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Boosting Regulator Current/ATX power supply

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I am building a power supply for ATX motherboard.
I have two 12V batteries which I want to use for this purpose.
My plan is to use regulator ICs and add current boosting capability to them.
For this purpose consider the LM7805 IC. Im boosting its current capability like shown in the attached image

Here 3.3 ohm is serving as a current-sensing resistor that will switch on the transistor as soon as the current draw increases 212 mili amperes, hence according to theory 212 mA current shud flow through 7805 and the rest through the power transistor, BUT this isnt whats happening (atleast with me).
The voltage across the current sesing resistor shows 3.3 V!
The resistor heats up rapidly! The transistor doesnt heat up,
The 7805 heats up!
I cant understand whats going on.
I am testing with a load of 1A in form of bulb.

Waiting for your help.


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You are using the tranistor backwards. The voltage will not be divided with the 3.3ohm resistor. You are not getting any VBE to turn on the transistor. You do have VCE and you do have the 7805 in a good position. I would use the standard setup and your only problem will be the voltage drop associated with the VCE. You could divide the voltage to ground at the base and at least turn your transistor on.

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Thank you Ante. I can see that the regulators current produces a VBE. The resistor then allows you to set the current by providing VBE. Isn't it interesting how this configuration for the transistor is not the standard seen in power supplies. My guess is that the regulator has the standard setup, while the external transistor is also used. Interesting. I bet the regulation is not as good because the input voltage to the regulator is changing and this directly affects the majority of the current to the load. There is not feedback to the external transistor.

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Thanks for the replies guys.

The current boosting technique I followed is exactly as it is given in the book "Electronic Principles" by Malovino 6th edition.

The value for the current sensing resistor is calculated according to the desired current we want 7805 to pass.

Like if I want 7805 to only pass 70mA and the rest of the current I want to pass through the pass transistors then I would have the current sensing resistor of 10 ohms.
Actually a little over 0.7 volts is required at the base of power transistor 2955 to turn it on. So If i use the 10 ohm current sensing resistor then whenever current through the 7805 increases 70 mA then the transistor turns on and allows the rest of the current to flow through it (70mA * 10 ohm = 0.7 volts).
This was its mentioned in Malvino and in some other books and websites.

Here is the latest update.
I attached a heat sink to the 2955. used the 10ohm current sensing resistor (5W; anything below 5W was burning like hell) and ran it.
I could see that a little over 70mA was flowing through 7805 and the rest through 2955, i had a total current drain of 1.3 A (in the form of 6V bulbs). Thats fine!
BUT the 2955 along with the heat sink was getting very hot, infact very very hot! the data sheet of TIP2955 says that it can handle peak current of 15A otherwise 10A continous current, but it was already very hot on 1.xx A!! Am i doing something wrong?

I added another 2955 to the setup as another pass transistor now the current would divide between them, this time the pass transistors werent as much hot.

Im still confused as when 2955 can support 10A continous current emitter to collector then why is it getting soooo much hot on only 1 A?

Any comments on this whole setup would be more than welcomed.

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Hello Kevin,

Currently the my setup is working fine with 10 ohm 5 W resistor.
I am just worried about the TIP2955 getting very hot!
And since datasheet mentions it can easily support currents of 10A then why is it getting so hot on only 1.5A current (even with an adequate heatsink installed) ?


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It's definitely a low beta. I have it. At the current you are using, the VCE should be greater. Get rid of the input resistor. You now have 7 volts VCE. Reduce your 3.3 ohm resistor to get a lower VBE. You now have what is called a limitation of the device. You may not be able to get the current you want because you don't have the VCE. You have to keep the beta in an acceptable range.

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Hello Kevin,
Sorry i cudnt understand you, theres no input resistor that Im using.
Only the 10 ohm 5W current sensing resistor at the input of 7805 and base.
Im not using 3.3 ohms anymore, im using 10 ohm, which turns on my 2955 when ever load current increases 70mA.

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What you have to do to make the circuit work is satisfy a couple of conditions. The output of 5v sets the load current. The VBE also sets the load current. They must work in conjunction. Now you have a VCE that is in question.

Try adjusting your VBE down while maintaining 5v to the load. This is done by reducing your 10ohm back to 3.3 ohms. This will increase your beta. It does this by reducing your base current.

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Hi Coder,
If your battery is a fully-charged lead-acid at 13.2V, and your 7805 regulator has its minimum voltage of 4.75V, then the transistor has 8.45V across it. With your total current of 1.3A, minus the 70mA through the 7805, the current through the PN2955 is 1.23A.
Therefore the power (V X I) dissipated in the transistor is only 10.4W, like Ante says. If your heatsink is adequate then it shouldn't heat much. Possible problems:
1) Regulator is oscillating because bypass caps are poor quality or too far away.
2) Regulator's input or output bypass cap is leaking (conducting) badly.
3) Although the heatsink heats-up, it has a poor mechanical connection to the transistor due to burrs or lack of thermal grease.
4) Heatsink is too small or fins are missing.
5) Current is actually more than 1.3A due to the light-bulbs operating on a voltage that is lower than their rating (their resistance has a positive temperature coefficient, so they draw more current at reduced voltage).

BTW, the transistor will heat the same amount with 10A through it but only 1.04V across it. Don't try 10A with 10V or more across it.

How did you fix your problem of 3.3V Vbe?

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Ante, it's not just the power that is of concern. You can't go below the rated beta of the transistor. This is normally associated with a base current that is too high compared to the collector current. This is because you don't have the VCE to increase the collector current beyond the base current.

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Sorry, but what are you doing with a PNP transistor with its collector positive?
Shouldn't the current-limit circuit be like this:
1) Normally, Q2 is off.
2) The 7805 passes 70mA through R1, like before.
3) Q1 passes the remainder of the load current, minus the 70mA.
4) When the current through R2 reaches 2.1A, then Q2 turns-on and bypasses the current through Q1 to pass through the regulator, which current-limits, and the load current is limited to 2.1A.

Now you will have plenty of heat! The regulator will shutdown and pulse on and off if it isn't properly heatsinked.


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After checking my sketch again I discovered one mistake, I forgot to connect the collector with the VO of the regulator. Concerning the rest of the circuit I stand firmly! Posting the correct drawing here. I think a PNP transistor must have higher voltage on the emitter than on the base to be able to conduct. This is what I have learned as the forward biasing of the emitter base diode. But it was a long time since I was in school so things can be different now I am not sure.

Ante ::)


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