Hoffman Amplifiers Tube Amplifier Forum
Amp Stuff => Tube Amp Building - Tweaks - Repairs => Topic started by: kagliostro on March 17, 2023, 08:54:12 am
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This days I spent some time to find documentation about the use of Mixed Bias on Tube Amps (Fixed Bias + Cathode Bias)
Now I'm asking help for the math to use
As I'm not very comfortable with calculations usually to choice a Fixed Bias Voltage or the value of a Cathode Resistor I look to similar plans and plagiarize or, at least, I use an online Bias calculator
This time I must do an effort and learn some math
I would like to learn which are the calculations envolved to establish the negative Fixed Bias voltage and the value of the RK resistor if I want to realize say ... 35% of fixed Bias and 65% cathode bias, 35% and 65% are an example, may be I would like to use different percentages of each different Bias
Values that I have to use are the 12W of dissipation of the el84 tubes, I'll use a pair, so total 24W max plate dissipation, 330v B+ (may be anode cathode voltage around 315/325v, may be) and a load of 8Ka-a
Many Thanks
Franco
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You seem to be sharply focused on mixed bias these days. Why?
This is what I would try. I'll use my AC-15 (http://sluckeyamps.com/VAC15/Vox_AC15.pdf) amp as an example since the voltages are pretty close to your numbers. I have decided that 12V will be my magic bias number. Remember, bias voltage is the difference between grid and cathode voltages. So, I want 35% of that 12 volts to be a negative voltage placed on the grid. 35% of 12 is 4.2V. I adjust my negative bias supply to put -4.2V on the grid. That leaves 7.8V to be put on the cathode. So, I experimentally determine the value of cathode resistor that will put 7.8V on the cathode. The combined effect will give a total bias voltage of 12V, which is what I wanted.
This may not be the exact answer you seek and my example may be overly simplistic, but this is what I would try. I don't know if it's even possible to completely solve your problem in the classroom. Sometimes you must also go to the lab to get the total answer you seek.
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Ciao Steve, Grazie
Yes, this days I'm trying to understand better what is around Mixed Bias
The reason ? The purpose is, as I'm drawing a conversion plan for a Geloso G1 1020 A for a guy, to use the existing Fixed Bias and to experiment with something new to me and at DIYItalia
Your suggestion usually will be my approach, but as I haven't the amp under hands (it is of a guy that lives far from me) I'm trying to give him the more support I can (he isn't so skilled) and if along the way I learn something new is all gained for me
Franco
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Use a plate characteristics chart to plot a load line and bias point (to calculate total cathode current) and determine the bias voltage split (35% grid 65% cathode or whatever) and (having calculated the cathode current already) use ohms law on the cathode voltage to get the cathode resistor value you need.
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Ciao Tubeswell
Thanks
Plot a load line ... :help:
OK let's start further :sad2:
Say I know that the current that must flow on cathodes must be 60mA (330v B+ - 2 x EL84 Tubes - 8Ka-a load)
Cathode Bias = 65% of 60mA = 39mA
Fixed Bias = 35% of 60mA = 21mA
To obtain 7.8v on the Cathode Resistor @ 39mA flow the resistor value must be 200R
And to have the other 21mA flowing on the catodes, to obtain the 60mA, as Steve say -4.2v must be supplied to G1
:dontknow: :dontknow: :dontknow:
Total Crazy :w2: :w2: :w2: or Correct :w2: :w2: :w2:
I really don't know
Franco
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Keep it simple.
Say that a 13QU6 typically wants 234 ohms self-bias or -13V fix-bias.
Try 117 ohms with -6.5V for 50:50 mix-bias.
You want mostly self-bias? Try 0.65*234r with 0.35*13V.
Pre-compute a safe cathode current for the B+ voltage you expect. Have a 1-Ohm series resistor and a DVM connected for smoke-check and be ready to shut-down fast if current goes too high.
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Ciao PRR Many Thanks
So, if I understand you correctly
I start from the specifications of the el84 and check, for the same voltages, the required negative voltages and the RK value
(https://i.imgur.com/XcgMvt8.png)
in AB1 PP using 300v B+ and a load of 8Ka-a application data are
(https://i.imgur.com/PPutEv3.png)
Fixed Bias = -14.7v
Cathode Bias = 130R
to set Fixed 50% + Cathode 50% is -14.7v / 2 = -7.35v and 130R / 2 = 65R
if I want Fixed 35% + Cathode 65% is 0.35*-14.7v = -5,145v and0.65*130R = 84.5R
This seems a simpler way to consider things
my only incognita is about the fact B+ isn't 300v, it is 330v
But proceeding with caution there shouldn't be any problems getting to a correct setting
Thanks Again
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:think1: :think1: :think1:
Here is late and so strange thoughts cross my mind
a voltage of -10v bias hotter than a voltage of -15v
a 100R resistor bias hotter than a 150R resistor
if this is correct I am in confusion ....
did you see what I see or am I completely lost ???
Franco