Biasing Math/Terminology Question

[phyrexia]

Hi. This is probably a silly question.

In the bias supply of my 60W Mark III, will more resistance = 'less' negative voltage (ie closer to zero) or will more resistance give a higher negative voltage (ie further from zero)?

"Hotter" bias means higher negative value, correct? EL34 needs hotter than 6L6, etc. Right?

Also, maybe some of the techheads will know if I'm phrasing all this right as far as 'less' vs 'more' since we're dealing with negative voltage and it's been a while since my math classes.

Thanks lots,
Victor

 

[Buddy]

The concept is the same with either neg or positive voltages .
You are using a voltage divider to get the desired Vo .
http://people.clarkson.edu/~svoboda/eta/designLab/VoltageDividerDesign.html
Circuit Design Tutor
This cool little tool should demonstrate .

 

[stokes]

Better to think in terms of plate current.If you increase neg voltage,tube current will decrease,if the negative voltage is decreased,current goes up.So to give an example if -52v on the grid gives you 25ma plate current, then -42 will give you more than 25ma's plate current.You must also take into consideration the plate volts and determine the watts at idle.You get idle watts by plate current times plate volts.So,350vX .060a's=21 watts idle.Be aware that 60ma's is .060 in this formula.

 

[JOEY B.]

In terms of a fixed bias Boogie, replacing the bias resistor(s) with one(s) of a lower value will decrease the amount of resistance in the circuit and move the negative voltage at pin 5 closer to zero, thus making the tubes pull more current (hotter). And I'm no techhead, for sure. :lol: . Hope this helps. This webpage is very helpful at calculating the wattage and such.

http://www.webervst.com/tubes/calcbias.htm

 

[phyrexia]

There Joey goes, dumbing it down just enough so I can understand it.

thanks for everyone's input so far.

 

[JOEY B.]

Again, I am no tech, for sure. Just call me the wordsmith.