High Voltage and measurement instruments

[sara39]

Hi all,
Can anybody help me on HVs, high voltages up to 60Kv? Actually i've got this HV supply , then i came across with an old ammeter in my garage which i thought could help in showing me the exact voltage my supply is transferring but i have a poblem in matching the resistances. actually i wanna make an exact ratio of 1/(10^6). the approximate ratio of the ammeters input resistance and the supplies output resistance is 1/10^6, how can i make sure that putting a potentiometer in series or parallel can give me an exact ratio? thanks for your help!
Sara

 

[Nigel Goodwin]

Hi all,
Can anybody help me on HVs, high voltages up to 60Kv? Actually i've got this HV supply , then i came across with an old ammeter in my garage which i thought could help in showing me the exact voltage my supply is transferring but i have a poblem in matching the resistances. actually i wanna make an exact ratio of 1/(10^6). the approximate ratio of the ammeters input resistance and the supplies output resistance is 1/10^6, how can i make sure that putting a potentiometer in series or parallel can give me an exact ratio? thanks for your help!
Sara

You need to know the FSD (Full Scale Deflection) of the meter, plus it's internal resistance (although for measuring 60KV you can ignore it's internal resistance).

You specify 'ammeter' rather than 'milli-ammeter' or 'micro-ammeter', if it's FSD is in the amps range it's going to draw that amount of current from the supply - and require substantially sized resistors to dissipate the power!.

To calculate the resistance required is simple ohms law, assuming the meter is 1mA FSD, and you want to read 60KV FSD, simply divide the voltage by the current, so 60,000/.001 = 60,000,000 ohms. With a 100uA meter, it would (obviously) be 600,000,000 ohms. You should subtract the internal resistance of the meter from this value - but I think it's fairly obvious that a couple of 100 ohms from those values is a waste of time!.

You also need to be concerned with the voltage and wattage ratings, with a 1mA meter it will dissipate 60W, and with a 100uA meter it will dissipate 6W. Also resistors have voltage ratings, usually only a few hundred volts, so you need to use multiple resistors in series to withstand the voltages, or a smaller number of special high-voltage resistors in series. These special resistors are usually very high values as well, designed for this type of use.

Layout is VERY important, it needs to be well spaced out, so it can't arc anywhere - probably best NOT to use a PCB. Also it needs to be VERY well insulated.

 

[chemelec]

At That 60 Kilo-Volts, It is Nearly Impossible to make an Accurate Divider using Normal Resistors.

Depending on What your meter is, I have some values of High Voltage, High Value resistors on my website.

**broken link removed**

It is also possible to make a High Voltage Probe for use with a Digital Multimeter, but you need to know the "Ohm per Volt" rating on the meter.

Take care..........Gary

 

[David Bridgen]

It is also possible to make a High Voltage Probe for use with a Digital Multimeter, but you need to know the "ohm/ per Volt" rating on the meter.

Most, if not all, DMMs have a fixed input resistance of 10M. (And some f.e.t. input analogue meters too.)
Using them exacerbates the selection of suitable multiplier resistors.

 

[chemelec]

Most, if not all, DMMs have a fixed input resistance of 10M.

I Agree, But Not All, My Beckmans are 11 Megohm.

But its Best to Check anyway.

 

[David Bridgen]

Ah, yes Gary. I remember now that you mention it. Thanks.

 

[sara39]

To calculate the resistance required is simple ohms law, assuming the meter is 1mA FSD, and you want to read 60KV FSD, simply divide the voltage by the current, so 60,000/.001 = 60,000,000 ohms. With a 100uA meter, it would (obviously) be 600,000,000 ohms. You should subtract the internal resistance of the meter from this value - but I think it's fairly obvious that a couple of 100 ohms from those values is a waste of time!.

You also need to be concerned with the voltage and wattage ratings, with a 1mA meter it will dissipate 60W, and with a 100uA meter it will dissipate 6W. Also resistors have voltage ratings, usually only a few hundred volts, so you need to use multiple resistors in series to withstand the voltages, or a smaller number of special high-voltage resistors in series. These special resistors are usually very high values as well, designed for this type of use.

Hi,
"The resistances in series with a meter" well, I did try making a 30Mohm resistor by connecting 30 1Mohm resistors in series, to readout the high voltage with my old mili-ammeter but a friend told me that each resistor has a specific voltage tolerance and the voltage tolerance of a series combination of 30 resistors is not the sum of the tolerance of each resistor.That means, the resistors combination tolerance is equal to the tolerance of only one resistor, although the power dissipated in the circuit is split up between the 30 resistors. So there is no way of reading the high voltage with a regular ammeter. Is my friend making a mistake, or is it really so?
thanx
Sara

 

[Nigel Goodwin]

"The resistances in series with a meter" well, I did try making a 30Mohm resistor by connecting 30 1Mohm resistors in series, to readout the high voltage with my old mili-ammeter but a friend told me that each resistor has a specific voltage tolerance and the voltage tolerance of a series combination of 30 resistors is not the sum of the tolerance of each resistor.That means, the resistors combination tolerance is equal to the tolerance of only one resistor, although the power dissipated in the circuit is split up between the 30 resistors. So there is no way of reading the high voltage with a regular ammeter. Is my friend making a mistake, or is it really so?

As with most things, simply apply ohms law!.

This will show that identical resistors in series share the voltage equally between them - so your 30 resistors in series will each have 1/30th of the applied voltage across each of them.

You will find resistors used in this way in MANY commercially produced items, it's a perfectly standard technique.

 

[zevon8]

Nigel is correct. However, a "garden variety" 1/4 Watt resistor will likely only have a voltage rating of around 250 to 500 volts, depending on its body construction and materials used. This is far less than the 2000 volts you are placing on each of the 30 resistors you have in series. You run the risk of "flash-over" or arcing, leading to a domino type effect across the series chain.

You would need to use resistors capable of withstanding the required voltage applied to each one to do it this way.

 

[Nigel Goodwin]

Nigel is correct. However, a "garden variety" 1/4 Watt resistor will likely only have a voltage rating of around 250 to 500 volts, depending on its body construction and materials used.

Most likely the lower end of that range!.

A quick check in the Maplin catalogue shows 0.6W metal film resistors rated at 250V max, and 1W metal film at 500V max.

However, they do a small range of high voltage metal film resistors, these are rated at 3500VDC or 2500VAC for 1M to 33M, and at 10000VDC or 7000VAC for 47M.

Incidently, their wirewound resistors are rated at 200V for 3W, 350V for 7W, and 500V for 10W (with peak voltage ratings of double those).

 

[chemelec]

Your 30 Resistors connected in Series to make 30 M isn't going to work.

At 60 KV that is a load of 2 mA. Resulting in 120 watts going through the resistors String. (4 watts into each one.)

Besides that 2 mA may LOAD DOWN your 60KV which will result in an Incorrect Reading, Each Resistor will be Frying.

MOST High Voltage Probes are up into the Gig Ohm of Resistance and use resistors with appropriate voltage ratings as in the link I posted previously.

Take care..........Gary