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I am trying to make a power supply which gives a constant 5V at output with a current limit of 1 Amps. I am using D1047 NPN transistor in common collector configuration, with 5V (low current) voltage from zener diode at the base, and 20 rectified AC from the transformer at the collector.

The problem that i face is that the voltage across the load varies with the change in current, even though the common collector configuration is supposed to give unity voltage gain.

Edit: Circuit Updated. Voltage across the zener/base of the transistor stays the same. But Voltage at emitter decreases when current through increases.

Input is 40V peak to peak sine. enter image description here

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  • \$\begingroup\$ Please add the "with 5V (low current) voltage from zener diode at the base" to your circuit diagram and a link to the datasheet \$\endgroup\$ – jippie Nov 27 '13 at 8:04
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First of all, I'm going to assume your complete circuit looks like this:

enter image description here

[BTW, you should post your complete circuit if you expect to get any meaningful answers.]

Secondly, the unity voltage gain of the common collector refers to AC, not DC.

From the image above, you can see that the output voltage will be \$V_Z-V_{BE}\$.

And \$V_{BE}\$ will have some variation with the collector current, but not too much: \$V_{BE}\propto ln(I_C)\$.

On the other hand, \$I_B\$ is not negligible, it could be up to 20mA (for the transistor's minimum \$h_{FE}\$ of 50), and you don't really show how you are biasing your zener, so it could be that the base is sucking more than you are providing and the voltage across the zener will drop, and this drop will be directly reflected at the output voltage of this circuit.

By the way, from the 2SD1047 datasheet, \$V_{BE}\$ at 1A will be about 0.7V, so your output should be about 4.3V (not 5V), and like I said, will vary a bit with \$I_C\$. At 1A, it will dissipate quite a bit: \$1A(20V-4.3V)\approx 16W\$. The transistor should be able to thermally handle it though, since its thermal resistance is only 1.25°C/W.

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  • \$\begingroup\$ Circuit Updated. voltage across the zener stays the same, but the Vbe changes with change in load. With a zener of 6V2, voltage at the emitter is 6.2 when the load is low and changes linearly to 6.2-0.8 with increase in load. \$\endgroup\$ – Hassan Nadeem Nov 27 '13 at 12:54
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You are confusing first order approximations or rules of thumb with exact reality. Yes, the emitter voltage of a emitter follower with a fixed base voltage stays constant to first approximation, but when you're pushing the limits you have to look more carefully than that.

There are a number of problems with your circuit.

The zener voltage won't be exactly constant over its current range. Note that its current range will be considerable, since the base needs anywhere from 0 to about 20 mA over the 0-1 A output load range. You still need a little current thru the zener at full output. Let's say 1 mA. That means you need to feed it with 21 mA at no load. Can this zener even do that? 21 mA x 5.6 V = 120 mW, which is within the plausible range for some zeners.

The 2.2 kΩ resistor can't supply the necessary zener current. You have 20 V sine in, which is 28.3 V peak. Let's say 1.5 V lost in the full wave bridge, so you have 27 V max on the cap. (27 V - 5.6 V) / 2.2 kΩ = 9.6 mA, which is way too little to sustain 1 A of output current.

The transistor will get quite hot. If you were able to feed its base to keep the output at 5V, it would drop 22 V, which means it will dissipate 22 W. That's going to reqire a serious heat sink.

You could fix this circuit to provide more base drive and more active feedback to regulate the output voltage, but then you'd just be reproducing a 7805 regulator at best. Those have all that stuff built into a single 3-pin part, regulate the voltage well, can handle 1 A, and even have thermal shutdown to prevent damage in case of overheating.

However, what you really want is a switching power supply. That will be simpler and cheaper than having to deal with the heat from a linear regulator.

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Check the DC current gain in the datasheet for D1047. For a 1A collector current, the DC current gain is 60 worst case. This means that the base current will be \$I_B = \dfrac{I_C}{h_{FE}}=\dfrac{1}{60} \approx 17\text{mA} \$. This means that your zener circuit will be loaded with 0..17mA, depending on the output load.

Now check how your zener circuit behaves with this varying load:

  • what will be the zener circuit's output voltage when drawing 0mA;
  • what will be the zener circuit's output voltage when drawing 20mA (rounded up).
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With the addition of an op-amp you'll get a lot more performance. The emitter, under load conditions will always be less that thbase. The BJT output will sag with load currents because it has only finite current gain whereas if you add an op-amp to drive the BJT you'll get better performance all-round: -

enter image description here

This is just an example circuit from the internet. The output on \$R_L\$ will be, under quite severe limits of load always virtually be equal to the voltage \$V_{IN}\$. The op-amp power rails can be tied to the output of the bridge and ground as per your circuit.

The way it works is simple - any difference between output voltage (measured on -Vin of op-amp) and input voltage (+Vin of op-amp) will result in a correction voltage applied to the base of the transistor. This will regulate the output voltage from the emitter,

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