Bleeder resistor and meter question

[KC2TAU]

I have a power supply that I use with my Hallicrafters R-45/ARR-7 that I would like to install some bleeder resistors in after receiving a rather unpleasant wake up call. The supply sits at 210vdc under load and must supply a maximum of 150ma of current.

I'm unsure if this is the correct formula but given:

T(in secs.) = R(in ohms) * C(in farads)

So, 15/0.00004 = 375k ohm. What is the equation for calculating the wattage required for the resistor as well as the current draw of the bleeder resistor itself?

Secondly,

I'm working on creating a new front panel for my ART-13 power supply. I want to use matching meters for all three functions (total current draw, HV voltage and AC input voltage). If I can't find three matching meters in the right ranges I'd be fine with using matching meters but recalibrating the ranges such as using a 15vdc or 150vdc meter to measure from 0-1500vdc. I know this is done by using meter shunts but what is the calculation for this? This shunt is placed in parallel with the meter, correct? Also, I'd imagine there is a current calculation for this to make sure the resistor is adequately sized for the current it is expected to shunt, correct?

[AB2EZ]

It will take 3 x T for the capacitor to discharge (bleed down) to 5% of its fully charged voltage.

Therefore if you set T=15 seconds, you must wait at least 45 seconds for the capacitor to discharge to .05 x 210V = 10.5V. It is always a good safety practice to verify that the capacitors are discharged by using an insulated-handled shorting stick to short the B+ to ground after the expected 3 x T discharge time.

The current that flows through the bleeder resistor, when the power supply is under load, will be V/R, where V is the power supply output voltage under load (210V), and R is the value of the bleeder resistor, in ohms. The current that flows through the bleeder resistor, when the power supply is not under load, will be V(no load)/R, where V(no load) is the power supply output voltage when the power supply is not under load, except for the load associated with the bleeder resistor (a value that will be greater than 210V), and R is the value of the bleeder resistor, in ohms

The wattage rating of the resistor should be at least 2x the actual power that the resistor will have to dissipate when the power supply is at its highest output voltage (i.e. not under load). The actual power that the resistor will have to dissipate is V(no load) x V(no load)/R, where R is the value of the bleeder resistor in ohms.
For example: if V(no load) = 250V and R= 375k ohms, then the resistor will be dissipating 0.17 watts when there is no load (other than the bleeder resistor) on the power supply. Therefore the wattage rating should be 0.34 watts or more. A 0.5 watt resistor would be a good, conservative choice.

When using a meter that is designed for a given voltage in a higher voltage application (e.g. using a 15V full scale meter to measure a voltage whose maximum value is 1600V), one has to be very concerned about safety. E.g. will the insulation of the components (resistors, wires, terminal strips) in the voltage divider be able to withstand the high voltages they will be exposed to? Will anyone or anything be able to accidently come in contact with the high voltages that appear on the terminals of these components? The usual approach is to construct a voltage divider consisting of a N resistors in series, each of value R (ohms), placed between the high voltage test point and the "+" side of the meter; and connecting the "-" side of the meter to ground. To determine the values of N and R, measure the resistance of the meter from the "+" terminal to the "-" terminal. Call this value R(meter).

Set the value of N = to the integer whose value is greater than or equal to V(max)/500V. For example, if V(max) = 1600V, then N= 4 (i.e. 1600V/500V = 3.2; and the next larger integer is 4).

Set the value of R so that N x R = R(meter) x {[V(max)/V(meter full scale)]-1}

For example, if V(max) = 1600V, and V(meter full scale) = 15V, then {[V(max)/V(meter full scale)]-1}= {[1600V/15V]-1} = {106.667 -1} = 105.667.

If R(meter) = 10,000 ohms, and N=4, then R = 10,000 ohms x 105.667 / 4 = 264,168 ohms.

Since 264,168 ohms is not a standard value, you can use a combination of 5 or 6 resistors in series, none of which has a value larger than 264,168 ohms, which add up to 4 x 264,168 ohms = 1,056,670 ohms. For example, you can use 4 resistors of value 220,000 ohms, and 1 resistor of value 180,000 ohms. These five resistors in series add to produce a total of 1,060,000 ohms.

For wattage, calculate the maximum current that will flow through the resistors and the meter: I(max) = V(max) / [(N x R) + R(meter)].

Use the formula: power dissipated = I(max) x I(max) x R(actual) to calculate the actual maximum power that will be dissipated in each resistor having value R(actual). Select resistors whose individual dissipation ratings are at least 2x the actual maximum power that will be dissipated in each of those resistors.

For example, if I(max) = 1600V/[1,060,000 ohms + 10000 ohms] = 1.5mA, then the actual maximum power dissipated by each 220,000 ohm resistor will be 0.0015A x 0.0015A x 220,000 ohms =  0.5 watts. Therefore each of the 4 resistors whose value is 220,000 ohms should be a 1 watt (or higher wattage) resistor.

Remember, this set of series resistors is connected to 1600V at one end (in this example). Extreme caution must be exercised to ensure that no one (and no thing) comes into electrical contact with any of these resistors or the wires and terminal strips they are attached to. Never touch the meter terminals until the power is turned off, and the power supply capacitors have discharged via their bleeder resistors. If the meter coil opens up, or if the wire between the "-" side of the meter and ground were to become disconnected or open... then the full value of V(max) will appear on one or both of the terminals of the meter.

Stu

[N2DTS]

Since masth makes my head hurt, I tend to experiment with meter dropping ressistors and shunts.
None of the meters on my equipment were for the correct values, and I make shunts using low value resistors and/or enameled wire wound on something till I get the right reading using an adjustable power supply with a current meter.

For voltage dropping resistors, I do the same thing, using a calibrated source, I add/subtract resistance till I get the right reading.
If you are measuring high voltage, it helps to tap off a bleeder resistor on the ground side.
I often use resistors in series to get the bleeder value I need, or you can insert a low value of resistance between the bleeder and ground and tap off between them, that way you do not have 2000 volts on the meter.

You can also use a calibration pot in series with the meter.

Resistors under high voltage will do odd things, and pull more power then you think.
They make special high voltage resistors, but I never had any of the right value.
If you try and measure 2500 volts on a low voltage meter, you need a LOT of resistors in series, and at a good wattage, depending on the meter sensitivity.
Some of my early stuff was built with the meter on the high voltage side, and the resistors often change value a lot.
Oddly, they tend to read lower over time.


Bleeder resistors should always be way overkill in wattage, that way they do not open up and kill you.


 

[DMOD]

Hi Michael,



Below is a quick and dirty method I use for first order medium voltage power supplies:

If you are using a solid state Full Wave rectifier, you may want to place a series resistor of 10 Ohms 5 Watts in between the rectifier output and the first filter capacitor.

Added:

Page 2 is a MV Power supply with a PI filter for additional ripple reduction.

[W7TFO]

For your Voltage meters, an 0-1 mA movement is easiest to use.

Use a 1.5 meg resistor in series.  Cal the scale to 1500VDC.

If you need more meters contact me via PM.

73DG

[N8ETQ]

Yo'

   the Harbach replacement P. Sup. board for the Collins
30L-1 takes 45 sec. to bleed down. The normally rock solid
1.6kv Ep rises to 1950, almost pegging the meter. I have
NEVER been real excited bout' pokein' around in a HV supply
without a screwdriver taking the leed..  Wear Shoes! Turn
it off, have a puff, get back to it in while... Still, let
the screwdriver take the leed.. Diode protect the meter.

    I don't think you get "Carbon Credits" even if you are one!

73

/Dan

[wa3dsp]

If you don't want to dissipate all that power and heat when it is turned on you could always use a relay (normally open contacts) to switch in a parallel lower resistance bleeder when the supply is turned off and the relay opens. For lower voltage supplies this could also be done at the power switch (DPDT)