Capacitance multiplier: which Darlington to choose?

[abbey road d enfer]

OK. These values were not mentioned in the datasheet I had.
So that means that the current drawn by the base of the Darlington is about Iout/600+0.28mA.
For 50mA output, it amounts to 0.36mA. For 150mA output 0.53mA.
This is the value to use in the calculation.
R1=(Vunreg-Vreg)/Ib

I made a simulation of the circuit. I don't have a Spice model for the 2SD1049, but I hacked one from the datasheet.
It shows that the current that the R1 resistor must supply is dominated by the current drawn by the 2.5k base-to-emitter resistor, so the output voltage does not vary much with load. So one could say it's a regulator in addition to a capacitance multiplier. Load regulation is pretty good, 0.7V for Iout varying from 30 to 300mA.
Of course the value for achieving 320V at the output depends very much on the input voltage.
With 400V input and 320V output I see a 300k resistor.
With 350V input I see a 100k resistor.

Actually, the output voltage follows the input voltage with an almost constant difference.
With 100k, the difference is 30V, with 300k it's 80V.
When you think of it, it is not so surprizing, since the current drawn from R1 is almost constant.

 

[Walter66]

Is it 50mA or 150mA?

Because the calculation is sbject to too much unknowns, I suggest you experiment with it.
190 ohms is certainly way too low, but not likely to result in failure of teh circuit. Il may result in failure of teh circuit is is connected to, so I would suggest caution and using a dummy load.
320V at 50mA is 16W, 48W at 150mA, so this dummy load should be commensurate.
Actually, with such a low value of R1, the output voltage would be close to the unregulated voltage, which you haven't specified.

It's 50mA current draw. hfe Ic should be 400 for this working point. I use a dummy load for experimenting.
My concerns with the low 190 Ohm is, when the cap is discharged, there is the complete Ub of approx. 280V on the circuit and only this small resistor limits the current, means high current at point zero of the time scale. Will it destroy the circuit?

 

[Walter66]

OK. These values were not mentioned in the datasheet I had.
So that means that the current drawn by the base of the Darlington is about Iout/600+0.28mA.
For 50mA output, it amounts to 0.36mA. For 150mA output 0.53mA.
This is the value to use in the calculation.
R1=(Vunreg-Vreg)/Ib

I made a simulation of the circuit. I don't have a Spice model for the 2SD1049, but I hacked one from the datasheet.
It shows that the current that the R1 resistor must supply is dominated by the current drawn by the 2.5k base-to-emitter resistor, so the output voltage does not vary much with load. So one could say it's a regulator in addition to a capacitance multiplier. Load regulation is pretty good, 0.7V for Iout varying from 30 to 300mA.
Of course the value for achieving 320V at the output depends very much on the input voltage.
With 400V input and 320V output I see a 300k resistor.
With 350V input I see a 100k resistor.

Actually, the output voltage follows the input voltage with an almost constant difference.
With 100k, the difference is 30V, with 300k it's 80V.
When you think of it, it is not so surprizing, since the current drawn from R1 is almost constant.

Thank you so much for the simulation. It could be interpreted that if someone wants low voltage loss on the "regulator" circuit, he chooses the 190 Ohm resistor. For higher input voltages, the resistor value can be choosen higher. Higher input voltage in combination with a higher value of R1 should improve the noise supression, too.

 

[abbey road d enfer]

It's 50mA current draw. hfe Ic should be 400 for this working point. I use a dummy load for experimenting.
My concerns with the low 190 Ohm is, when the cap is discharged, there is the complete Ub of approx. 280V on the circuit and only this small resistor limits the current, means high current at point zero of the time scale. Will it destroy the circuit?

It shouldn't destroy anything. There will be a temporary current surge in the resistor (your R1) but not enough to destroy the resistor (unless you use a 1/10th W), but anyway 190 ohms is way too low.
As the simulation and the calculation show, the resistor value should be in the order of kiloohms. Unless you want to have a very small voltage drop, which is not a good choice.

 

[Walter66]

Thanks very much! Will start within that kiloohms range for experimenting.