I have been reading through the safety practices with voltage sticky and I can't wrap my head around a couple things. I have odered both tube amp books and they are enroute so I will eventually get the answers there, but since I am impatient, here goes...
1) Is there a calculation for bleed resistors for dummies? Seems to me like the resistors should be timed at about 2-3 seconds after power is removed.
2) If you install a bleed resistor, do you need to consider increasing the cap values? Another math for dummies.
1) Is there a calculation for bleed resistors for dummies? Seems to me like the resistors should be timed at about 2-3 seconds after power is removed.
2) If you install a bleed resistor, do you need to consider increasing the cap values? Another math for dummies.
Bleed resistors are a compromise between low values (discharges quickly, but gets hots and wastes power) and high values (opposite).
The voltage will reduce to safeish values within 3 or 4 time constants - especially if the amp is still drawing current for part of this time, as will usually be the case. CR of around a few seconds should be fine, unless you are impatient.
A bleed resistor should not take enough current to require a PSU redesign.
The voltage will reduce to safeish values within 3 or 4 time constants - especially if the amp is still drawing current for part of this time, as will usually be the case. CR of around a few seconds should be fine, unless you are impatient.
A bleed resistor should not take enough current to require a PSU redesign.
I still need some calculations
Thank you for your response, but I still need formulas for both my questions.
Thank you for your response, but I still need formulas for both my questions.
Just to spell out what DF96 is saying, a formula would be:
R = time constant / C.
Make the time constant equal to a couple of seconds, say. It will take three or four times that to discharge to a very low voltage, assuming nothing else is drawing current.
After you have calculated R, work out how much current is going to be wasted in it when the amp is running normally (and if it's too much, increase the resistance!). You probably don't want it to take more than a few milliamps at most, or it will waste power and drag down the smoothing of the power supply.
I prefer to do things the other way around and start by choosing a resistor that consumes a couple of milliamps, say, and then maybe work out what time constant that gives me. Often I won't bother with that bit; I know they're going to discharge within minutes if not seconds, and that's all I really care about. If I'm going to work on the amp, I will discharge the caps manually whether or not a bleeder is present. The bleeder is more like a mild long-term bonus rather than a case of "they must be discharged this instant or I'm going to lose my mind with worry!!"
R = time constant / C.
Make the time constant equal to a couple of seconds, say. It will take three or four times that to discharge to a very low voltage, assuming nothing else is drawing current.
After you have calculated R, work out how much current is going to be wasted in it when the amp is running normally (and if it's too much, increase the resistance!). You probably don't want it to take more than a few milliamps at most, or it will waste power and drag down the smoothing of the power supply.
I prefer to do things the other way around and start by choosing a resistor that consumes a couple of milliamps, say, and then maybe work out what time constant that gives me. Often I won't bother with that bit; I know they're going to discharge within minutes if not seconds, and that's all I really care about. If I'm going to work on the amp, I will discharge the caps manually whether or not a bleeder is present. The bleeder is more like a mild long-term bonus rather than a case of "they must be discharged this instant or I'm going to lose my mind with worry!!"
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It will take three or four times that to discharge to a very low voltage, so your time constant of 3 seconds might be "safe to work on" after 9 seconds, say.
The exact figure depends what voltage they started at, and whether the circuit is still drawing current after the switch has opened, and what value you consider safe. 50 volts? 5 volts?....
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Is for tubes = HV, voltage started at 338VDC,the SSHV2 regulator CCS is adjusted to 30mA, the preamp draws less than 10mA = mu follower two ECC81, consider safe 5V.....
Suggestion: Get a calculator with exponential notation... You're off by three orders of magnitude.
R = 3/440E-6 = 6.8E3 = 6.8 kOhm.
Now, let's say the B+ voltage is 400 V. Let's calculate the power dissipated on the resistor:
P = E^2/R --> P = 400^2/6.8E3 = 23.5 W.
That's an awful lot of power to dissipate. Even if the B+ is "only" 250 V, you'd still be burning nearly 10 W in the resistor.
I tend to set up the bleeder up such that it draws about 2 mA at nominal B+. I install a high-efficiency LED in series with the bleed resistor to remind me when the power is on. With just the resistor there, the discharge time is a few minutes. But with an amp connected to the supply, the reservoir caps are normally discharged to safe working levels (< 10 V) after 10~15 seconds. You can see an example of this type of circuit in the supply for my 300B amplifier (schematic is on my website). I use two 3 W resistors in series. I could probably use 2 W types as they only dissipate 400 mW each. I don't like running the resistors at more than 1/4~1/3 of their rated power as they get screaming hot.
~Tom
R = 3/440E-6 = 6.8E3 = 6.8 kOhm.
Now, let's say the B+ voltage is 400 V. Let's calculate the power dissipated on the resistor:
P = E^2/R --> P = 400^2/6.8E3 = 23.5 W.
That's an awful lot of power to dissipate. Even if the B+ is "only" 250 V, you'd still be burning nearly 10 W in the resistor.
I tend to set up the bleeder up such that it draws about 2 mA at nominal B+. I install a high-efficiency LED in series with the bleed resistor to remind me when the power is on. With just the resistor there, the discharge time is a few minutes. But with an amp connected to the supply, the reservoir caps are normally discharged to safe working levels (< 10 V) after 10~15 seconds. You can see an example of this type of circuit in the supply for my 300B amplifier (schematic is on my website). I use two 3 W resistors in series. I could probably use 2 W types as they only dissipate 400 mW each. I don't like running the resistors at more than 1/4~1/3 of their rated power as they get screaming hot.
~Tom
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If the time constant for low power dissipation is too long for you, and the power dissipation is too great for an acceptable time constant, use an NC contact relay to switch in the high dissipation resistor in when power is removed, and use a low dissipation resistor for additional safety.
Power the relay such that the NC contact will open while HT is starting to build, and drop out as quickly after power is removed.
Power the relay such that the NC contact will open while HT is starting to build, and drop out as quickly after power is removed.
I'm a dummy...
Thanks for the explanation everybody.
DF96: Thank you, but as I stated in the beginning, I needed a "dummies" version of the formulas. The first response had an implied formula that made perfect sense to you, but it was still confusing to me. Now that I have read through all the responses, I understand what I need to know.
Tomchr and MerlinB: As for the high-efficiency LED, is there a specific part number you would recommend? I expect it would have to withstand the B+ voltage, yet still glow at very low <10V. As for a resistor or series of resistors, you recommend a cumulative wattage rating of 4 times the demand (due to heat). In the case above, 23.5 X 4 = 94 watts for single resistor if I want a 3 second draindown. That thing would be the size of a roll of dimes and would also be quite a drain on the electricity when the amp is running. I see what you two are talking about! You said you would rather just increase the time and decrease the resistor size so the demand is say 400mW each. Makes perfect sense! Tomchr, I am going to have a look at your 300B schematic.
Gimp: I like the idea of a contact relay too. If budget and space allow, that would be great.
Thanks for the explanation everybody.
DF96: Thank you, but as I stated in the beginning, I needed a "dummies" version of the formulas. The first response had an implied formula that made perfect sense to you, but it was still confusing to me. Now that I have read through all the responses, I understand what I need to know.
Tomchr and MerlinB: As for the high-efficiency LED, is there a specific part number you would recommend? I expect it would have to withstand the B+ voltage, yet still glow at very low <10V. As for a resistor or series of resistors, you recommend a cumulative wattage rating of 4 times the demand (due to heat). In the case above, 23.5 X 4 = 94 watts for single resistor if I want a 3 second draindown. That thing would be the size of a roll of dimes and would also be quite a drain on the electricity when the amp is running. I see what you two are talking about! You said you would rather just increase the time and decrease the resistor size so the demand is say 400mW each. Makes perfect sense! Tomchr, I am going to have a look at your 300B schematic.
Gimp: I like the idea of a contact relay too. If budget and space allow, that would be great.
Any LED will do- they all light up bright enough with a couple of milliamps (doesn't even need to be high efficiency). And it isn't exposed to the high voltage because the voltage is dropped across the resistor.
I don't know what I was thinking. I suppose I was going to wire it parallel to the resistor. 😱
In addition to discharging reservoir caps., another important use of bleeder resistors is to ensure that the critical current is drawn, when choke I/P filtration is employed. The critical current, in mA., is approx. = to V/L. "Massage" the numbers and you will find that the bleeder needed is 1,000 Ω X L. As the power dissipated in the bleeder = V2/R, truly high wattage parts are needed. Use a well ventilated resistor rated for at least 3X the power to be dissipated.
Minute and a half . . .
I see you use Ohm's law to calculate a needed resistor for a given LED at 50% capacity based on the voltage. You also arrive at the 4 watt by using I^2R, or you could also use IV.
Based on the above information and using a 440uF/400V cap, could we use the following calculation to determine the drain time if we use the 4 watt cap and LED method calculated in your attached PDF?:
time = RC = 40,000 / 440 = 90.9 seconds with 4 watts drain you would use a 16 watt 40K resistor or better
I see you use Ohm's law to calculate a needed resistor for a given LED at 50% capacity based on the voltage. You also arrive at the 4 watt by using I^2R, or you could also use IV.
Based on the above information and using a 440uF/400V cap, could we use the following calculation to determine the drain time if we use the 4 watt cap and LED method calculated in your attached PDF?:
time = RC = 40,000 / 440 = 90.9 seconds with 4 watts drain you would use a 16 watt 40K resistor or better
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Permanently connected bleed resistors increase supply ripple.
For this reason I recommend relay activated bleeders.
A relay with the bleeders connected to the normally closed contacts and some other useful function activated by the normally open contacts gives the benefits of quick voltage dissipation and cool running and lower ripple.
For this reason I recommend relay activated bleeders.
A relay with the bleeders connected to the normally closed contacts and some other useful function activated by the normally open contacts gives the benefits of quick voltage dissipation and cool running and lower ripple.
Bleed resistors should not take enough current to materially increase ripple. If a relay-based bleeder is used then a small fixed bleeder should still be in place too - best not to trust mechnical devices for safety.
Ensuring sufficient current draw for a choke input supply is a separate issue. A string of zeners can do that.
Ensuring sufficient current draw for a choke input supply is a separate issue. A string of zeners can do that.
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