http://myelectronis-lab.blogspot.com/2013/05/dual-rail-variable-dc-power-supply.html

I just built this little circuit and it works wonderfully, giving me exactly half the supply voltage in a + and -

can someone tell me how this works, I am intrigued.

 

R1 and R2 form a potential divider. Since R1 = R2, pin3 of IC1a will be at half the input supply voltage, i.e. at Vcc/2 with reference to the input 'ground'.
IC1a can't provide much output current, so Q1 and Q2 are used as voltage-followers and boost that current. Their coupled emitters have a voltage output labelled Gnd (not to be confused with the input 'ground'). This voltage is fed back to pin 2 of IC1a. An op-amp configured as shown will drive its output such that the difference between the voltages on its input pins (2 and 3) is close to zero. Hence the output voltage on pin 1 must be half way between the two supply rails labelled Vcc and -Vcc. This half-way point is referenced as the output Gnd.

 

http://myelectronis-lab.blogspot.com/2013/05/dual-rail-variable-dc-power-supply.html

I just built this little circuit and it works wonderfully, giving me exactly half the supply voltage in a + and -

can someone tell me how this works, I am intrigued.

I don't think the regulation will be very good under varying loads. I would guess it will have a deadband of +/- 0.7V (Vbe of the transistors).

 

I don't think the regulation will be very good under varying loads. I would guess it will have a deadband of +/- 0.7V (Vbe of the transistors).

No dead band. The opamp takes care of that. There may be some cross-over noise when the output switches from sink to source, and the extent will depend on the speed of the amp.

This circuit is good enough for amateurs, though.

 

No dead band. The opamp takes care of that. There may be some cross-over noise when the output switches from sink to source, and the extent will depend on the speed of the amp.

This circuit is good enough for amateurs, though.

Isn't the dead band ~= Vbe / Aol ? Actually 2*Vbe / Aol ?

But of course that would be buried in the noise.


Regards, Dana.

 

Isn't the dead band ~= Vbe / Aol ? Actually 2*Vbe / Aol ?

But of course that would be buried in the noise.


Regards, Dana.

It'll definitely become noticeable at higher frequencies as Aol drops, and the answer would be a faster amp.

As all things, the component selection depends on the application.