I'm sorry if this topic has been covered in another thread. How do I calculate the resistor values in a tube rectified power supply? I have a Hammond 269jx transformer, 500V CT . I'm using a 6x4 rectifier. The unloaded voltage after the 6x4 is approximately 330V. I need to supply a circuit with 295V @ 30 milliamps. The trans is rated 250-0-250 @ 69ma. How do I calculate the resistance needed to drop the 330V to 295V?
Tube Rectified Power Supplies
- Thread starter pk63
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OK John,
You state "Ohm's Law"....How about expanding that a little? How do you use Ohm's law if you know the starting voltage of 330V, and know that 295V is required? I know my question may seem obvious to others, but I am new at building tube circuits. In fact this is my very first hands on prototype build. I'm not building anything special, just a simple tube rectified power supply. Eventually when I get through the learning curve, I will use the power supply in a preamp. As far as using PSUD2, its not all that straight forward for me.
If you could please show me the actual calculation it would be a great help in me learning how to calculate how much voltage I would wind up with after adding the caps and resistors to the circuit. Thanks in advance, PK
You state "Ohm's Law"....How about expanding that a little? How do you use Ohm's law if you know the starting voltage of 330V, and know that 295V is required? I know my question may seem obvious to others, but I am new at building tube circuits. In fact this is my very first hands on prototype build. I'm not building anything special, just a simple tube rectified power supply. Eventually when I get through the learning curve, I will use the power supply in a preamp. As far as using PSUD2, its not all that straight forward for me.
If you could please show me the actual calculation it would be a great help in me learning how to calculate how much voltage I would wind up with after adding the caps and resistors to the circuit. Thanks in advance, PK
It may (or may not...) be quite as simple as ohms law.
330 volts starting voltage and 295 required. So you need to 'loose' 35 volts across the series resistor.
If (and does it
) your circuit draws 30 ma then its a simple case of R=V/I which is 35/0.03 giving 1166 ohms.
Possible complications come because these glass bottle things are a bit unpredictable in a predictable kind of way. You say the supply is 330 volts unloaded. Glass bottles being what they are and all means that that value could actually fall a bit when you draw 30 ma and so the calculated value would be to high.
So calculate and get a starting point and then test that in real life modifying the result based on real measurement. The transformer voltage may also fall when delivering 30ma to a load. So you need to gather all the evidence to arrive at a final value.
Ohms law will work if all the variables are known... but they are not. You have the regulation factor of the transformer and the internal 'resistance' of the rectifier to consider.
330 volts starting voltage and 295 required. So you need to 'loose' 35 volts across the series resistor.
If (and does it
) your circuit draws 30 ma then its a simple case of R=V/I which is 35/0.03 giving 1166 ohms. Possible complications come because these glass bottle things are a bit unpredictable in a predictable kind of way. You say the supply is 330 volts unloaded. Glass bottles being what they are and all means that that value could actually fall a bit when you draw 30 ma and so the calculated value would be to high.
So calculate and get a starting point and then test that in real life modifying the result based on real measurement. The transformer voltage may also fall when delivering 30ma to a load. So you need to gather all the evidence to arrive at a final value.
Ohms law will work if all the variables are known... but they are not. You have the regulation factor of the transformer and the internal 'resistance' of the rectifier to consider.
It is possible that the PSU will drop below 295V under load. It will certainly drop, so bringing it well within 10% of the required voltage. Therefore it is almost certainly fine as it is. Once you have added some smoothing (RC or LC) you may come back to us and ask how you can boost the DC output of the PSU!
You won't destroy anything in the normal sense although you must make sure things like capacitors are rated for the maximum voltage they could see under all conditions.
For example, when you power up a tube (even got me saying it now... valves they are valves) circuit, then initially the filaments are cold and so any rail voltage could be at a maximum. OK, so a valve rectifier mitigates that because it to has to warm up...
If you had a solid state rectifier then the rails would appear immediately and would be say 330 volts in your example. When the valves warm and operate normally the voltage at the end of the series dropper resistor would settle to the 295 volts. During warm up though everything would see 330 volts. So everything must be rated accordingly.
For example, when you power up a tube (even got me saying it now... valves they are valves) circuit, then initially the filaments are cold and so any rail voltage could be at a maximum. OK, so a valve rectifier mitigates that because it to has to warm up...
If you had a solid state rectifier then the rails would appear immediately and would be say 330 volts in your example. When the valves warm and operate normally the voltage at the end of the series dropper resistor would settle to the 295 volts. During warm up though everything would see 330 volts. So everything must be rated accordingly.
You could use a resistor. Lets do the maths.
A single resistor to load the supply would be 330 volts divided by the 30 ma draw that is expected. So that is 11,000 ohms. The wattage needed would be W=V squared/R or simply W= I*V. So that is (330*330)/11000 which is 9.9 watts. A pretty hefty resistor.
As far as your circuit goes (based on the numbers you have given) then nothing bad is going to happen if you apply 330 volts DC to a circuit designed to work at 295 volts DC. Valves don't self destruct like that. Choose your calculated resistor and then tweak it if need be.
Going back to the previously calculated 1166 ohms (so 1k2 in practice) the wattage would need to be (35*35)/1200 which is 1.02 watts. We would probably go with a 3 or 5 watt part here as we are in valve territory as that would allow a big safety/reliability factor.
You really shouldn't start this with the unloaded voltage. You should begin by assuming you will load the PS enough to get the transformer into its normal self regulating voltage output range which is 250v then estimate the peak based on the type of filter input, and then the voltage drop through the rectifier. If you plan on a capacitor input then you can expect about 25% above the tranny regulated output = 312v. Then subtract the typical voltage drop through your rectifier which might be about 20v. = 292v. You should get close to this at the first filter directly off the rectifier if you load the tranny at least 50% of its rating.
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Trobbins is exactly right. PSUD2 is very accurate if given accurate data. Measure the unloaded voltage of the transformer, measure the primary resistance, secondary resistance, plug in those values, along with the choice of rectifier, caps, inductors, etc.. Then look at the output values, using a current tap as a proxy for the amp. Not only will it give you the resulting voltages, but also determine whether you fall within the peak voltages of your rectifier.
Sheldon