G7 PSU voltage divider + RC filter values?

[pasarski]

I have hard time calculating right values for these components in G7 PSU.

http://www.flickr.com/photos/47781016@N06/4382228346/

If it was just a voltage divider it would be easy. And I can use RC filter calculator also. But in this circuit they are sort of combined. Do I just ignore the R1 and C1 when calculating the voltage divider, and ignore R2 and R3 when calculating R1 (and C1)?

(I want 100 V B+ cause I'm using 5840 tube)

 

[riggler]

Here's my two cents:

I think that since you're concerned with DC for your B+ you can ignore the C of the RC network. R1, however, will drop voltage.

What I would do is take a voltage measurement at the node of R1 and C1, and the node of R3 and GRND. Use this voltage as your starting point.

Then consider R2 and R3 as a voltage divider, and do your math from there.

Someone else will probably make a correction to my attempt!

 

[pasarski]

Thanks for your two cents. That's what i thought also, if I understand you correctly. In other words the RC filter drops voltage only after the cap, and R1 is not a part of the voltage divider, I see it now. Damn this was easy after all.

EDIT: No, voltage drops across R1?

 

[gemini86]

Yes, voltage does drop across r1. Use ohm's law, where your tube ma draw and the polarity divider ma draw will be the load, you can calculate what r1 should be to drop you down to 100 volts.

 

[riggler]

Pasarki, (And someone PLEASE correct me if I'm wrong, okay? I am still learning as well.)

The RC network is more ineffective as frequency drops. The reason it is here is to help filter AC noise (120Hz and harmonics, etc.) from the DC current. As the AC current *SWINGS* the capacitor charges and discharges, removing the swing, leaving you with a signal that's pretty much DC.  The higher the frequency of the swing, the faster it occurs, the more efficient the cap is at removing it. DC current = 0Hz so the RC filter has no effect at DC because there is no swing. However, that does not mean that R1 itself has no effect on DC! It does, and it's part of a bigger voltage divider as you see. So yes, R1 does drop DC, as do the rest of the resistors in your diagram.

I said to use the node after R1 as a starting point because I assumed (not knowing the rest of the circuit) that at that point you had a known voltage level to work from. You showed 100V B+ already coming out as drawn. If this is not really the case then yes you will have to recalc R1 too. To play with your polarization voltage, change R2 and R3.

Hope that makes more sense!

 

[pasarski]

Thank you guys, now I get it!

And riggler: Yes, 100 V in my drawing is just the goal I'm aiming at.