How can this supply work 2

LucasBS · May 25, 2017

Here's a schematic:

https://encrypted-tbn0.gstatic.com/...Tr5stnyC-Occsj3gSU2OLPVISGzxgmgdWiigYeCITiYIM

It's a very popular power supply

But I wonder: how can a PNP transistor conduct if the voltage at the emitter is the same as the base

One could argue that the resistor between the capacitors cause a voltage drop on the base, but the same resistor can be found on the emitters.

The regulator only lowers the voltage on the output... So... Answer not here, I supoose

What am I missing

Skogsgurra · May 25, 2017

Someone else missed something. Not you.
The LM7812 isn't supposed to deliver more than 1 A. Maximum. So, the .1R going to it will not produce more than 100 millivolts. Which obviously can't even start to turn the 2N2995 on. If that resistor were 1R, it would produce something like 1 V, maximum. But still not enough to drive the bases hard enough to make Ic even close to 5 A. OK, if the .1R resistor in the emitters were removed. But then you get current hogging, so that would be a bad idea. I think that the resistor should be more like 2.2R or somewhere near.

Gunnar Englund

http://www.gke.org

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Half full - Half empty? I don't mind. It's what in it that counts.

IRstuff · May 26, 2017

The obvious answer is that someone at

http://www.circuitdiagram.org

mis transcribed the schematic. The resistor is more likely to be a 5 or 10 ohm. It needs to be high enough that the majority of the LM7812 input current is coming through the bases of the TIP2955s. You really don't want to be burning current through this resistor and making the LM7812 running at its max current just to burn it in the resistor.

A quick web search shows:

https://www.eeweb.com/blog/circuit_...rent-power-supply-by-tip2955-pass-transistors

which has a much higher valued input resistor, 100R, which is consistent with my initial thought of a 10 ohm resistor.