Need a little help with a power supply I designed.

[bill mountain]

I have been playing around with tubes for a few years but I have always used published designs. This time around I am trying to design a power supply for various 12A_7 preamps that I plan to build.

Here is the power supply:



The B+ is a little higher than I would like. I thought the voltage doubler was going to double the AC voltage but it appears that it has actually doubled the rectified voltage (I must have missed that when I was reading about voltage multipliers). I bread-boarded a common cathode stage last night to test it out. It seems to work fine but I am trying to think of ways to lower it 70-100 volts. The 12A_7 tube data sheets all seem to have a max plate voltage around 300V. I'm not sure if the max voltage is measured after the plate resistor or not. If my B+ is safe for these tubes then I won't worry about it.

The winding for the heaters is supposed to be 12.6V/ 300mA. What scares me is that it is actually putting out almost 16 volts! The tube worked fine last night but I'm worried about long term stability. I tried rectifying it and then using a regulator but there wasn't enough current left to power the heaters.

I plan on putting trimmers on the heater windings and adjusting down to 12.6V. Is there a better way to do this?

Thanks for reading! Any advise or criticism will be greatly appreciated.

[M Gregg]

Originally Posted by bill mountain

The B+ is a little higher than I would like. I thought the voltage doubler was going to double the AC voltage but it appears that it has actually doubled the rectified voltage (I must have missed that when I was reading about voltage multipliers).

Remember your meter reads RMS not peak AC. You will charge the caps to peak not RMS.

Just a thought, what is the difference between the voltage after the Anode load with no current draw and with the heaters working IE with current draw. If you exceed the flash over voltage of the tube? I have seen this on EL34's when the cathode resistor goes "pop". With no current draw the volt drop is 0 across the resistor.

Regards
M. Gregg

[bill mountain]

Originally Posted by M Gregg

Remember your meter reads RMS not peak AC. You will charge the caps to peak not RMS.

Regards
M. Gregg

So what you're saying is that my voltage is even higher? I understand RMS and Peak in relation to power amp output. Is it the same for DC voltage?

[Michael Bean]

Transformers are rated to deliver the specified voltage at the specified current, so with no load the voltage readings are going to be higher. Put a resistor equal to the intended load on the transformer and power supply output, then measure the voltage.

Mike

[bill mountain]

Originally Posted by Michael Bean

Transformers are rated to deliver the specified voltage at the specified current, so with no load the voltage readings are going to be higher. Put a resistor equal to the intended load on the transformer or power supply output, then measure the voltage.

Mike

This sounds good. How do I figure the load for the heaters?

[Fenris]

V = I R
12.6 = .3 R
R = 42

[M Gregg]

Originally Posted by Michael Bean

Transformers are rated to deliver the specified voltage at the specified current, so with no load the voltage readings are going to be higher. Put a resistor equal to the intended load on the transformer or power supply output, then measure the voltage.

Mike

As above the voltage out of the Tx is different on load.

Regards
M. Gregg

[bill mountain]

Originally Posted by M Gregg

Just a thought, what is the difference between the voltage after the Anode load with no current draw and with the heaters working IE with current draw. If you exceed the flash over voltage of the tube? I have seen this on EL34's when the cathode resistor goes "pop". With no current draw the volt drop is 0 across the resistor.

Regards
M. Gregg

You'll have to forgive me but I don't follow what you're saying. It takes me a while to understand this stuff so I'll need to read it a bunch of times to give you an appropriate response!

[bill mountain]

Originally Posted by Fenris

V = I R
12.6 = .3 R
R = 42

I never think of OHM's law. It would solve so many problems!

If I'm not mistaken at 12.6 the current is 150mA so it might actually be:

12.6 = .15 R
R = 84

[M Gregg]

Originally Posted by bill mountain

You'll have to forgive me but I don't follow what you're saying. It takes me a while to understand this stuff so I'll need to read it a bunch of times to give you an appropriate response!

When you first switch on, the tubes are not drawing current.
So the voltage drop across any resistors in the anode have no effect on the applied voltage.
If the B+ is higher than the flash over voltage of the tube- it will flash over. ie "short out".