Please review this power supply.

[Lightium]

Hello,

Please review this power supply for ion-thruster experiments.
View attachment 137891

 

[Nigel Goodwin]

My immediate impression is the poor drive to Q5, using a darlington configuration is a very poor idea, drastically reducing drive voltage and current. Reconfiguring the circuit with Q4 as PNP would be a big improvement.

I'm also not very convinced by the use of a single ended class A output stage, which will be very inefficient and get very hot.

Does your simulation allow you to calculate the dissipation in Q5?.

 

[Lightium]

Thank you for the reply Nigel.

Yes I can view the dissipation.
I am running it high for maximum current output.

I am not very good at amps, I did have trouble with it up to this point. For 30 watts in I am getting 130 watts out. What would you suggest to amplify the osc?

 

[Nigel Goodwin]

Thank you for the reply Nigel.

Yes I can view the dissipation.
I am running it high for maximum current output.

I am not very good at amps, I did have trouble with it up to this point. For 30 watts in I am getting 130 watts out. What would you suggest to amplify the osc?

Need less to say, for 30W in you can't get 130W out, that would be impossible, do you mean 300W in for 130W out?.

What exactly are you trying to do, and to what purpose?.

 

[Lightium]

R8 and C8 form a spark gap. When I compare the output of the battery (V2) with output from the spark gap. I get 30 watts in and @130 watts out.

 

[Lightium]

View attachment 137893

 

[Les Jones]

I don't understand how you can consider that R8 and C8 can be considered to behave like a spark gap. Can you explain ?
I was wandering how you worked out that 30 KV at 4 mA was 130 watts. I think that should be 120 Watts. You have not given any details of the load. As you have not shown any details of the load I wondered if you considered the 200 meg resistor (R8) to be the load. But with 30 Kv across it it only dissipates 4.6 watts.
I agree Nigel that you can't get more power out than you put in.

Les.

 

[crutschow]

Don't confuse the instantaneous peak energy at the output with the continuous (average) power in.
You have to average the power out for that.

You can display the integrated average by doing a CTL left-click on the plot display power node name.

 

[Lightium]

Thanks Crutschow, that's going to be very helpful.

 

[Lightium]

O.K. now, according to crutschows help I get 3.7868 Watts on the output.

And thanks to Nigel I am considering a class D amp.

 

[crutschow]

I get 3.7868 Watts on the output.

One small point I didn't mention is, you need to take the average over a integral number of signal periods for the best accuracy.
Thus, if the signal period is 1ms, the integration period must be multiples 1ms.

 

[atferrari]

Both links apear broken.