Using low voltage lamp in power supply

[Darkstar52]

Could the secondary of the xfmr be carefully unwound til you reach your target voltage?

[kevinkr]

A small power resistor will work just fine and will make a CRC input filter circuit which is going to be required to knock the ripple down sufficiently for reasonable performance..

Anything between 220 - 270 ohms should work. Use a 5W resistor..

[Joshua_G]

Originally Posted by Frank Berry

You may be able to put a 24 volt zener in series with the output of the supply.
That will give you a 24 volt drop.

Zeners are noisy, not recommended here.

[Frank Berry]

A few millivolts of zener noise in a 140 volt supply?
Why would this be a problem?

[cbdb]

Even if it was a good thing who would buy an amp that needs a lightbulb replaced regularly.

[cbdb]

And it wouldnt compress the signal, just decrease the supply voltage with increased current demand.

[EddieRich]

Originally Posted by Darkstar52

Could the secondary of the xfmr be carefully unwound til you reach your target voltage?

To much trouble, not enough knowledge or skill.

Originally Posted by Frank Berry

You may be able to put a 24 volt zener in series with the output of the supply.
That will give you a 24 volt drop.

Originally Posted by kevinkr

A small power resistor will work just fine and will make a CRC input filter circuit which is going to be required to knock the ripple down sufficiently for reasonable performance..

Anything between 220 - 270 ohms should work. Use a 5W resistor..

Just found some 243 ohm 3W resistors. I think I'll skip the light bulb idea for now. I will definitely replace the resistor with a light bulb once I'm done because my curiosity has been tweaked and I need to know.

Originally Posted by cbdb

Even if it was a good thing who would buy an amp that needs a lightbulb replaced regularly.

Pilot lights on old radios used to be wired in series with the filaments. The pilot light burns out, the radio stops working. Or even worse, pilot light shorts and causes excess current through the filaments and burns the tubes out as well.

I understand your point, but I'm not selling anything. I'm making one, and only one of these for use exclusively in my house. Like most things I've made, it will end up in the basement after a year or so of use, or when I move on to a push-pull design.

[ChrisA]

Originally Posted by EddieRich

I'm designing a two channel tube amp using 60FX5's. I am using an isolation transformer and when rectified it gives me 161 VDC. I want to run the tubes with 140V on the tube and a cathode bias of 3V for a total of 143VDC. So, I need to drop 18V. The Q point of the circuit will draw 36mA plus 10mA from the screen per channel, for a total of 92mA.

If my math is correct, a 24V 3W bulb will drop 17.6V at 92mA.

So my question is, can I use a 24V 3W lamp to drop the voltage?
I know it's not a plain old resistor, so does this effect the circuit in any way?

A light bulb is exactly the wrong thing to use. Their resistance is a function of temperature but it goes the "wrong" way. It is low when cold then the resistance builds. You'd be better off with a sandstone resistor.

If you are going to use a resister to drop voltage why not put that drop to good use? Add in an extra section of CRC filter. As long as you are forced to use the resistor adding just one more small 40uF cap will greatly reduce the 120Hz ripple on the B+

[Frank Berry]

If you put an electrolytic capacitor after the lamp, there is a good chance that the lamp will blow before the capacitor charges.
Initially, you will have the full power supply voltage across the lamp.

[Tom Bavis]

You may NOT need to drop voltage at all - in fact you may net even get as much as 140V. Was 161V measured under load or open circuit? Expect it to drop considerably under load.

If it IS voltage at full load, one trick that should work is to reverse the isolation transformer primary and secondary. They're wound with a small "step up" in ratio to compensate for the voltage drop under load. Reverse it and you'll get a small step down plus the normal voltage drop.