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Old 11th April 2011, 05:55 PM   #1
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Default Bleeder Resistor (Need Help)

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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Old 11th April 2011, 06:11 PM   #2
M Gregg is offline M Gregg  United Kingdom
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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.


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Old 11th April 2011, 06:18 PM   #3
DF96 is offline DF96  England
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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.
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Old 11th April 2011, 06:27 PM   #4
M Gregg is offline M Gregg  United Kingdom
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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!

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Old 11th April 2011, 06:29 PM   #5
PWR RYD is offline PWR RYD  United States
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Could the bleeder resistor be placed across C2 instead of C5? That way the extra current drawn by the bleeder resistor wouldn't result in a larger voltage drop across the series filter resistors under normal "on" state?
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Old 11th April 2011, 06:55 PM   #6
DF96 is offline DF96  England
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Yes, any cap will do.
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Old 11th April 2011, 07:03 PM   #7
Sheldon is offline Sheldon  United States
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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..

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Old 11th April 2011, 08:19 PM   #8
tomchr is offline tomchr  United States
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Quote:
Originally Posted by M Gregg View Post
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.
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
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Old 11th April 2011, 08:31 PM   #9
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Quote:
Originally Posted by fingerboy21 View Post
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).
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.
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Old 11th April 2011, 08:35 PM   #10
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Well considering I already purchased the capacitors, I would like to use them. I would like a safe unit that I wont have to worry about while using it, though.

I have multiple solutions to the bleeder resistor. I have many 100k and 680k 2.5W resistors, a single 15k 10W, two 3.5k 5W.
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