Power Supply - Schematic attached

[EFK]

I'm still attempting an education in understanding schematics and how things work - just for the sake of interest - and I've learned a lot on this forum so thank you to everyone.

Take a look at the attached schematic - this is the power supply in one of my amps (Gibson GA42RVT). This is ss rectified and I believe D1/D2 are the actual rectifying diodes, correct? Then I see the capacitors at C41/C42 and C39/C40 on the other side of the standby switch. What I really want to know is: what is the purpose of the 220K resistors which appear in (parallel?) to these caps? Is this some kind of simulated sag? Any info appreciated - thanks!

[defaced]

They are to balance the voltage across the caps. If they're omitted, the voltage drop across each cap could be different, and if things went out of whack could exceed the rating of one of the caps. They also drain your caps when the power is turned off. Very nice safety benefit for this type of circuit.

I think I know the technical reason why this can happen, but I'm not positive so I'll let someone else explain why.

[mooreamps]

That is absolutely correct ; on both counts.


-g

[defaced]

What I'm fuzzy on is what causes the imbalance. Is it the cap-to-cap difference in leakage current?

[jrfrond]

ESR (Equivalent Series Resistance) is different from cap-to-cap, even with modern manufacturing and close tolerances. If they are seriously unbalanced (especially the bottom one of the pair), a cap could draw too much charge current and pop. Adding external resistors that you KNOW are balanced takes the ESR out of the equation. It might be just fine with no resistors, and they are not used by all manufacturers, but I HAVE seen a series cap pop a number of times when no resistors were used. Coincidence? I don't know.

In the same manner, you will often see a small cap across each rectifier diode to squelch switching transients.

[EFK]

OK I understand this perfectly! Thank you! Now, if I wanted to insert a resistor for a tube rectifier sag simulation, would I install it into the circuit at the circled area (see attachment)?

[tedmich]

the resistor would work; would you want to switch it in and out? I ran across this nice method to drop B+ volts cleanly and cheaply with a zener and mosfet (the zener alone will wok but costs more). the switch will be much less strained with the zener /mosfet circuit.

its from RG Keen:http://www.geofex.com/Article_Folder...osfetfolly.htm

[black_labb]

the resistor would be in the right spot. if you want to get more similar to a tube rectifier sound then the resistor is a good bet. a zener ot a zener and mosfet would reduce the voltage a certain level, so V=IR doesnt create a sagging b+. both will work.

[EFK]

Thanks guys! If I just use a resistor, do you think one of those aluminum-housed 25W would be appropriate? The amp is cathode biased and they are used for the cathode resistor, and consequently I have a lot of them in many different values. Or should I step up the watt rating to maybe 30? Actually 25W seems like waaaay overkill but I'm not an expert. ???

[black_labb]

depends on the current going through it. V=IR to find voltage it will drop. P=IV to find the power it will dissipate. make sure you use the current the tubes pull at full voltage.

i used a 10w 270ohms for a 5e3ish build, which seems fine, but i havent put the amp through a huge amount of playing yet. mine is switchable (more voltage for when i use a zener for the cathode bias to get closer to fixed bias.)