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Pwr Amp- Pwr Supply

8/10/1999 2:00 AM
Bill R.
Pwr Amp- Pwr Supply
I'm working on a 60's 50wt rackmount power amp with 2- 6L6's. The power supply is conventional except that after the screen supply resistor and before the driver B+ there is a LARGE (30 watt?) 10k power resistor to chassis ground. This amp has a 2-prong AC pwr cord. Is this simply a "bleeder" resistor to drain off any residual DC from the caps or is there another function? Can I convert this to a grounded 3-wire AC supply? Bill R.  
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8/10/1999 5:06 PM
ken gilbert
Any hefty resistors to ground are there to load the supply down. Is the amp a choke-input or cap-input? IOW, what is connected to the output of the rectifiers?  
This makes a difference as to the function of the resistor. With a choke input there is a minimum current which must be drawn at all times for the supply to function as designed. Also, the bleeder may be there to induce a voltage drop so as to avoid over-voltaging the supply caps downstream.  
And yes, you can easily convert it to a three pronger. It DOES have a power transformer, right?  
8/11/1999 3:39 AM
Bill R
Ken: thanks for the response. Yes it has a pwr transformer and is a cap input design. Uses a 2K resistor instead of a choke btwn pwr tube B+ and screen supply. I havent powered it up to test voltages yet. Bill R.
8/11/1999 6:00 PM
Ken Gilbert
Bill, it sounds like that big resistor is there to cause a constant current draw on the power supply. This could be done for a few reasons:  
a) this will ensure that the voltage at the node after the series dropping resistor (the screen node) is NEVER as high as the raw B+ voltage. A voltage divider is formed between these two resistors which will ensure this effect, which may be to protect the screens from overvoltage, or the filter caps downstream.  
b) believe it or not, causing a constant current to be drawn from a power supply results in BETTER regulation. This assumes that you have enough capacitance to reduce the ripple voltage sufficiently for your application. If that holds true, then by adding a constant current draw to the rail, you will decrease the voltage drop from no load to full load (since the change between those two loads, percentage-wise, is decreased).  
Hope this helps,  

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