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How to design a HV start-up circuit

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refugee

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I'd like to design a HV start-up circuit with zero static current. Does it possible? Somebody told me that it's an impossible task.
 

I guess the fact that you don't get an answer would show that nobody understands your problem!? At least I don't. High Voltage? 5V? 1kV? Start-up what? A current mirror? An oscillator? A company? You better provide a little bit more info! ;-)
 

An ideal inductor will work this way.

But it will also cause an DC source to short after a short while.
 

erikl said:
I guess the fact that you don't get an answer would show that nobody understands your problem!? At least I don't. High Voltage? 5V? 1kV? Start-up what? A current mirror? An oscillator? A company? You better provide a little bit more info! ;-)

Hello erikl: I have same problem, the posting doesn't provide sufficient info to give useful reply.
 

erikl said:
I guess the fact that you don't get an answer would show that nobody understands your problem!? At least I don't. High Voltage? 5V? 1kV? Start-up what? A current mirror? An oscillator? A company? You better provide a little bit more info! ;-)

Please forgive me I did not make it clear.
Let's say that I need to design a boost converter. Vin=5~30V. So, I need design an internal power supply block. In this block, I designed a constant gm to generate current. So, I need a start up circuit.
But I don't know how to design a start up circuit with zero static current. The attachment is my circuit.
The only way to reduce the static current is to increase the resisitor. If R=2M, Vin=20V, the static current is 10uA. It's too large.
 

refugee said:
... The only way to reduce the static current is to increase the resisitor. If R=2M, Vin=20V, the static current is 10uA. It's too large.
Why not use a capacitor instead of the resistor? It's a short circuit at start, then loads up, and finally the current tends to zero. After switching off, you'd have to wait for a while before a new start.
 

Hello,

When this circuit is for start up only, can you disconnect the resistor from ground after the main process runs?

Other method is to store some energy in a capacitor and use an external pulse or thyristor/diac like circuit to generate the current pulse required for starting the main process.

The thyristor like approach works independently of input dV/dt as you need a minimum stored charge before the thyristor fires. I use this approach in power oscillators to start them.
 

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