Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.
Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.
I have a 3.5Kw transformer that is connected to a load. The load is drawing a max current of 10A. If I would like to measure the AC voltage across the load (Which should not exceed 2kVac) with a DMM, how can I do that? Is there any schematic that I can reference to?
I do have a Fluke 80K-40 high voltage probe. However the reading is not very accurate. I am expecting a voltage of around 900V but it only read 600V. I have attached a picture of how I connect up the probe. Anyone can advise?
A standard 1/2W resistor is only rated at most a few hundred volts. With most of the HV across the 8.2Meg resistor, it will arc over, putting the HV directly into the DMM!!!
You can use a voltage divider, but the divider should be constructed either out of a series string of about 10ea 1/4W resistors placed inside a piece of poly tubing, or get a a special High Voltage Resistor which has an end-to-end voltage rating of 5KV or higher.
A standard 1/2W resistor is only rated at most a few hundred volts. With most of the HV across the 8.2Meg resistor, it will arc over, putting the HV directly into the DMM!!!
You can use a voltage divider, but the divider should be constructed either out of a series string of about 10ea 1/4W resistors placed inside a piece of poly tubing, or get a a special High Voltage Resistor which has an end-to-end voltage rating of 5KV or higher.
Can you provide me some form of schematic on the series string of resistor? How shall I wrap them up? For the ene to end resistor, can you point to to some website who sell them? I have no idea where to find these resistor.
Top pick off results from "**broken link removed**"
Simple divider as suggested by Boncuk will work.
Think two resistors R1 and R2.
R2 shunts the DMM +/- terminals.
R1 goes from HV to DMM +.
DMM- goes to the other end of the HV supply.
Say you want a 10:1 divider. That requires R1 = 9*R2.
Now, you want R2 to be as high as possible, but < 1% of the input impedance of the DMM, which is likely 20MegΩ. 1% of 20MegΩ is 200K. so use 200K for R2 (nothing special, 1/2W, 1% metal film resistor)
That makes R1 = 9*R2 = 9*200K = 1.8MegΩ
We must buy a High Voltage 1.8Meg, or break R1 into a string of say 9ea 200Ks the same as the one above.
So this way, you get to buy 10ea 200K 1% 1/2W resistors
The total voltage across the series string of 1.8MegΩ + 0.2MegΩ = 2MegΩ is 2000V, so the power dissipated by the string is P=E^2/R = 2000^2/2e6 = 4e6/2e6 = 2W. The series string is capable of 10*1/2W = 5W, so a total of 10 resistors will work.
The highest voltage seen by any one individual resistor is 1/10th of 2KV = 200V.
You can solder the 9 resistors end-to-end and pull some teflon tubing over them to brace them.
The probe you have should work. The main reason I have seen these probes read low is if the input to the meter is less than 10 MΩ. I would check the AC input impedance of your DMM. That failing to work and if things are right then consider building a divider as was covered by MikeMl and just keep in mind the voltage you are working with so isolation and safety is paramount.
Top pick off results from "**broken link removed**"
Simple divider as suggested by Boncuk will work.
Think two resistors R1 and R2.
R2 shunts the DMM +/- terminals.
R1 goes from HV to DMM +.
DMM- goes to the other end of the HV supply.
Say you want a 10:1 divider. That requires R1 = 9*R2.
Now, you want R2 to be as high as possible, but < 1% of the input impedance of the DMM, which is likely 20MegΩ. 1% of 20MegΩ is 200K. so use 200K for R2 (nothing special, 1/2W, 1% metal film resistor)
That makes R1 = 9*R2 = 9*200K = 1.8MegΩ
We must buy a High Voltage 1.8Meg, or break R1 into a string of say 9ea 200Ks the same as the one above.
So this way, you get to buy 10ea 200K 1% 1/2W resistors
The total voltage across the series string of 1.8MegΩ + 0.2MegΩ = 2MegΩ is 2000V, so the power dissipated by the string is P=E^2/R = 2000^2/2e6 = 4e6/2e6 = 2W. The series string is capable of 10*1/2W = 5W, so a total of 10 resistors will work.
The highest voltage seen by any one individual resistor is 1/10th of 2KV = 200V.
You can solder the 9 resistors end-to-end and pull some teflon tubing over them to brace them.
The probe you have should work. The main reason I have seen these probes read low is if the input to the meter is less than 10 MΩ. I would check the AC input impedance of your DMM. That failing to work and if things are right then consider building a divider as was covered by MikeMl and just keep in mind the voltage you are working with so isolation and safety is paramount.
How do I measure the AC input impedance of my DMM? Do you mean that if the input impedance is lower than 10Mohms, the high voltage probe cannot be used?
A high voltage probe is nothing but R1 in my discussion. It relies on R2 being the input impedance of the meter. If you down load/read the manual on your DMM, it will tell you the input impedance in the Specifications section. If the High Voltage probe was made and sold to be used with your DMM, then it should match the input impedance of the meter. If it came from another manufacturer, your guess is as good as mine...
The input impedance of some meters isn't accurately controlled, it's often just specified as 10M maximum so just connecting a series resistor will often not work.
You could try measuring the impedance with another meter, then adding a series resistor to give the appropriate dividing ratio but I wouldn't recommend this on an auto-randing meter as the impedance could change as the range is switched.
I do have a Fluke 80K-40 high voltage probe. However the reading is not very accurate. I am expecting a voltage of around 900V but it only read 600V. I have attached a picture of how I connect up the probe. Anyone can advise?
The spec for the Fluke 80K-40 says "The probe provides high accuracy when used with a voltmeter having 10 Megohm input impedance."
What DMM are you using it with? If it's a Fluke, one would think it should give good results.
A possibility you don't seen to be considering is that maybe the voltage IS 600 volts, rather than the 900 volts you expect.
Here's another way to proceed. Assuming your DMM can measure 1000 volts directly, reduce the voltage you're applying to the primary of your transformer with a variac to something like 1/3 of the normal grid voltage. Measure the voltage applied to the primary and the voltage at the secondary. Then use the ratio of those two voltages to calculate what the secondary voltage would be with full grid voltage applied.
Furthermore, if the voltage is only 600 volts, and you're using a Fluke, you should be able to measure it directly, without a high voltage probe. Most Fluke DMMs can withstand a surge substantially above the maximum voltage they can measure, which is probably 1000 VAC. You could touch your probe momentarily and if the voltage is less than 1000 VAC (you said you're expecting 900 volts), you can measure it without a high voltage probe.
The spec for the Fluke 80K-40 says "The probe provides high accuracy when used with a voltmeter having 10 Megohm input impedance."
What DMM are you using it with? If it's a Fluke, one would think it should give good results.
A possibility you don't seen to be considering is that maybe the voltage IS 600 volts, rather than the 900 volts you expect.
Here's another way to proceed. Assuming your DMM can measure 1000 volts directly, reduce the voltage you're applying to the primary of your transformer with a variac to something like 1/3 of the normal grid voltage. Measure the voltage applied to the primary and the voltage at the secondary. Then use the ratio of those two voltages to calculate what the secondary voltage would be with full grid voltage applied.
Furthermore, if the voltage is only 600 volts, and you're using a Fluke, you should be able to measure it directly, without a high voltage probe. Most Fluke DMMs can withstand a surge substantially above the maximum voltage they can measure, which is probably 1000 VAC. You could touch your probe momentarily and if the voltage is less than 1000 VAC (you said you're expecting 900 volts), you can measure it without a high voltage probe.
Actually I have tried to measure with the multimeter directly and it reads 900+ volts but with the fluke HV probe it only read 600. IK have 2 HV probe and both give similar result.
What kind of DMM are you using? Is it a Fluke? What model?
Actually, you can check to see if the multiplier resistor in the HV probes are ok.
Use the ohmmeter function on the DMM to measure the resistance between the tip of the HV probe and the banana terminals; one of them should measure something like 1000 megohms. To measure this (if you have a Fluke DMM) you would need to use the Siemens range.
This is a transformer operating of the the grid, right? Should be 50/60 Hz. It isn't running on 400 Hz or something unusual, is it?
And, it's just pure AC you're trying to measure? Not unfiltered rectified, with a lot of ripple, right?
What kind of DMM are you using? Is it a Fluke? What model?
Actually, you can check to see if the multiplier resistor in the HV probes are ok.
Use the ohmmeter function on the DMM to measure the resistance between the tip of the HV probe and the banana terminals; one of them should measure something like 1000 megohms. To measure this (if you have a Fluke DMM) you would need to use the Siemens range.
This is a transformer operating of the the grid, right? Should be 50/60 Hz. It isn't running on 400 Hz or something unusual, is it?
And, it's just pure AC you're trying to measure? Not unfiltered rectified, with a lot of ripple, right?
I am using a Fluke processmeter 789 and a Fluke 187 True RMS meter. The transformer is running at 50Hz. I don;t quite understand what you mean by siemen range. Can elaborate. Yes it pure AC without any rectification.
I am using a Fluke processmeter 789 and a Fluke 187 True RMS meter. The transformer is running at 50Hz. I don;t quite understand what you mean by siemen range. Can elaborate. Yes it pure AC without any rectification.
On the 187 when you set the selector switch to ohms, you will see above the Ω symbol, in blue, the designation "nS". That means nanosiemens (which is a measurement of conductance, instead of resistance. Conductance is the reciprocal of resistance). After selecting ohms, press the blue button twice to enter the nanosiemens range.
Using the ordinary probes, connect them to the tip of the HV probe, and the banana plug that goes into the jack on the far right, where normally your red probe goes. The reading will be slow to settle down, but eventually it should reach a value of 1.00 (plus or minus probably 10% or so), which is equivalent to 1000 megohms (1 gigaohm). I think the HV probe you have has a 1000 megohm multiplier resistor inside. If not, then this technique will give you a mesurement of its actual resistance. You should be able to find out what the intended resistance of the internal multiplier is from the spec sheet of the HV probe.
If you plug both banana plugs of the HV probe into the 187 while in nanosiemens range, with the ground clip of the HV probe not connected to anything, the reading should approach zero (more probably 0.01 which is 100 gigaohms); there's not supposed to be much leakage between those two banana plugs.
It occurs to me that if you have the HV probe connected to the 187 in the normal way to make a measurement, you could connect the ground clip to the probe tip, and with the 187 in nanosiemens range, you would see the value of the internal multiplier resistor (in nanosiemens, of course).
Since you have two meters, for a really definitive check, you could connect the HV probe to the 187, and measure your voltage with the 789 directly at the same time as you use the HV probe on the 187. Then you have a known good reading from the 789 to compare with the HV probe reading at the same time.
Perhaps the 80K-40 isn't specified to work with AC.
Even if the HV probe isn't specified for AC, you can use the two meter trick I just described to determine a calibration factor for the HV probe when measuring AC volts, which you can then use for higher voltages if need be.
While making the simultaneous readings, divide the 789 known good reading by the HV probe reading. This gives you a factor by which you can multiply a subsequent HV probe reading to get the actual voltage (which might then be greater than 1 kV) being measured.
Also, once you have both meters making the measurement, using the HV probe on the 187, watch the reading on that meter (the one with the HV probe) when you disconnect the other one, and see if the reading changes. This will tell you if the 10 megohms input impedance of the 789 is loading your circuit significantly.
On the 187 when you set the selector switch to ohms, you will see above the Ω symbol, in blue, the designation "nS". That means nanosiemens (which is a measurement of conductance, instead of resistance. Conductance is the reciprocal of resistance). After selecting ohms, press the blue button twice to enter the nanosiemens range.
Using the ordinary probes, connect them to the tip of the HV probe, and the banana plug that goes into the jack on the far right, where normally your red probe goes. The reading will be slow to settle down, but eventually it should reach a value of 1.00 (plus or minus probably 10% or so), which is equivalent to 1000 megohms (1 gigaohm). I think the HV probe you have has a 1000 megohm multiplier resistor inside. If not, then this technique will give you a mesurement of its actual resistance. You should be able to find out what the intended resistance of the internal multiplier is from the spec sheet of the HV probe.
If you plug both banana plugs of the HV probe into the 187 while in nanosiemens range, with the ground clip of the HV probe not connected to anything, the reading should approach zero (more probably 0.01 which is 100 gigaohms); there's not supposed to be much leakage between those two banana plugs.
It occurs to me that if you have the HV probe connected to the 187 in the normal way to make a measurement, you could connect the ground clip to the probe tip, and with the 187 in nanosiemens range, you would see the value of the internal multiplier resistor (in nanosiemens, of course).
Since you have two meters, for a really definitive check, you could connect the HV probe to the 187, and measure your voltage with the 789 directly at the same time as you use the HV probe on the 187. Then you have a known good reading from the 789 to compare with the HV probe reading at the same time.
Perhaps the 80K-40 isn't specified to work with AC.
Even if the HV probe isn't specified for AC, you can use the two meter trick I just described to determine a calibration factor for the HV probe when measuring AC volts, which you can then use for higher voltages if need be.
While making the simultaneous readings, divide the 789 known good reading by the HV probe reading. This gives you a factor by which you can multiply a subsequent HV probe reading to get the actual voltage (which might then be greater than 1 kV) being measured.
Also, once you have both meters making the measurement, using the HV probe on the 187, watch the reading on that meter (the one with the HV probe) when you disconnect the other one, and see if the reading changes. This will tell you if the 10 megohms input impedance of the 789 is loading your circuit significantly.
1. The Fluke HV probe doesn't have near the accuracy spec of the meters you use it with. As I recall, they're only good to maybe 5% or so at best.
2. Looking at the numbers in the original post -- I see the potential for death here. If voltages are in the kilovolt range and currents are in the 10a amp range, that's scary. You're talking power company line stuff here.
3. Also, as I recall, unlike the HV probes of the old vacuum tube voltmeters, the Fluke probe has a built-in termination resistor so that it doesn't depend upon the meter load being exactly 10M ohms as the second resistor of the voltage divider scheme.
4. In general, HV circuits (not the one in the OP) tend to also be high-impedance circuits, so meter loading can become a problem. Take circuit impedance into account when the voltage reading of 600v is far below the expected 800v
This site uses cookies to help personalise content, tailor your experience and to keep you logged in if you register.
By continuing to use this site, you are consenting to our use of cookies.
