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Old 21st September 2010, 07:48 AM   #1
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Default GM70 bias & IT ratio question

After listening to some valuable advices, I have changed my design from a 3-stage C3g:GM70:GM70 to a simple 2-stage C3g:GM70.

Click the image to open in full size.

All filaments are DC.

B+ for GM70 is set at 800V as I am using two RCA83 for rectification. The max plate voltage for 83 is only 400V.

Plate dissipation = 120W
HT = 800V
120W / 800V = 150mA
Anode current = 150mA
Bias voltage = 60V
60V / 150mA = 400ohm
Bias resistance = 400ohm

Click the image to open in full size.

Is my biasing resistor value for GM70 correct? What wattage should I use? Will it be running very hot?
Will this schematic run in pure class A1?

Im not too sure how to calculate the IT ratio. If I want to load the C3g with around 10K, what will be the suggested ratio for the IT?

- Louis
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Last edited by louischia; 21st September 2010 at 07:54 AM.
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Old 21st September 2010, 09:54 AM   #2
Arnulf is offline Arnulf  Europe
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Originally Posted by louischia View Post
B+ for GM70 is set at 800V as I am using two RCA83 for rectification. The max plate voltage for 83 is only 400V.
Tell us, what voltage do rectifiers see if you want get 800V DC out ?

Quote:
Is my biasing resistor value for GM70 correct? What wattage should I use? Will it be running very hot?
Will this schematic run in pure class A1?
This is hardly the design that would be appropriate for somebody who doesn't understand the basics - on the other hand they say there is no education like life's experience, it's just that tuition gets expensive sometimes. This is going to be one of those occassions.

You should look up some elementary school physics book and focus on its electrics section. There you will find explained Ohm's law, Kirchoff's laws and the relation between voltage across a load & current through it and power dissipated on said load. Once you establish maximum dissipation, you should go for component that is rated higher than the projected maximum, including all the positive deviations (mains voltage variation). A factor of 2 is a good safety margin.

Quote:
Im not too sure how to calculate the IT ratio. If I want to load the C3g with around 10K, what will be the suggested ratio for the IT?
Transformer works as impedance converter with the ratio that is square of number of winding turns ratio ([Np / Ns]^2). Since you set up approximately infinite grid leak for the output tube whatever ratio you pick will result in approximately infinite load for the driver tube.

You really should look up tube basics first, then do a simple project where youll learn how things work and only then consider something like what you're attempting now. Failing that you're just going to spend alot of money for what could only avoid being a huge dissapointment by sheer coincidence. If you feel lucky play lottery instead, the relationship between investment and reward will be far more favorable to you.
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Last edited by Arnulf; 21st September 2010 at 09:58 AM.
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Old 21st September 2010, 10:55 AM   #3
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Quote:
Originally Posted by Arnulf View Post
Tell us, what voltage do rectifiers see if you want get 800V DC out ?
sim with duncan psu designer and you will find that 83 can't go up to KV even with twin tubes voltage doubling.

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Originally Posted by Arnulf View Post
You should look up some elementary school physics book and focus on its electrics section. There you will find explained Ohm's law, Kirchoff's laws and the relation between voltage across a load & current through it and power dissipated on said load. Once you establish maximum dissipation, you should go for component that is rated higher than the projected maximum, including all the positive deviations (mains voltage variation). A factor of 2 is a good safety margin.
thanks for you kind advice. Im not an electronics or electrical student but ohm's law and kirchoff's law are not too new to me. Based on my limited understanding, the bias resistor will end up being a huge heater and I can hardly find a precision (multi-turn) variable pot of that wattage to fit in! Knowing that 5W "might" not work for the bias resistor (more like a 50W), I still print it in my schematic because that is the "biggest" multi-turn pot that I can get my hands on. All I need is to have some experts like you to confirm my calculations or point out my newbie errors.

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Originally Posted by Arnulf View Post
Transformer works as impedance converter with the ratio that is square of number of winding turns ratio ([Np / Ns]^2). Since you set up approximately infinite grid leak for the output tube whatever ratio you pick will result in approximately infinite load for the driver tube.
I do understand how a transformer work. Maybe I should re-phrase my question: Should I parallel a resistor with the IT as load for the driver stage and use a 1:1 ratio IT or the GM70 grid will have some loading that I can convert with a different ratio IT?

Quote:
Originally Posted by Arnulf View Post
You really should look up tube basics first, then do a simple project where youll learn how things work and only then consider something like what you're attempting now. Failing that you're just going to spend alot of money for what could only avoid being a huge dissapointment by sheer coincidence. If you feel lucky play lottery instead, the relationship between investment and reward will be far more favorable to you.
thanks for you suggestions. I believe I have did enough "clonings" of tested tube schematics and all I learnt so far is drilling, wiring, soldering, etc. Like you just said, "there is no education like life's experience". Maybe it's just the time to pay some tuition fees.

- Louis
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Old 21st September 2010, 11:55 AM   #4
Arnulf is offline Arnulf  Europe
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Originally Posted by louischia View Post
sim with duncan psu designer and you will find that 83 can't go up to KV even with twin tubes voltage doubling.
But you've stated the maximum allowed anode voltage for each rectifier is 400V ? There is bound to be some drop across the rectifier to begin with. Now you're also mentioning voltage doubler (which imposes twice the voltage onto rectifier). Since you haven't posted schematic of your supply I can only guess as to what you're planning to do but things don't seem quite right.

As for wasted power for cathode biasing: why bother ? Set up a separate bias supply, it only takes a tiny current to bias the tube so it can be really cheap.

Quote:
I do understand how a transformer work. Maybe I should re-phrase my question: Should I parallel a resistor with the IT as load for the driver stage and use a 1:1 ratio IT or the GM70 grid will have some loading that I can convert with a different ratio IT?
Grid of the output tube presents pretty much infinite load as long as it stays in class A1 (= Vg never comes close to 0V, let alone above 0V). Besides all tubes have suggested maximum grid leak resistance, usually in the area of 1M or more (depending on bias arrangement) so you should have included it there already. Lower resistance means heavier load for preceeding stage but if you want 10K load you can get 10K easily with suitable resistor once you know what the impedance transformation ratio of your transformer is (it's more difficult to find different ratio of transformer than it is to find a suitable resistor to get desired value).

Quote:
thanks for you suggestions. I believe I have did enough "clonings" of tested tube schematics and all I learnt so far is drilling, wiring, soldering, etc. Like you just said, "there is no education like life's experience". Maybe it's just the time to pay some tuition fees.
Well, be careful along the way, 800V DC is no joke
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Old 21st September 2010, 12:15 PM   #5
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Your biasing scheme is wrong, the filament supply will keep the right side of the heater/cathode at a solid +20V, no matter what the current through the tube is.

The left side of the cathode/heater is a 0V, so you have actually a fixed bias of (averaged) +10V. Boom!!
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Old 21st September 2010, 03:32 PM   #6
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Originally Posted by Arnulf View Post
Since you haven't posted schematic of your supply I can only guess as to what you're planning to do but things don't seem quite right.
Sorry for not attaching my PS schematic earlier.
Click the image to open in full size.
Increasing the 2uf capacitors might be able to further alter the output voltage for another 50~100V.

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Originally Posted by Arnulf View Post
it's more difficult to find different ratio of transformer than it is to find a suitable resistor to get desired value
I am custom building all the irons. Given the choice, what ratio do you suggest?

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Originally Posted by Arnulf View Post
Well, be careful along the way, 800V DC is no joke
Thanks for the reminder.
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Last edited by louischia; 21st September 2010 at 03:35 PM.
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Old 21st September 2010, 03:45 PM   #7
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Originally Posted by kavermei View Post
Your biasing scheme is wrong, the filament supply will keep the right side of the heater/cathode at a solid +20V, no matter what the current through the tube is.

The left side of the cathode/heater is a 0V, so you have actually a fixed bias of (averaged) +10V. Boom!!
Thanks for pointing that out!

In fact, I think it looks weird too. The initial schematic was like this:
Click the image to open in full size.
But I read somewhere that since Im using DC for filament, hum pot don't make much difference.
So I took it out and changed the schematic to this:
Click the image to open in full size.
Then I thought since Im going to share a common ground for the filament and B+, the other side will be grounded anyway.
Thats why I linked it down to ground.

So what is the correct way of connecting it?

- Louis
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Old 21st September 2010, 04:05 PM   #8
kevinkr is offline kevinkr  United States
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Frankly I would use a (couple of) 5U4GTB or even 5R4 to power this off of a single 600V winding using a hybrid (Graetz) bridge with solid state diodes on the negative half of the bridge.

I would also use fixed bias which would allow you to tune operating point over some range and eliminate a costly cathode bypass capacitor which could impart its own unnecessary flavor to the audio path.
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Old 21st September 2010, 04:23 PM   #9
Arnulf is offline Arnulf  Europe
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Louis, few things:

1: Your rectifiers have their filament setup incorrectly (same problem Kavermei pointed out), you can't have cathode at same potential as anode

2: Your capacitors (the 2 uF ones) are a bit small. Are you sure your rectifiers can only handle 2 uF ? This totals 1 uF per supply, almost negligible value. You will probably want at least 10 uF (or 20-47 uF for quieter PSU) there.

3: Regarding IT ratio: this is something I haven't dealt with before so I'm sure others wuill be able to provide far more helpful tips than I can. However common sense dictates that you obey all the rules that pertain to transformers in general:
- wire must be thick enough to withstand DC current (quescent condition of driver stage)
- core must be airgapped or sufficiently large to withstand DC current without entering saturation
- inductance must be high enough not to cause LF rolloff
- leakage inductance must be low enough not to cause substantial loss
- parasite capacitance is usually low enough not to be worth the concern, but you can keep it in mind
You can read alot about these with regards to the output transformers (OPTs). Everything that applies to OPTs applies to these as well.

4: Regarding winding ratio: if your output tube never exits class 1 (= Vg is alwats negative enough for no grid current to flow) you can use your transformer as voltage step-up ("amplification") element, either getting some extra gain or getting rid of some gain without NFB. 1:1 is most straightforward as you only need one thickness of wire and everything is more straioghtforward but since you're winding your own you can tweak it to your specific goals.

5: Regarding cathode bias - it's so obvious but I haven't looked at your schematic long enough to notice it like Kavermei did: cathode resistor causes voltage drop. This drop is used to bias the tube, employing the self-adjusting relationship between grid voltage and cathode current (at some point the equilibrium is reached and if tube is stable there without melting down that's your operating point). So you obviously have to connect your bias to the correct side when using cathode bias with directly heated tubes (you can't make this mistake with indirectly heated ones because only one wire goes from the cathode). I still think you should be using fixed bias though.
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Old 21st September 2010, 05:29 PM   #10
45 is offline 45  Italy
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Quote:
Originally Posted by Arnulf View Post
2: Your capacitors (the 2 uF ones) are a bit small. Are you sure your rectifiers can only handle 2 uF ? This totals 1 uF per supply, almost negligible value. You will probably want at least 10 uF (or 20-47 uF for quieter PSU) there.
..also to calculate the resonant frequency of that filter so that one makes sure it is as low as possible (this one clearly has a resonance in the bass region!) and with an appropriate merit factor.


Overall 800V at 150 mA for the GM70 is not a good choice for me because of very poor efficiency.....120W plate dissipation for few watts of Pout!!
For example, one could go for 800V/100 mA using a fixed bias at approx. -70V for 15-18W output with a 4-5K primary load. With 800V anode voltage and that driver I don't think one can get much more.

Cheers

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