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.

 

[abbey road d enfer]

Hello, now I have build, measured and did variations to this capacitance multiplier circuit.

The resistor R1 was choosen from 180 Ohm (low filtering) to 3K6 Ohm (higher SNR). Current flow is 40mA.
With high value of R1, Uce becomes 8.75V.

That's because of the built-in resistors that draw more current than the actual bases.

What astonished me is the constant Ube of the darlington if values of R1 changes. When the voltage drop across R1 becomes greater, Uce becomes greater in the same way so the transistor has its fixed working point which can't be altered.

This is the way transistors work. Their op point is governed predominantly by collector current.

When changing R1 to 33K, the darlington died.

Probably because the dissipated power exceeded nominal. With such a high value as 33k, all the voltage drop is dissipated in the transistor.

When changing R1 from low value to high filtering, the sound becomes more and more squeaky.

Have you measured the actual output voltage vs. resistor value. I would think that with the highest value, the output was very low.

Is it the wrong working point for this IC?

Yes.

Is there anything left to improve the sound or filtering performance of this PSU?

Either use a discrete Darlinton or a MOSFET.

 

[Walter66]

Still searching for a Darlington TO-220 IC with high voltage (400V) and high DC current gain at 60mA load.

Most of those units are designed for other purposes and have their max. hfe @ currents far greater than 100mA.

I really think that the working point for this 2SD1409 is correct.
Ube= 0.7V and Uce=3.2V.

And yes, with higher values of the R1 resistor the filtering becomes more effective, but the audible sound decreases, too.

I really don't know why this is happening? From a technical view, what working point is to choose when the highest SNR doesn't improve the sound most? How does other audio- manufacturers solve this problem?

We all learned that the best technical solution delivers the best possible sound. True or false?

It does it's job but I'm wondering on how to find the best bipolar Darlington for this task?

Are there any other fields of application besides ignition coils and horizontal screen deflection circuits?
Doesn't anybody needs a high current gain Darlington with low current AND high voltage?

I'd like to try another Darlington for sound comparisons.

 

[abbey road d enfer]

I really think that the working point for this 2SD1409 is correct.
Ube= 0.7V and Uce=3.2V.

No. Vbe in normal operation is about 1.4V. 0.7V shows that only one junction is correctly biased.

And yes, with higher values of the R1 resistor the filtering becomes more effective, but the audible sound decreases, too.

Did you do your homework? measuring Vout as a funtion of R1?
I don't think so.

I'd like to try another Darlington for sound comparisons.

As I said earlier, use a discrete Darlington, so you can get rid or tweak the resistors.

 

[Cqwet Dbdfte]

Use a MOS FET, lots of choices and basically no current draw thru gate.
Protect the gate with a Zener diode.
Lots of HV circuit designs to choose from, that also avoids big (expensive, noise prone) chokes, and and many big caps.
The ideal power supply has a fixed voltage, regardless of load, and no noise. While "zero" is an an objective, microvolts is possible.
The circuit diagram above looks wrong.

 

[abbey road d enfer]

The circuit diagram above looks wrong.

Can you elaborate?

 

[Cqwet Dbdfte]

0,48V Vcb

 

[abbey road d enfer]

0,48V Vcb

That's because the base resistor is too small, thus the Darlington is not properly biased, but the principle is good, it works with a much larger base resistor.

 

[Walter66]

No. Vbe in normal operation is about 1.4V. 0.7V shows that only one junction is correctly biased.

Did you do your homework? measuring Vout as a funtion of R1?
I don't think so.

As I said earlier, use a discrete Darlington, so you can get rid or tweak the resistors.

Ube= 0.68V seems to be a constant, whatever value of R1 is choosen.

R1 varied from 180 Ohm to 1K Ohm or even 3.6K, the value of Ube isn't changing at all.
If R1 goes much higher (33K), the Darlington died (Uce became too high).

Ucb and Uce do vary as R1 has different values.
Uce= 8.75V with 3.6K value of R1
Uce= 3.2V with 1K value of R1

This means, with a very small value of R1 the Darlington goes into saturation and is fully open, even with Ube=0.7V?

What astonished me is the constant Ube of the darlington if values of R1 changes. When the voltage drop across R1 becomes greater, Uce becomes greater in the same way so the transistor has its fixed working point which can't be altered.

"This is the way transistors work. Their op point is governed predominantly by collector current."

So I can never reach 1.4V=Ube with a current of 60mA when now using 40mA and the result is 0.7V?

 

[Walter66]

That's because the base resistor is too small, thus the Darlington is not properly biased, but the principle is good, it works with a much larger base resistor.

How should this work when Ube= constant and it's only dependend from the current flow and not from variation of R1?
I did change the value of R1 but Ube doesn't changed at all. Just Uce changed and that's what I don't understand:
How can the Darlington become more or less open for current flow when base voltage doesn't change?

 

[abbey road d enfer]

Ube= 0.68V seems to be a constant, whatever value of R1 is choosen.

R1 varied from 180 to 1K Ohm or even 3.6K, the value of Ube isn't changing at all.
If R1 goes much higher (33K), the Darlington died (Uce became too high and after death: zero!)

Ucb and Uce do vary as R1 has different values.
Uce= 8.75V with 3.6K value of R1
Uce= 3.2V with 1K value of R1

There is something very wrong. If the documentation for the 2SD1409 is correct, Vbe should be 1.3-1.4V even with a 27k resistor, and Vce should be about 10V.
I suspect your Darlington is shot.

 

[Walter66]

There is something very wrong. If the documentation for the 2SD1409 is correct, Vbe should be 1.3-1.4V even with a 27k resistor, and Vce should be about 10V.
I suspect your Darlington is shot.

It's not shot. I put a new one into the circuit and still the exact same Ube=0.7V.
From the datasheet, it would need 1.2V to open at a current of 40mA, but this value isn't achievable, as the base voltage remains constant.

With R1=2K7, Uce becomes 7.2V, this voltage becomes smaller with smaller value of R1.
So the resistor determines how much this Darlington is open or closed.

If the base voltage remains constant, how can the Darlington change Uce with different values of R1?
Why the base voltage doesn't change with different values of R1? Please explain.

 

[ruffrecords]

Probably because all you are seeing is the action of the 2K7 resistor.

Cheers

Ian

 

[JohnRoberts]

It's not shot. I put a new one into the circuit and still the exact same Ube=0.7V.
From the datasheet, it would need 1.2V to open at a current of 40mA, but this value isn't achievable, as the base voltage remains constant.

With R1=2K7, Uce becomes 7.2V, this voltage becomes smaller with smaller value of R1.
So the resistor determines how much this Darlington is open or closed.

If the base voltage remains constant, how can the Darlington change Uce with different values of R1?
Why the base voltage doesn't change with different values of R1? Please explain.
View attachment 119118

283.2v-272.5v=0.7v which is too little voltage for two forward base-emitter junctions.

289.7v -283.2v= 6.5v 6.5v/2.7k= 2.4mA the 560k is unlikely to drain off much of that current so where is it going?

the 2.5k across the top darlington base-emitter will only account for roughly 0.2mA so over 2 mA is going into the base?

When properly forward biased there should be two diode junctions across the full darlington device.

I suspect the darlington device is faulty (or voltage measurements are).

JR

 

[Cqwet Dbdfte]

Yep, a working Darlington would show two Vbe drops 1.4V. You can check this with a DMM before it goes into the circuit.
Big inductors can kick out some nasty surprises. Better check that on/off action on a scope, DC in 10X probe.
Protection diodes are in order.
That PSU circuit is a lot of effort to get rid of some hum for a pre amp some RC network, or a HV MOSFET+LM317 would easily deal with.
Channel separation in preamps with next to constant current draw not much of an issue. One good regulator should do ok.
Channel separation is also limited by the space between your ears, in an acoustic environment that blends the two together. I think this is old school "figure of merit" of practical little value.

 

[abbey road d enfer]

It's not shot. I put a new one into the circuit and still the exact same Ube=0.7V

Which again shows that the rest of the circuit is wrongly dimensioned.

Why the base voltage doesn't change with different values of R1? Please explain.

You must understand that junction transistors are current devices. E.G. for a variation of 100x of the base current, Vbe increase only by about 120mV.
So when you say the base voltage does not change, it actually does, but probably not by what you expect. Transistors are not tubes.
Anyway, less than 1.2V in a Darlington results in one of the transistors not being properly biased.