Hey guys, I have a power supply design for my B+ supply. I would like help figuring out a suitable bleeder resistor for it. By the way, this supply is only powering a single 12AX7 (both sides).
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Any value that does not exceed max power is O.K.
I use 500k across the PSU output some use alot lower value. However it depends on how fast you want the B+ to drop. Remember it will drop quite quickly with the tubes connected at power off. The danger is if the heater supply fails and you open it up to find out why! The B+ will remain high without a bleeder for a while. Just remember that the resistor must have a working voltage higher than the B+. Some 2 Watt have 500V working. To work out wattage use Ohms law for permanent connection V/R = I ....VxA = W.
Regards
M. Gregg
I use 500k across the PSU output some use alot lower value. However it depends on how fast you want the B+ to drop. Remember it will drop quite quickly with the tubes connected at power off. The danger is if the heater supply fails and you open it up to find out why! The B+ will remain high without a bleeder for a while. Just remember that the resistor must have a working voltage higher than the B+. Some 2 Watt have 500V working. To work out wattage use Ohms law for permanent connection V/R = I ....VxA = W.
Regards
M. Gregg
That is a lot of smoothing for a single 12AX7! The bleeder resistor has to discharge the supply within a reasonable time. A time constant of 20-30s may be OK. For your 1320uF this means 15-20K at 10W. The snag is that the bleeder will take much more current than the circuit - this is the price to be paid for excessive smoothing! Alternatively, use a higher value and accept a much longer discharge time.
Just a thought!
I agree with DF96 you have a lot of capacitance in the circuit for a couple of 12AX7! Remember the smoothing is one issue. The other is the more caps on the PSU the more current is available under initial fault conditions, that includes shock factor (discharge into you to Gnd), High volts low current is bad enough - high volts high current is bad news!
Regards
M. Gregg
I agree with DF96 you have a lot of capacitance in the circuit for a couple of 12AX7! Remember the smoothing is one issue. The other is the more caps on the PSU the more current is available under initial fault conditions, that includes shock factor (discharge into you to Gnd), High volts low current is bad enough - high volts high current is bad news!
Regards
M. Gregg
Agree with the comments on smoothing overkill. Model your supply in PSUD. Play with the values and you will find that you can get a very quiet supply with much less capacitance.
Sheldon
For example: If you run a sim on your supply, you will get ripple of around 5nV. Reduce all the caps to 10uF and you will get on the order of 5mV. Use 50uF in the input position and 10uf for the others, and you get about 1mV, etc..
Sheldon
For example: If you run a sim on your supply, you will get ripple of around 5nV. Reduce all the caps to 10uF and you will get on the order of 5mV. Use 50uF in the input position and 10uf for the others, and you get about 1mV, etc..
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Dissipating the rated power in a power resistor will cause the resistor to become screaming hot. As in 250~300 deg C. Unless the ambient temperature (i.e. the temperature inside the chassis) can be guaranteed not to exceed 25 deg C, it will also lead to resistor failure within a rather short amount of time.
I strongly advise you to use a 3~5x margin. I.e. P = 5*(E^2/R).
I tend to use a 2 W or 3 W rated resistor in the 100 kOhm to 470 kOhm range.
Note that the time constant of the circuit (tc = R*C) will set the discharge time for the circuit. Normally, you can consider the circuit to be fully discharged after five time constants, hence, [Discharge Time] = 5*R*C. After this discharge time, about 0.6 % of the starting voltage will be left on the caps. All units are SI units. I.e. resistance in ohm, voltages in volt, capacitance in Farad, time in seconds, power in Watt.
Some low-power resistors have a max voltage rating of 300-ish volts. Use two in series if that's the case. Most modern power resistors are limited by the dissipated power and not the voltage. I.e. max power dissipation will be reached before the voltage across the resistor reaches the dielectric breakdown of the resistor.
~Tom
Definite filter overkill there. As to the choice of suitable bleeder resistance, if you have no other considerations (like a minimum load to keep an L-input filter operating like, well, an L-input filter, or using a voltage divider to do double duty) then the usual criterion is 100R per volt.
Just be sure to be extra generous with bleeder resistor power ratings, as a burnt out bleeder is more dangerous than no bleeder at all. Also, keep an eye on voltage ratings to avoid the possibility of flash-over. With 250Vrms of AC input, it wasn't much of a problem in the good ol' days. These days, components have gotten smaller and smaller. Some 1W metal oxide resistors are smaller than 1/4W C-comps. Might not make much of a difference with SS electronics, but that's just asking for trouble in HS designs where voltages run a good deal higher than with normal SS practice.
Tom,
I was not saying he should use the calculation to give the wattage to use! Sorry if this was taken as such. It was to see what he would dissipate! I agree he needs at least 3 times the rating or more!
Regards
M. Gregg
Burning more than a few Watt in the bleeder would be overkill. Remember that you have to get rid of the heat somehow. If you have many 680 kOhm, 2.5 W resistors, you could put two or three resistors in parallel across each cap. Unfortunately, the time constant would still be a couple of minutes, hence, expect the circuit to be discharged in about 10 minutes without any other load than the bleeders.
I second the opinions of the way overkill filtering. If you want an ultra-clean supply, I strongly suggest that you look at regulated supplies.
~Tom
I second the opinions of the way overkill filtering. If you want an ultra-clean supply, I strongly suggest that you look at regulated supplies.
~Tom
Unfortunately, my power supply is already designed and it needs to be sent in to a manufacturer soon. It is for my senior project in Electronic Engineering Technology Big changes can't be made at this time, but the bleeder resistor was placed in the circuit.
An externally hosted image should be here but it was not working when we last tested it.
It doesn't matter too much. This is an Associate's degree project (2 year degree). I'll be sure to mention this forum. Thanks for all the help.
I take R3 is your bleeder. It'll cook those electrolytics. Leave at least 500 mil between the resistor and the caps and drill some ventilation holes below the resistor (I use 150 mil dia). Also leave space between the B+ connector and the hot resistor.
I don't see any reason why you'd want the caps spaced so far apart.
I'm assuming that you'll be using this project after its completion and not just throw it on the scrap heap. Hence, you might as well design it right regardless of deadline pressure.
Actually I have another idea for your bleeder. You could use a CCS like the IXYS IXCP10M45. If you set it up for 10 mA current, your caps would discharge in about 42 seconds. Recall, C = Q/V, Q = i*t <--> t = C*V/i. You could have an LED in series with the CCS. This would be your DANGER indicator. LED on --> high voltage present. You'd have to provide a heatsink for the CCS, but one of the PCB mounted ones could be used.
~Tom
I don't see any reason why you'd want the caps spaced so far apart.
I'm assuming that you'll be using this project after its completion and not just throw it on the scrap heap. Hence, you might as well design it right regardless of deadline pressure.
Actually I have another idea for your bleeder. You could use a CCS like the IXYS IXCP10M45. If you set it up for 10 mA current, your caps would discharge in about 42 seconds. Recall, C = Q/V, Q = i*t <--> t = C*V/i. You could have an LED in series with the CCS. This would be your DANGER indicator. LED on --> high voltage present. You'd have to provide a heatsink for the CCS, but one of the PCB mounted ones could be used.
~Tom
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