#1

hello, I built a preamp last year which I am planning to add a power amp to. In the first reservoir cap in the power supply I have 450V. Tube bias calculator at ax84 says with this voltage, the proper 6l6gc bias current (70% for AB) is 46.67mA. The power amp I will be adding to the preamp will be the SLO 50W power amp section.

I have no clue whatsoever how I will bias the amp when its done since I have no oscilloscope nor do I know anyone with a scope. I also might not be able to raise the volume to just before clipping since I might not invest in building a dummy load(budgets real tight right now). My question is what would the ballpark voltage at the bias supply output have to be to get a decent bias.

I have bought all the necessary parts to build the power amp section. All I have available for the bias is a multimeter.

I have no clue whatsoever how I will bias the amp when its done since I have no oscilloscope nor do I know anyone with a scope. I also might not be able to raise the volume to just before clipping since I might not invest in building a dummy load(budgets real tight right now). My question is what would the ballpark voltage at the bias supply output have to be to get a decent bias.

I have bought all the necessary parts to build the power amp section. All I have available for the bias is a multimeter.

#2

Ill just bump this once.

#3

The voltage in the filter caps isn't going to tell you how the tubes are biased. They just aren't. You can measure one half of the equation all you want, you can't get an answer without the other half.

Get some 1 ohm resistors and calculate the bias current using those. If you're really really tight on cash you can use whatever resistors you have around and just do a bit more math to determine the current. But you can't measure dissipation by only measuring voltage no matter how many ways you slice it.

Get some 1 ohm resistors and calculate the bias current using those. If you're really really tight on cash you can use whatever resistors you have around and just do a bit more math to determine the current. But you can't measure dissipation by only measuring voltage no matter how many ways you slice it.

#4

I have 2 1ohm resistors for this purpose. The thing is I always read online about how you need to input a sine wave and raise the volume until just before clipping. I dont ever see mention of any other methods of biasing. Since i dont have a oscilloscope, I cant exactly bring the output to just before clipping. I dont have a waveform generator either, but those are easy enough to make.

And I think youre wrong about the method of calculating bias. The plate voltage is going to tell you one part of the equation "P = I * E" which is "P = I * 450". the P Im aiming at is 21W(for 70% bias), so the I have 2 parts for the equation, "21W=I*450V" which is then "I=21W/450V". the equation tells me that I need to bias the tubes with 46.6mA. this is the voltage I would have to look for through the 1R at the cathode.

And I think youre wrong about the method of calculating bias. The plate voltage is going to tell you one part of the equation "P = I * E" which is "P = I * 450". the P Im aiming at is 21W(for 70% bias), so the I have 2 parts for the equation, "21W=I*450V" which is then "I=21W/450V". the equation tells me that I need to bias the tubes with 46.6mA. this is the voltage I would have to look for through the 1R at the cathode.

*Last edited by bustapr at Aug 28, 2014,*

#5

You're playing around with the equation thinking that somehow you can tease out an answer using insufficient data. The fact that you know what the answer should be is meaningless if you do not measure what your variable actually

You can measure your scale length all you want, and you can know down to eight decimal points the tension you need for proper pitch, but if you can't measure the pitch your string is actually producing, you're not tuning your guitar. All the math in the world isn't going to save you if you cannot measure a necessary part of the equation. You're basically trying to tune a string without knowing its pitch. There's no substitute for that variable. You can measure it in any number of different ways, but you

All that said, I'm not sure what you were reading but you do not need a signal generator to bias your amp. Guitar amps are usually biased at idle, so you don't need any kind of signal at all. The amp just needs to be on. Plenty of tutorials out there for how to do this.

http://www.duncanamps.com/technical/lvbias.html

*is*. You need two measurements; you have one. Manipulating the equation is not going to magically yield another measurement that you do not have. It that were possible, it would be unnecessary to have that variable in the equation and we'd all just bias off of plate voltage.You can measure your scale length all you want, and you can know down to eight decimal points the tension you need for proper pitch, but if you can't measure the pitch your string is actually producing, you're not tuning your guitar. All the math in the world isn't going to save you if you cannot measure a necessary part of the equation. You're basically trying to tune a string without knowing its pitch. There's no substitute for that variable. You can measure it in any number of different ways, but you

*have to measure it.*All that said, I'm not sure what you were reading but you do not need a signal generator to bias your amp. Guitar amps are usually biased at idle, so you don't need any kind of signal at all. The amp just needs to be on. Plenty of tutorials out there for how to do this.

http://www.duncanamps.com/technical/lvbias.html