If diode tubes are run in parallel I am guessing they would have the same problem as power tubes in parallel? One tube running away. Would doing this require current regulation on each tube? Would small plate inductors on each tube solve this problem well enough? What kind of filter cap max increase could be applied?
The tube specs say that if you run higher than 40uf filter cap after the tube you need a plate inductor that is why I was thinking inductor rather than cathode resistors to balance the current. I would love to run a 100uf, 33ohm ,100uf Pi filter to get maximum voltage and filtration but I would be happy with 47uf 33ohm 100uf pi running 5 ohm cathode resistors. I don't want to use silicon. I don't know how to figure out what inductors would be required for the 100uf option or if it even a good idea. I think I will be happy with your advice and run the 47uf option with 5 ohm cathode resistors.
Why is it that there is none of these design tools that run on current operating systems?
I have a lot of 117n7 tubes so I was thinking of using two of them as diodes. I have a 115-120 transformer and one tube gives me 90V under load after being rectified and filtered but I really need 100V DC PS. A little stiffer first filter cap raises the voltage and two of them in parallel should also cut the voltage drop in half? I just need 100VDC using the parts I already have.
I have a lot of 117n7 tubes so I was thinking of using two of them as diodes. I have a 115-120 transformer and one tube gives me 90V under load after being rectified and filtered but I really need 100V DC PS. A little stiffer first filter cap raises the voltage and two of them in parallel should also cut the voltage drop in half? I just need 100VDC using the parts I already have.
That's a great idea. Thanks.
I would like an answer, just out of curiosity, guess it could run a test to find out, but would two tube diodes in parallel cut the dropping voltage in half or would I just get the same dropping voltage but twice the current? I think the latter is more likely huh?
I would like an answer, just out of curiosity, guess it could run a test to find out, but would two tube diodes in parallel cut the dropping voltage in half or would I just get the same dropping voltage but twice the current? I think the latter is more likely huh?
You'd need the rated diode resistance, and the expected variance, to decide on the
minimum needed value of the series resistor. Don't forget the needed power rating
of the resistors, which can be significant.
Just as a calculation example, say the diode is 100R, and it may vary by +/- 20%.
The worst case is for one diode to be 120R, and the other diode to be 80R.
If you want the current to share within 10%, add equal resistors in series
with each diode, so that the total values are within 10% of each other.
(120 + R) = (80 + R) ) x 1.1 ..... Solving for the value of R gives 320R for each balancing resistor.
If we relax the requirement to currents balancing within 20%, (120 + R) = (80 + R) ) x 1.2
and now each balancing resistor = 120R.
minimum needed value of the series resistor. Don't forget the needed power rating
of the resistors, which can be significant.
Just as a calculation example, say the diode is 100R, and it may vary by +/- 20%.
The worst case is for one diode to be 120R, and the other diode to be 80R.
If you want the current to share within 10%, add equal resistors in series
with each diode, so that the total values are within 10% of each other.
(120 + R) = (80 + R) ) x 1.1 ..... Solving for the value of R gives 320R for each balancing resistor.
If we relax the requirement to currents balancing within 20%, (120 + R) = (80 + R) ) x 1.2
and now each balancing resistor = 120R.
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I'm not sure why this is posted in the subwoofers forum. I'm going to move it.
You want to balance the rectification pulses, which have high peaks, so the resistors have to be
located before the input capacitor. The Pi resistor won't have an effect on the balance.
If you had run away on one diode in the bottle how would you know? You might be damaging one of the diodes, or am I overthinking this? Should I not worry about it? I have read this happens in PSE amps if some kind of current balancing is not applied, sharing one cathode resistor, over-driving one of the tubes even burning them out. If it can happen to a triode or pentode, why wouldn't it happen in diodes?
Then again I have seen a lot of 12ax7 tubes in parallel and no balancing there, maybe because the current is so low in pre-amp tubes? Still if one side is over driven wouldn't that mean added distortion?
Then again I have seen a lot of 12ax7 tubes in parallel and no balancing there, maybe because the current is so low in pre-amp tubes? Still if one side is over driven wouldn't that mean added distortion?
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With 100 Hz mains power, and full wave rectification:
47 uF at 100Hz has 33.9 Ohms of capacitive reactance.
A balancing resistor of 320 Ohms would have to drive that 33.8 Ohms of capacitive reactance, yes?
It is going to take some time to charge that cap.
Now, once it is charged, you have to consider how much current the amplifier load takes from that 47uF cap.
That can be a voltage divider that makes the B+ be lower than you want it to be.
It might be a loosing battle.
A higher secondary voltage might be needed.
The 12.6V filament transformer that was suggested to use as a way to get the B+ transformer put out more volts was a good idea.
But your 117N7 rectifier tube is only rated for maximum plate volts of 117V rms.
You probably need to rethink what B+ transformer you use, and what rectifier tube you use.
Start with the exact B+ voltage you need, and the maximum current that the load is (all amplifier stages, at quiescent and maximum output power, and also allow for the current of a proper bleeder resistor across B+ to ground.
By the way,
47 uF at 100Hz has 33.9 Ohms of capacitive reactance.
A balancing resistor of 320 Ohms would have to drive that 33.8 Ohms of capacitive reactance, yes?
It is going to take some time to charge that cap.
Now, once it is charged, you have to consider how much current the amplifier load takes from that 47uF cap.
That can be a voltage divider that makes the B+ be lower than you want it to be.
It might be a loosing battle.
A higher secondary voltage might be needed.
The 12.6V filament transformer that was suggested to use as a way to get the B+ transformer put out more volts was a good idea.
But your 117N7 rectifier tube is only rated for maximum plate volts of 117V rms.
You probably need to rethink what B+ transformer you use, and what rectifier tube you use.
Start with the exact B+ voltage you need, and the maximum current that the load is (all amplifier stages, at quiescent and maximum output power, and also allow for the current of a proper bleeder resistor across B+ to ground.
By the way,
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I was thinking about that but I don't have nearly your understanding of it. Since I can only run 117V on the 117n7 and I am only getting 90V out after the filter, I would have to cheat a little and run the 12.6V filament transformer in it's own circuit, maybe with a silicone bridge and a high value cap, then put the two voltages in series to get to 100V. Guess it would add a 12.6V option as well. Not sure what issues that could add if any. I have lots of that kind of parts laying around. Do you think the balancing caps are necessary, I'm getting different opinions on it, but I think it might be a bad idea not to.
This would only be HW rectification because I can find no way to get to full wave given that in the 117n7 the heater is directly attached to the diode plate. No way to reverse it. If there was a way to use the two as a full wave, say a hybrid bridge with two silicon diodes I could get to 100V no problem.
What issues does charging time create? I'm looking at a max of about 200mA of expected output current. The power transformer is only 120V 250mA
This would only be HW rectification because I can find no way to get to full wave given that in the 117n7 the heater is directly attached to the diode plate. No way to reverse it. If there was a way to use the two as a full wave, say a hybrid bridge with two silicon diodes I could get to 100V no problem.
What issues does charging time create? I'm looking at a max of about 200mA of expected output current. The power transformer is only 120V 250mA
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We need more information about your proposed amplifier.
1. The 117N7 tube has 75mA maximum DC rating, 2 in parallel would have 150mA maximum, 3 would have 225mA, etc.
2. Your transformer is rated at 120V. And 250 mA into a resistor (AC load rating)?
Depending on the rectifier (full wave, half wave, bridge, etc.), the maximum permitted DC load current is always far less than the maximum AC current, especially when you use a cap input B+ filter.
Be sure that your transformer is rated for 250mA DC, before you load it with 200mA DC.
If it is rated for 250mA AC, then rectifying it, and cap filtering it, and drawing 200 mA DC off of it will cause the transformer to be Smoking Hot!
Please post a schematic of your amplifier circuits (that your power supply is going to have to power).
And let us know more details of your power transformer (is it High voltage B+ only, or does it have separate low voltage filament windings too?).
And, does it have to power the 117N7 filaments?
1. The 117N7 tube has 75mA maximum DC rating, 2 in parallel would have 150mA maximum, 3 would have 225mA, etc.
2. Your transformer is rated at 120V. And 250 mA into a resistor (AC load rating)?
Depending on the rectifier (full wave, half wave, bridge, etc.), the maximum permitted DC load current is always far less than the maximum AC current, especially when you use a cap input B+ filter.
Be sure that your transformer is rated for 250mA DC, before you load it with 200mA DC.
If it is rated for 250mA AC, then rectifying it, and cap filtering it, and drawing 200 mA DC off of it will cause the transformer to be Smoking Hot!
Please post a schematic of your amplifier circuits (that your power supply is going to have to power).
And let us know more details of your power transformer (is it High voltage B+ only, or does it have separate low voltage filament windings too?).
And, does it have to power the 117N7 filaments?
I don't have a schematic yet, though I have a few ideas. At this point I am trying to understand what parallel diodes tubes in a power supply would entail. I haven't seen it done before and so I expect some surprises. A generic 100VDC PS seemed like a good prototype and the 117n7 means I don't need a separate heater supply. Now, as always, you have pointed out something else I had no idea about. The Transformer is a Triad F6-120 and so it is rated 250mA AC. What would be a safe output in DC for this transformer? Is there some formula for this, say for given current rating AC you can run X amount of DC current? Does the capacitance of the filter add into this?
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