setting the dropping resistor in fixed bias -- ?
Moderators: pompeiisneaks, Colossal
setting the dropping resistor in fixed bias -- ?
I'm getting ready to start on the new build utilizing a pair of 6L6's in fixed bias. I'm using a Hammond 273BX PT and it has no bias tap, which means I have to tap off the HV. How do I calculate the size of the dropping resistor to generate the desired negative voltage?
Re: setting the dropping resistor in fixed bias -- ?
I use a 220k going to the diode. then a 10uf /150 volt filter cap, a 15k resistor, then another 10uf/150 volt cap. this feeds the bias point. Also a 33k in series with a 10k trimmer to ground. This is pretty much the 50 watt marshall setup. I think they use a 47k and 25k trimmer.
For 6l6's you will have to jump the 220k with a 1 meg (in parallel) to get the voltage up where you need it. It seems to work out just about right for a 220k for EL34's and jump with a 1 meg for 6l6's.
See a marshall 1987 schematic for more details.
For 6l6's you will have to jump the 220k with a 1 meg (in parallel) to get the voltage up where you need it. It seems to work out just about right for a 220k for EL34's and jump with a 1 meg for 6l6's.
See a marshall 1987 schematic for more details.
Re: setting the dropping resistor in fixed bias -- ?
Or you could find a cheap transformer that has 50v secondaries.
Tom
Don't let that smoke out!
Don't let that smoke out!
Re: setting the dropping resistor in fixed bias -- ?
Why would you put a 1M in parallel with a 220K resistor?
If 180K is what you need in a circuit, put in a 180K resistor.
If 180K is what you need in a circuit, put in a 180K resistor.
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Re: setting the dropping resistor in fixed bias -- ?
This is a great question and one that I've thought about and worked with alot.
AC~-/\/\/\/\/\/\--<|---/\/\/\/\---|---/\/\/\/\/\/---|---/\/\/\/\--/\/\/\/-ground
.............................................cap.................cap................pot
..................................................................... |
....................................................................bias
sorry for the extra dots but the program won't allow spaces in text drawings
So we look at this as a voltage divider by calculating the current dropped by the entire string of resistors. Because current is constant in a series circuit I conclude that we set the voltage through the string at .707 times the AC voltage. So let's say that the resistors are valued, in order from left to right as 220K, 15K, 15K 33K and a 10K pot. Lets also assume that the voltage between the center tap and one side of the high voltage tap is 330VAC. Let's assume too that because the diode only conducts in one direction, that the AC exposed resistor still only drops voltage in accordance with the .707 formula. Now:
The total resistance of the string is 293,000 ohms. 330*.707/293000 = .0008 So now we can find out how many volts get dropped by each resistor. 220,000 * .0008 = 176 vdc 15000 * .0008 = 12vdc per 15K. So with the bias pot open all the way we should see .707 * 330 = 233Vdc - 200 drop across the resistors or 33 volts at the bias take off point. As you reduce the resistance at the bias pot you will change both total current in the circuit and the percentage of total voltage dropped by the initial three reisistors.
Sounds terrific eh? Only problem is it doesn't work. It's close, but you inevitably have to fight it for a couple of hours playing with the fourth resistor and the size of the pot. I don't know why. Maybe Andy le Blanc, or Ears, or Novosibir or one of the other brighter bulbs can tell us. It's too many for me.
AC~-/\/\/\/\/\/\--<|---/\/\/\/\---|---/\/\/\/\/\/---|---/\/\/\/\--/\/\/\/-ground
.............................................cap.................cap................pot
..................................................................... |
....................................................................bias
sorry for the extra dots but the program won't allow spaces in text drawings
So we look at this as a voltage divider by calculating the current dropped by the entire string of resistors. Because current is constant in a series circuit I conclude that we set the voltage through the string at .707 times the AC voltage. So let's say that the resistors are valued, in order from left to right as 220K, 15K, 15K 33K and a 10K pot. Lets also assume that the voltage between the center tap and one side of the high voltage tap is 330VAC. Let's assume too that because the diode only conducts in one direction, that the AC exposed resistor still only drops voltage in accordance with the .707 formula. Now:
The total resistance of the string is 293,000 ohms. 330*.707/293000 = .0008 So now we can find out how many volts get dropped by each resistor. 220,000 * .0008 = 176 vdc 15000 * .0008 = 12vdc per 15K. So with the bias pot open all the way we should see .707 * 330 = 233Vdc - 200 drop across the resistors or 33 volts at the bias take off point. As you reduce the resistance at the bias pot you will change both total current in the circuit and the percentage of total voltage dropped by the initial three reisistors.
Sounds terrific eh? Only problem is it doesn't work. It's close, but you inevitably have to fight it for a couple of hours playing with the fourth resistor and the size of the pot. I don't know why. Maybe Andy le Blanc, or Ears, or Novosibir or one of the other brighter bulbs can tell us. It's too many for me.
The Last of the World's Great Human Beings
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Seek immediate medical attention if you suddenly go either deaf or blind.
If you put the Federal Government in charge of the Sahara Desert, in five years time there would be a shortage of sand.
Re: setting the dropping resistor in fixed bias -- ?
Thank you; this is great stuff, skyboltone it's very much what I was seeking to find out. I hope others will chime in, too. I know I could use even more clarification.
Re: setting the dropping resistor in fixed bias -- ?
That PT is only rated at 100mA. Seems a little light to me...
Re: setting the dropping resistor in fixed bias -- ?
huh?dave g wrote:That PT is only rated at 100mA. Seems a little light to me...
273BX 182VA, sec. 350-0-350, DC ma 175, Fil.#1 (rct) 5.0V @ 3a, Fil.#2 (htr) 6.3V @ 5.0a ct.
Re: setting the dropping resistor in fixed bias -- ?
Skyboltone,
We multiply peak voltages by 0.707 to obtain RMS voltages.
Since the secondary voltages of power transformers is already given in V RMS, why are you multiplying it by 0.707?
This transformer has secondary RMS voltages of 350-0-350, when the primary is 115 VAC. It will have a Vpeak of about 495, probably higher as our line voltages these days are 120VAC and above.
VAC is the same as Vrms and not the same as Vpeak. 120VAC = 170Vpeak.
Duncan Amps has a power supply designer that people might find useful for this application:
http://www.duncanamps.com/psud2/index.html
We multiply peak voltages by 0.707 to obtain RMS voltages.
Since the secondary voltages of power transformers is already given in V RMS, why are you multiplying it by 0.707?
This transformer has secondary RMS voltages of 350-0-350, when the primary is 115 VAC. It will have a Vpeak of about 495, probably higher as our line voltages these days are 120VAC and above.
VAC is the same as Vrms and not the same as Vpeak. 120VAC = 170Vpeak.
Duncan Amps has a power supply designer that people might find useful for this application:
http://www.duncanamps.com/psud2/index.html
Re: setting the dropping resistor in fixed bias -- ?
This could explain why my scope has appeared out of calibration too me. I assumed the peak to peak on the scope should match what my Fluke DMM says. Now I'll scope my line and hope for 1.41 of what the Fluke says, right?modern wrote:Skyboltone,
We multiply peak voltages by 0.707 to obtain RMS voltages.
Since the secondary voltages of power transformers is already given in V RMS, why are you multiplying it by 0.707?
This transformer has secondary RMS voltages of 350-0-350, when the primary is 115 VAC. It will have a Vpeak of about 495, probably higher as our line voltages these days are 120VAC and above.
VAC is the same as Vrms and not the same as Vpeak. 120VAC = 170Vpeak..............
If it says "Vintage" on it, -it isn't.
Re: setting the dropping resistor in fixed bias -- ?
Actually, if you're comparing peak to peak vs. RMS, you should expect 2.82 times as much. Also, (and you may already know this) multimeters have frequency limits. I don't know which Fluke you're using, but I know mine poops out at about 400 Hz. Another thing that mine doesn't do well is measure AC superimposed on DC (like when measuring signal at the plate of a tube). In general, for signal measurements, I count on the 'scope. Depending on which meter you have, YMMV.jjman wrote:This could explain why my scope has appeared out of calibration too me. I assumed the peak to peak on the scope should match what my Fluke DMM says. Now I'll scope my line and hope for 1.41 of what the Fluke says, right?
W
Re: setting the dropping resistor in fixed bias -- ?
But we weren't talking about Vpp (peak to peak) we were talking about Vp (peak).
Thanks for confusing the matter even more ...
Thanks for confusing the matter even more ...
Re: setting the dropping resistor in fixed bias -- ?
Certainly wasn't my intention. I realize the discussion at hand referred to peak voltage - construction of a DC supply usually does, unless a voltage doubler is involved. I was addressing JJ's comment that specifically asked about peak to peak voltage.modern wrote:Thanks for confusing the matter even more ...
W
Re: setting the dropping resistor in fixed bias -- ?
Wayne,
I apologise, I didn't realise peak to peak had been brought.
I should have read JJman's post a little closer.
I apologise, I didn't realise peak to peak had been brought.
I should have read JJman's post a little closer.
Re: setting the dropping resistor in fixed bias -- ?
No problem.
'K, guys, guess you can have your thread back now . If you catch me doin it again, jab me in the ribs - or worse yet, cut me off at the bar!
W
'K, guys, guess you can have your thread back now . If you catch me doin it again, jab me in the ribs - or worse yet, cut me off at the bar!
W