Hoffman Amplifiers Tube Amplifier Forum
Amp Stuff => Tube Amp Building - Tweaks - Repairs => Topic started by: Vlada on September 16, 2021, 01:57:50 pm
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Hello guys, I am building JCM800 for myself from scratch and want to try some mods-PS based not tonewise
Saw this intersting bias supply at Merlins site where he sugests using full wave for bias supply
The question I have is can I get away with just one diode per side for bias rectification or they must be in pairs (two diodes in series per side)?
http://www.valvewizard.co.uk/bias_supply2.jpg
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One diode per side will do for the bias circuit.
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Thanks Sluckey :worthy1:
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These diodes have to be good for twice the "+HT" voltage. So a 499V main B+ needs 1,000V diodes (same as for the +HT side). In days when we paid good money for high voltage rating, that was not worth the very slight benefit of full-wave.
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The diodes must also have a Reverse Repetative Maximum (Vrrm) rating that exceeds the peak-to-peak AC voltage (measured from one end to the cente tap), twice the value needed for a bridge rectifier. This is equal to 2.8 × Vrms. A 1N4007 is rated for 1000V. This corresponds to an AC voltage of 1000V/2.8 = 357Vrms. Knock off 10% to allow for variation in mains voltage, plus another 10% for transformer regulation and we are left with about 290Vrms. In other words, we shoudln't use the 1N4007 with anything more than a 290-0-290V transformer.
What Merlin wrote about rectifier diodes (paragraph above) made me think that one diode isnt suifficient for bias because its taken from 350-0-350 winding :w2:
It differs from standard Marshall bias where resistor goes first and then diode. Cost isnt an issue in my case, but board space might be :dontknow:
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> resistor goes first and then diode
What difference does that make? Parts in series, with nothing hanging off the middle, can go in either/any order.
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Won’t the resistors will drop some voltage, and so reduce the PIV on the bias supply diodes, compared to the regular HT supply diodes?
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Won’t the resistors will drop some voltage, and so reduce the PIV on the bias supply diodes, compared to the regular HT supply diodes?
Thats what I thought :w2:
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me too.
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> resistor goes first and then diode
What difference does that make? Parts in series, with nothing hanging off the middle, can go in either/any order.
???
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Anyone?
Also what should power rating for series droping resistor be in full wave arrangement if used for EL34 push pull amp?
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I like to use 3 watt metal oxide but 1 watt is sufficient.
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???
is somewhat cryptic, it may be helpful to phrase your query differently :dontknow:
I use 1/2W MF droppers, 500V rated.
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hmm, strange :w2:
Lets say we use full wave rectifier bias with Marshall 50watt PT, we would get something like -450vDC which then goes thourgh series droping resistor (100k) and shunt (10k bias pot+10k idiot resistor), so we get voltage divider
Calculations
I=450/100000+20000=3,75mA
P(R1)=0,00375*0,00375*100000=1,40625watts for maximum bias voltage
I=450/100000+10000=4,09mA
P(R1)=0,00409*0,00409*100000=1,67watts for minimum bias voltage
Am I missing something or I am just overcomplicating all this :BangHead:
Refering to this schematic http://www.valvewizard.co.uk/bias_supply2.jpg
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Your series resistor will realistically be closer to 220K that the 100K shown in the example. Just clear your head and use a 3 watt resistor. :wink:
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Is -450VDC a measured value? As I see it, that would require a capacitor input rectifier, then the voltage dropper.
I used a 270k series resistor per diode, so 1W total.
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Your series resistor will realistically be closer to 220K that the 100K shown in the example. Just clear your head and use a 3 watt resistor. :wink:
Thanks
So I should use 220k? I dont understand :BangHead:
Is -450VDC a measured value? As I see it, that would require a capacitor input rectifier, then the voltage dropper.
I assume its going to be the same as HT supply, probably thinking to much and got everything messed up :dontknow:
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... Lets say we use full wave rectifier bias with Marshall 50watt PT, we would get something like -450vDC which then goes thourgh series droping resistor (100k) and shunt (10k bias pot+10k idiot resistor), so we get voltage divider ...
Your series resistor will realistically be closer to 220K that the 100K shown in the example. ...
... So I should use 220k? I dont understand
Merlin shows R1 = 100kΩ, P1 = 10kΩ, R2 = 10kΩ.
Now go look at a 50w Marshall schematic (https://el34world.com/charts/Schematics/files/Marshall/Marshall_50w.pdf): R1 is 220kΩ, P1 and R2 are combined as a 56kΩ resistor, and there's an extra 15kΩ besides (in an extra stage of filtering).
So that's 1.55mA from the 450v winding, rather than 3.75mA. Do the math again, you have roughly 1/2w dissipation in the 220kΩ so best to use a 1w resistor. You can simply use the 3w resistor you planned to use if it makes you feel better.
... So I should use 220k? ...
You "should" build a solid bias supply. 220kΩ or 100kΩ is a matter of choice about how to get there.
Some will opt for 220kΩ (and higher resistance for P1, R2) because those parts draw less current, dissipate less heat.
Others will decide that they want their bias supply to be "low impedance" * and shoot for lower-resistance throughout at the cost of higher wattage ratings on resistors.
* When output tubes are played clean, they do not draw any grid current from the bias supply and/or phase inverter. But when output tubes get overdriven, they draw grid current that can resulting in bias-shift and blocking distortion. Some folks advocate for a "low impedance bias supply" that can deliver a small amount of current to output tube grids to help recovery from grid-blocking/bias-shift.
What constitutes a "low impedance bias supply"? How low is low enough? Is that even the best way to deal with the grid-blocking issue? Does altering the bias supply impact the action of the phase inverter?
Turns out there are quite a few inter-related circuit elements that are affected when someone tries to tackle that issue. Changing the bias supply is but one of the possible methods that could be employed.
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So that's 1.55mA from the 450v winding, rather than 3.75mA. …
The winding should be around 350-0-350VAC. The reservoir cap is after the dropper resistor, so as I see it, the VDC at the rectifier output will be more like -350V than -450V. A capacitor input rectifier is needed for the VDC to charge up go the VACpeak level.
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The winding should be around 350-0-350VAC. The reservoir cap is after the dropper resistor, so as I see it, the VDC at the rectifier output will be more like -350V than -450V. A capacitor input rectifier is needed for the VDC to charge up go the VACpeak level.
Yes, except not-even. 350VAC, half wave, 100k series resistor, typical loading, will make 70V-60V due to half-wave and the very heavy loading.
This is not a thing to "design" by forum polling. This type bias supply can be anti-intuitive. Sluckey and HBP have some extended experience and can pick likely values by reading tea leaves; most of us are VERY much better served by STEALING a plan from a likely known-good amplifier.
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... Sluckey and HBP have some extended experience and can pick likely values by reading tea leaves; most of us are VERY much better served by STEALING a plan from a likely known-good amplifier.
No, I steal from a known amp! :icon_biggrin: And then check the bias d.c. volts before installing tubes for the first time, and adjust-on-test.