Simple shunt reg

[AndrewT]

Originally posted by Onvinyl
@Andrew: with 100mA, the real reg could have a voltage drop of only 1 Volt.

the lowest input voltage (Vimin) - Vdrop = max Vout.

If Vdrop goes up then a higher voltage transformer is required.
Go down that route and all the dissipations go even higher when mains voltage is running at maximum.
There is a big advantage in component size and heatsink requirement and internal heat generation by adopting the lowest Vdrop for the CCS. It should not affect the operation of the shunt or the circuit. It's simply down to heat.

[Onvinyl]

Hi Andrew,
If you have 100mA, R12 alone eats 0.47 volts. If you have 400mA, then it drops 1.8 volts and so on.
Rüdiger

[AndrewT]

Originally posted by Onvinyl
R12 alone eats 0.47 volts. If you have 400mA, then it drops 1.8 volts and so on.

NO!.
R12 is part of the CCS.
The voltage across R12 is set by the Vbe of the CCS transistor. A slight correction for the 10k on the base lead means that R12 always has near 600 to 650mV irrespective of the chosen CCS current.

[Onvinyl]

Originally posted by AndrewT
NO!.
R12 is part of the CCS.
The voltage across R12 is set by the Vbe of the CCS transistor.

...as long as it's working... Andrew, I have to check what's wrong there. Will take some days, though.
Rüdiger

[Onvinyl]

I couldn't wait. Well, both CS-Q's are easily destroyed. It wasn't obious, because the voltage drop across R12 when using it alone happens to be 0.6 Volts in my setup...
Fixed it. The CS-voltage drop is 0.7 volts. The reg as a whole is best used at 2 Vdrop minimum.
Rüdiger

[Lumba Ogir]

Hi Rudiger,
how are things going? Round and round? However, you are going for a nice project this time. Shunt regulators are popular in Germany, right? Well, they offer excellent sonic performance due to low deteriorating level, Class A operation, not much harmful global negative feedback generated effects here...
The active series pass element (instead of resistor), improves ripple rejection, further improvements are possible by using more adequate current sources and voltage references. Even the LTP would deserve and highly appreciate a better CCS for its function. 1KOhm is a very low impedance compared to the achievable several tens of MOhms. Light diodes are noisy having a zener-like behavior.

May I come up with a slightly different version?
Both series and parallel regulators need a lot of internal feedback. FETs are much more accurate for voltage to current conversion than bipolars.
The LTP is heavily simplified.

[Onvinyl]

Hi Lumba,
thanks for this *very* useful contribution!

Your T14, however, does not turn in my simulation, still keeping my upside-down error amp configuration. As I understand it, it should be part of the current mirror (T12, T13) but probably I'm wrong.

Your input CSS looks tricky, I'll chew on this a bit...
thanks,
Rüdiger

[Onvinyl]

Hm, I tried a more simpler configuration, think my initial post but with 2-bjt current mirror. Whatever I do, I could not make the error amplifier turn on with a current source at the tail. This is hold true with both my perfboard in reality and the simulation.
Rüdiger

[aparatusonitus]

Originally posted by Onvinyl
Your T14, however, does not turn in my simulation, still keeping my upside-down error amp configuration. As I understand it, it should be part of the current mirror (T12, T13) but probably I'm wrong.

T14 is part of CM (only "funny" drown), it should only supplay current for the bases of T12/13.

[Onvinyl]

Leaving T14 for now.

I have the problem, that with a current source in the diffpair tail, the error amp does (often) not work, if the output compound pair (Q1, M1) is attached. With resistor in the tail, this is no problem. I first had this problem with the real circuit, but the sim shows the same behaviour with a non-ideal current source.

So I inserted Q8, an E-follower which also helps Zout in the high freq region a lot.
But, since I did not check if it works in reality, and there might stability issues as well, *and* it was an attempt out of pure desparation: why doesn't it work without Q8?

Rüdiger