Power supply RC (ohm's law) question -NEWBIE ALERT

[Blue Jinn]

OK,

Trying to get my head around this. I want about 100VDC out from a simple bridge and RC filter network with a 100v zener diode across the output as a sort of regulator. I'm stuck with this topology, because that is all the PCB is setup for. I have 120 volt transformer winding feeding the bridge. I have room on the PCB for another resistor (maybe) on the anode of the zener.

I can't seem to calculate the value of the R to knock down the 160or so volts after teh diodes to 100v. The tube plate and screen is drawing maybe 10mA, and the plate resistor is 100k.

Caps are 220uF.

[tomchr]

If you want the zener to provide any regulation, you need some current running in it. You can look up the recommended current in the datasheet for the zener diode, or figure that you'll need a few mA. So if the circuit this supply is driving is drawing 10 mA, then you need to size the resistor for a current draw of 10 mA plus the zener current.

So let's assume you want 5 mA in the zener. Note that at 100 V, this equals 100*0.005 = 0.5 W dissipated in the zener. It needs to be able to handle this.

Then as outlined above, use ohm's law. R = (Vin-Vzener)/I --> R = 60/0.015 = 4000 ohm (= 4 kOhm). I'd use 3.9 kOhm as that's the nearest lower standard value. If you want slightly lower current in the zener, go with 4.3 kOhm or 4.7 kOhm.

~Tom

[DF96]

You need to allow for the current drawn by the zener diode too. This is often 5mA minimum, but check the zener datasheet. If 5mA then the resistor will be 4000 ohms.

[Palustris]

I was going to answer this but the numbers don't add up: you can't have a B+ of 100V with a plate load of 100k and 10mA going through the tube. It has to be 1mA going through the tube.

[dgta]

Yup, that set of numbers ain't gonna happen. That pesky Ohm guy again

[DF96]

Oops! I should have spotted that.

[Blue Jinn]

Thanks to all. The tube is a 5840 pentode. Screen is connected to plate. Internal resistance is 260k and the plate resistor is 100k, I thought that you paralleled the two to obtain 72k (Or am I thinking of something else?)

I took the 10mA figure from the data sheet for screen and plate current at average operating conditions.

[DF96]

You parallel the anode/plate resistance and the external resistor to get the AC output impedance. For determining the working point you just use the 100k. You have to work things out at your working point, not the one given in the data sheet, unless they are the same.

I doubt if the plate resistance will be 260k with the screen grid connected to the plate. This is triode connection, which has a much lower impedance.

More reading needed?

[Blue Jinn]

More reading, and less posting while traveling....:-) To recap, I should be use the 100k plate resistor as the load. From there it should be pretty straightforward to figure the value of R as indicated above.