (Note, at 5A, the diode forward drop is larger than the textbook 0.6 V)
15 - 12 = 3 V, voltage drop across the resistor
3 V / 100 ohms = 30 mA, resistor current
5 A collector current / 30 mA base current = 167. This is the required gain of a single output transistor like a 2N3055

A power darlington can easily do this, but the output voltage will be about 10.5 V instead of 12. You can compensate for this by putting two 1n4004 diodes in series with the 12 V zener diode.

To stay with a single output transistor like a 2N3055, with a gain of 30, then:
5 A / 30 = 167 mA base current.
3 V (resistor voltage drop) / 0.167 = 18 ohms, the series resistor
9 (resistor voltage squared) / 18 = 0.5 W, the worst case resistor power dissipation. A 2 W resistor is sufficient.
0.028 (worst case zener current squared) x 12 = 0.33 W. A 1 W zener will work, but it will get warm.

ak

Should I use a 15V zener to bring the voltage up?
 
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