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High Voltage Measuring System
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I am looking to set up a voltage-measuring system for the output of a recently purchased HV power source. The source supplies around 0 - -60 kV DC and 10 mA to my load. Unfortunately, the source does not have a voltage measuring system, so I will be tasked in its creation. This will be of particular difficulty because I do not have enough money to purchase sufficient HV probes that could simply be attached to a multi meter. Thus, I will need a way to lower the incoming voltage from -60 kV to something manageable for the multi meter.
Currently, this is the schematic that I am considering to use in order to lower the voltage:
simulate this circuit – Schematic created using CircuitLab
The multi meter that I currently own can handle up to 660 VDC. Using my apparatus above coupled with the multi meter, would the multi meter be able handle the voltage that is delivered to it and give out an accurate reading of the line voltage (assuming that a series of calculations are performed)?
high-voltage multimeter
asked 22 hours ago
Super Nerds Team
394
 |Â
show 1 more comment
up vote
4
down vote
favorite
I am looking to set up a voltage-measuring system for the output of a recently purchased HV power source. The source supplies around 0 - -60 kV DC and 10 mA to my load. Unfortunately, the source does not have a voltage measuring system, so I will be tasked in its creation. This will be of particular difficulty because I do not have enough money to purchase sufficient HV probes that could simply be attached to a multi meter. Thus, I will need a way to lower the incoming voltage from -60 kV to something manageable for the multi meter.
Currently, this is the schematic that I am considering to use in order to lower the voltage:
simulate this circuit – Schematic created using CircuitLab
The multi meter that I currently own can handle up to 660 VDC. Using my apparatus above coupled with the multi meter, would the multi meter be able handle the voltage that is delivered to it and give out an accurate reading of the line voltage (assuming that a series of calculations are performed)?
high-voltage multimeter
asked 22 hours ago
Super Nerds Team
394
Maybe look into a capacitive divider. I know they've been mentioned here when discussing how to measure high DC voltages.
– JRE
21 hours ago
7
This is extremely dangerous. You have to make sure that all of your connections will not leak current at these high voltages. Don't plan to use a PCB. Please, please find a way to buy a good bit of commercial test equipment.
– Elliot Alderson
20 hours ago
2
Any high voltage like this will also create tiny quantities of static charge (as part of a gradient from one end to the other of a galvanic path) that immediately set themselves up in order to cause the current to flow. Those charges can be independently measured. That's another approach to consider. How often are the measurements? What kind of resolution do you require? Is there a separate accuracy requirement?
– jonk
20 hours ago
What is (1) the output impedance and (2) output capacitance of this -60 kV 10 mA DC PS? A schematic would help.
– Mike Waters
18 hours ago
Even wire is not easily arranged at high voltage; instead of 'a multimeter', consider using an electrostatic meter (a moving-vane capacitor, or even a piezoelectric crystal or bimorph) that draws no current.
– Whit3rd
7 hours ago
 |Â
show 1 more comment
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I am looking to set up a voltage-measuring system for the output of a recently purchased HV power source. The source supplies around 0 - -60 kV DC and 10 mA to my load. Unfortunately, the source does not have a voltage measuring system, so I will be tasked in its creation. This will be of particular difficulty because I do not have enough money to purchase sufficient HV probes that could simply be attached to a multi meter. Thus, I will need a way to lower the incoming voltage from -60 kV to something manageable for the multi meter.
Currently, this is the schematic that I am considering to use in order to lower the voltage:
simulate this circuit – Schematic created using CircuitLab
The multi meter that I currently own can handle up to 660 VDC. Using my apparatus above coupled with the multi meter, would the multi meter be able handle the voltage that is delivered to it and give out an accurate reading of the line voltage (assuming that a series of calculations are performed)?
high-voltage multimeter
asked 22 hours ago
Super Nerds Team
394
I am looking to set up a voltage-measuring system for the output of a recently purchased HV power source. The source supplies around 0 - -60 kV DC and 10 mA to my load. Unfortunately, the source does not have a voltage measuring system, so I will be tasked in its creation. This will be of particular difficulty because I do not have enough money to purchase sufficient HV probes that could simply be attached to a multi meter. Thus, I will need a way to lower the incoming voltage from -60 kV to something manageable for the multi meter.
Currently, this is the schematic that I am considering to use in order to lower the voltage:
simulate this circuit – Schematic created using CircuitLab
The multi meter that I currently own can handle up to 660 VDC. Using my apparatus above coupled with the multi meter, would the multi meter be able handle the voltage that is delivered to it and give out an accurate reading of the line voltage (assuming that a series of calculations are performed)?
high-voltage multimeter
high-voltage multimeter
asked 22 hours ago
Super Nerds Team
394
asked 22 hours ago
Super Nerds Team
394
asked 22 hours ago
Super Nerds Team
394
asked 22 hours ago
Super Nerds Team
394
asked 22 hours ago
Super Nerds Team
394
394
Maybe look into a capacitive divider. I know they've been mentioned here when discussing how to measure high DC voltages.
– JRE
21 hours ago
7
This is extremely dangerous. You have to make sure that all of your connections will not leak current at these high voltages. Don't plan to use a PCB. Please, please find a way to buy a good bit of commercial test equipment.
– Elliot Alderson
20 hours ago
2
Any high voltage like this will also create tiny quantities of static charge (as part of a gradient from one end to the other of a galvanic path) that immediately set themselves up in order to cause the current to flow. Those charges can be independently measured. That's another approach to consider. How often are the measurements? What kind of resolution do you require? Is there a separate accuracy requirement?
– jonk
20 hours ago
What is (1) the output impedance and (2) output capacitance of this -60 kV 10 mA DC PS? A schematic would help.
– Mike Waters
18 hours ago
Even wire is not easily arranged at high voltage; instead of 'a multimeter', consider using an electrostatic meter (a moving-vane capacitor, or even a piezoelectric crystal or bimorph) that draws no current.
– Whit3rd
7 hours ago
 |Â
show 1 more comment
Maybe look into a capacitive divider. I know they've been mentioned here when discussing how to measure high DC voltages.
– JRE
21 hours ago
7
This is extremely dangerous. You have to make sure that all of your connections will not leak current at these high voltages. Don't plan to use a PCB. Please, please find a way to buy a good bit of commercial test equipment.
– Elliot Alderson
20 hours ago
2
Any high voltage like this will also create tiny quantities of static charge (as part of a gradient from one end to the other of a galvanic path) that immediately set themselves up in order to cause the current to flow. Those charges can be independently measured. That's another approach to consider. How often are the measurements? What kind of resolution do you require? Is there a separate accuracy requirement?
– jonk
20 hours ago
What is (1) the output impedance and (2) output capacitance of this -60 kV 10 mA DC PS? A schematic would help.
– Mike Waters
18 hours ago
Even wire is not easily arranged at high voltage; instead of 'a multimeter', consider using an electrostatic meter (a moving-vane capacitor, or even a piezoelectric crystal or bimorph) that draws no current.
– Whit3rd
7 hours ago
Maybe look into a capacitive divider. I know they've been mentioned here when discussing how to measure high DC voltages.
– JRE
21 hours ago
Maybe look into a capacitive divider. I know they've been mentioned here when discussing how to measure high DC voltages.
– JRE
21 hours ago
7
7
This is extremely dangerous. You have to make sure that all of your connections will not leak current at these high voltages. Don't plan to use a PCB. Please, please find a way to buy a good bit of commercial test equipment.
– Elliot Alderson
20 hours ago
This is extremely dangerous. You have to make sure that all of your connections will not leak current at these high voltages. Don't plan to use a PCB. Please, please find a way to buy a good bit of commercial test equipment.
– Elliot Alderson
20 hours ago
2
2
Any high voltage like this will also create tiny quantities of static charge (as part of a gradient from one end to the other of a galvanic path) that immediately set themselves up in order to cause the current to flow. Those charges can be independently measured. That's another approach to consider. How often are the measurements? What kind of resolution do you require? Is there a separate accuracy requirement?
– jonk
20 hours ago
Any high voltage like this will also create tiny quantities of static charge (as part of a gradient from one end to the other of a galvanic path) that immediately set themselves up in order to cause the current to flow. Those charges can be independently measured. That's another approach to consider. How often are the measurements? What kind of resolution do you require? Is there a separate accuracy requirement?
– jonk
20 hours ago
What is (1) the output impedance and (2) output capacitance of this -60 kV 10 mA DC PS? A schematic would help.
– Mike Waters
18 hours ago
What is (1) the output impedance and (2) output capacitance of this -60 kV 10 mA DC PS? A schematic would help.
– Mike Waters
18 hours ago
Even wire is not easily arranged at high voltage; instead of 'a multimeter', consider using an electrostatic meter (a moving-vane capacitor, or even a piezoelectric crystal or bimorph) that draws no current.
– Whit3rd
7 hours ago
Even wire is not easily arranged at high voltage; instead of 'a multimeter', consider using an electrostatic meter (a moving-vane capacitor, or even a piezoelectric crystal or bimorph) that draws no current.
– Whit3rd
7 hours ago
 |Â
show 1 more comment
3 Answers
3
active
oldest
votes
up vote
9
down vote
You are unlikely to find a single resistor that can handle 60kV. You will probably have to use many resistors in series. Make sure to consider the tolerance when calculating the voltage across each resistor.
The 100M resistor would be dissipating 36 W.
Be careful of the loading on the supply. This contraption draws 0.6mA.
answered 22 hours ago
Äεκ
2,8501915
add a comment |Â
up vote
4
down vote
From my knowledge, the accuracy of your reading will depend on the sensitivity of your multi meter. According to your divider,about 600uA flows through the resistors. So a small current like 100uA drawn into the multi meter will change the readings.
Almost every digital multi meter uses an electronic amplifier (op amp) in the input stage making the impedance fixed. So a good digital multi meter will do the job for you. But most of the analog multi meters will draw a varying current depending on the input voltage. So that will be a no go.
answered 21 hours ago
Arosha Dissanayake
517
As TEK mentioned, using multiple resistors in series rather than having a one big resistor will help the tolerance problem too. Negative errors may be cancelled out by positive errors.
– Arosha Dissanayake
21 hours ago
How many resistors should I use and what tolerance should they have?
– Super Nerds Team
21 hours ago
Try using 10x 39Mohms 1W 1% resistors in series instead of 100Mohms resistor this will decrease the max V output to ~150V and the heat dissipation too. Lesser tolerance gives better accuracy.
– Arosha Dissanayake
21 hours ago
Would I have to change R2 in the diagram (the 1 M ohm resistor)?
– Super Nerds Team
21 hours ago
5
@AroshaDissanayake Increasing accuracy when combining resistors doesn't usually work in practice. You are assuming that when you buy a batch of resistors that their mean value is the nominal value and their deviation has a normal distribution. It is extremely unlikely that you would get that in the real world.
– Elliot Alderson
20 hours ago
 |Â
show 15 more comments
up vote
3
down vote
A small change would be to use the multimeter in current measuring range, instead of voltage measuring.
This way you can reduce the measurement current to microamperes, and it would also be much less sensitive to the impedance of the multimeter. And the voltage on the wires going to multimeter will be less than a volt.
It is still a good idea to leave R2 in place to limit the voltage in case the meter is not connected. For µA level measurement current, your value of 1 Mohm is suitable to limit the voltage to quite safe ~60 volts.
And like other answers already mentioned, you will have to construct R1 from multiple resistors as a physically long chain, to avoid arcing. For example, 100x of 10Mohm resistors in series would give 1 µA = 1 kV response for the meter.
The advantage is that in current measurement mode, the multimeter's impedance is very low (at most a few kilo-ohms) compared to R2. This way all measurement current goes to good use in the meter, instead of being wasted as heat. Also the exact impedance matters less, because in any case it is much smaller than R2, whereas in voltage mode the impedance is several megaohms and is thus similar to R2 value.
By reducing the measurement current to microamperes, the power lost to heat is reduced to e.g. 60 µA * 60 kV = 3.6 W, which is only 0.036 W per resistor.
answered 19 hours ago
jpa
1,294611
So in this case, would I want to set up my connections to be measuring voltage or current? I initially designed the circuit from my post to reduce the 60 kV DC and then read a small voltage (around 150 V).
– Super Nerds Team
18 hours ago
@SuperNerdsTeam For measuring current. That way all the measurement current goes to good use, instead of being wasted as heat in voltage divider.
– jpa
8 hours ago
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
You are unlikely to find a single resistor that can handle 60kV. You will probably have to use many resistors in series. Make sure to consider the tolerance when calculating the voltage across each resistor.
The 100M resistor would be dissipating 36 W.
Be careful of the loading on the supply. This contraption draws 0.6mA.
answered 22 hours ago
Äεκ
2,8501915
add a comment |Â
up vote
9
down vote
You are unlikely to find a single resistor that can handle 60kV. You will probably have to use many resistors in series. Make sure to consider the tolerance when calculating the voltage across each resistor.
The 100M resistor would be dissipating 36 W.
Be careful of the loading on the supply. This contraption draws 0.6mA.
answered 22 hours ago
Äεκ
2,8501915
add a comment |Â
up vote
9
down vote
up vote
9
down vote
You are unlikely to find a single resistor that can handle 60kV. You will probably have to use many resistors in series. Make sure to consider the tolerance when calculating the voltage across each resistor.
The 100M resistor would be dissipating 36 W.
Be careful of the loading on the supply. This contraption draws 0.6mA.
answered 22 hours ago
Äεκ
2,8501915
You are unlikely to find a single resistor that can handle 60kV. You will probably have to use many resistors in series. Make sure to consider the tolerance when calculating the voltage across each resistor.
The 100M resistor would be dissipating 36 W.
Be careful of the loading on the supply. This contraption draws 0.6mA.
answered 22 hours ago
Äεκ
2,8501915
answered 22 hours ago
Äεκ
2,8501915
answered 22 hours ago
Äεκ
2,8501915
answered 22 hours ago
Äεκ
2,8501915
2,8501915
add a comment |Â
add a comment |Â
up vote
4
down vote
From my knowledge, the accuracy of your reading will depend on the sensitivity of your multi meter. According to your divider,about 600uA flows through the resistors. So a small current like 100uA drawn into the multi meter will change the readings.
Almost every digital multi meter uses an electronic amplifier (op amp) in the input stage making the impedance fixed. So a good digital multi meter will do the job for you. But most of the analog multi meters will draw a varying current depending on the input voltage. So that will be a no go.
answered 21 hours ago
Arosha Dissanayake
517
As TEK mentioned, using multiple resistors in series rather than having a one big resistor will help the tolerance problem too. Negative errors may be cancelled out by positive errors.
– Arosha Dissanayake
21 hours ago
How many resistors should I use and what tolerance should they have?
– Super Nerds Team
21 hours ago
Try using 10x 39Mohms 1W 1% resistors in series instead of 100Mohms resistor this will decrease the max V output to ~150V and the heat dissipation too. Lesser tolerance gives better accuracy.
– Arosha Dissanayake
21 hours ago
Would I have to change R2 in the diagram (the 1 M ohm resistor)?
– Super Nerds Team
21 hours ago
5
@AroshaDissanayake Increasing accuracy when combining resistors doesn't usually work in practice. You are assuming that when you buy a batch of resistors that their mean value is the nominal value and their deviation has a normal distribution. It is extremely unlikely that you would get that in the real world.
– Elliot Alderson
20 hours ago
 |Â
show 15 more comments
up vote
4
down vote
From my knowledge, the accuracy of your reading will depend on the sensitivity of your multi meter. According to your divider,about 600uA flows through the resistors. So a small current like 100uA drawn into the multi meter will change the readings.
Almost every digital multi meter uses an electronic amplifier (op amp) in the input stage making the impedance fixed. So a good digital multi meter will do the job for you. But most of the analog multi meters will draw a varying current depending on the input voltage. So that will be a no go.
answered 21 hours ago
Arosha Dissanayake
517
As TEK mentioned, using multiple resistors in series rather than having a one big resistor will help the tolerance problem too. Negative errors may be cancelled out by positive errors.
– Arosha Dissanayake
21 hours ago
How many resistors should I use and what tolerance should they have?
– Super Nerds Team
21 hours ago
Try using 10x 39Mohms 1W 1% resistors in series instead of 100Mohms resistor this will decrease the max V output to ~150V and the heat dissipation too. Lesser tolerance gives better accuracy.
– Arosha Dissanayake
21 hours ago
Would I have to change R2 in the diagram (the 1 M ohm resistor)?
– Super Nerds Team
21 hours ago
5
@AroshaDissanayake Increasing accuracy when combining resistors doesn't usually work in practice. You are assuming that when you buy a batch of resistors that their mean value is the nominal value and their deviation has a normal distribution. It is extremely unlikely that you would get that in the real world.
– Elliot Alderson
20 hours ago
 |Â
show 15 more comments
up vote
4
down vote
up vote
4
down vote
From my knowledge, the accuracy of your reading will depend on the sensitivity of your multi meter. According to your divider,about 600uA flows through the resistors. So a small current like 100uA drawn into the multi meter will change the readings.
Almost every digital multi meter uses an electronic amplifier (op amp) in the input stage making the impedance fixed. So a good digital multi meter will do the job for you. But most of the analog multi meters will draw a varying current depending on the input voltage. So that will be a no go.
answered 21 hours ago
Arosha Dissanayake
517
From my knowledge, the accuracy of your reading will depend on the sensitivity of your multi meter. According to your divider,about 600uA flows through the resistors. So a small current like 100uA drawn into the multi meter will change the readings.
Almost every digital multi meter uses an electronic amplifier (op amp) in the input stage making the impedance fixed. So a good digital multi meter will do the job for you. But most of the analog multi meters will draw a varying current depending on the input voltage. So that will be a no go.
answered 21 hours ago
Arosha Dissanayake
517
edited 18 hours ago
answered 21 hours ago
Arosha Dissanayake
517
answered 21 hours ago
Arosha Dissanayake
517
answered 21 hours ago
Arosha Dissanayake
517
517
As TEK mentioned, using multiple resistors in series rather than having a one big resistor will help the tolerance problem too. Negative errors may be cancelled out by positive errors.
– Arosha Dissanayake
21 hours ago
How many resistors should I use and what tolerance should they have?
– Super Nerds Team
21 hours ago
Try using 10x 39Mohms 1W 1% resistors in series instead of 100Mohms resistor this will decrease the max V output to ~150V and the heat dissipation too. Lesser tolerance gives better accuracy.
– Arosha Dissanayake
21 hours ago
Would I have to change R2 in the diagram (the 1 M ohm resistor)?
– Super Nerds Team
21 hours ago
5
@AroshaDissanayake Increasing accuracy when combining resistors doesn't usually work in practice. You are assuming that when you buy a batch of resistors that their mean value is the nominal value and their deviation has a normal distribution. It is extremely unlikely that you would get that in the real world.
– Elliot Alderson
20 hours ago
 |Â
show 15 more comments
As TEK mentioned, using multiple resistors in series rather than having a one big resistor will help the tolerance problem too. Negative errors may be cancelled out by positive errors.
– Arosha Dissanayake
21 hours ago
How many resistors should I use and what tolerance should they have?
– Super Nerds Team
21 hours ago
Try using 10x 39Mohms 1W 1% resistors in series instead of 100Mohms resistor this will decrease the max V output to ~150V and the heat dissipation too. Lesser tolerance gives better accuracy.
– Arosha Dissanayake
21 hours ago
Would I have to change R2 in the diagram (the 1 M ohm resistor)?
– Super Nerds Team
21 hours ago
5
@AroshaDissanayake Increasing accuracy when combining resistors doesn't usually work in practice. You are assuming that when you buy a batch of resistors that their mean value is the nominal value and their deviation has a normal distribution. It is extremely unlikely that you would get that in the real world.
– Elliot Alderson
20 hours ago
As TEK mentioned, using multiple resistors in series rather than having a one big resistor will help the tolerance problem too. Negative errors may be cancelled out by positive errors.
– Arosha Dissanayake
21 hours ago
As TEK mentioned, using multiple resistors in series rather than having a one big resistor will help the tolerance problem too. Negative errors may be cancelled out by positive errors.
– Arosha Dissanayake
21 hours ago
How many resistors should I use and what tolerance should they have?
– Super Nerds Team
21 hours ago
How many resistors should I use and what tolerance should they have?
– Super Nerds Team
21 hours ago
Try using 10x 39Mohms 1W 1% resistors in series instead of 100Mohms resistor this will decrease the max V output to ~150V and the heat dissipation too. Lesser tolerance gives better accuracy.
– Arosha Dissanayake
21 hours ago
Try using 10x 39Mohms 1W 1% resistors in series instead of 100Mohms resistor this will decrease the max V output to ~150V and the heat dissipation too. Lesser tolerance gives better accuracy.
– Arosha Dissanayake
21 hours ago
Would I have to change R2 in the diagram (the 1 M ohm resistor)?
– Super Nerds Team
21 hours ago
Would I have to change R2 in the diagram (the 1 M ohm resistor)?
– Super Nerds Team
21 hours ago
5
5
@AroshaDissanayake Increasing accuracy when combining resistors doesn't usually work in practice. You are assuming that when you buy a batch of resistors that their mean value is the nominal value and their deviation has a normal distribution. It is extremely unlikely that you would get that in the real world.
– Elliot Alderson
20 hours ago
@AroshaDissanayake Increasing accuracy when combining resistors doesn't usually work in practice. You are assuming that when you buy a batch of resistors that their mean value is the nominal value and their deviation has a normal distribution. It is extremely unlikely that you would get that in the real world.
– Elliot Alderson
20 hours ago
 |Â
show 15 more comments
up vote
3
down vote
A small change would be to use the multimeter in current measuring range, instead of voltage measuring.
This way you can reduce the measurement current to microamperes, and it would also be much less sensitive to the impedance of the multimeter. And the voltage on the wires going to multimeter will be less than a volt.
It is still a good idea to leave R2 in place to limit the voltage in case the meter is not connected. For µA level measurement current, your value of 1 Mohm is suitable to limit the voltage to quite safe ~60 volts.
And like other answers already mentioned, you will have to construct R1 from multiple resistors as a physically long chain, to avoid arcing. For example, 100x of 10Mohm resistors in series would give 1 µA = 1 kV response for the meter.
The advantage is that in current measurement mode, the multimeter's impedance is very low (at most a few kilo-ohms) compared to R2. This way all measurement current goes to good use in the meter, instead of being wasted as heat. Also the exact impedance matters less, because in any case it is much smaller than R2, whereas in voltage mode the impedance is several megaohms and is thus similar to R2 value.
By reducing the measurement current to microamperes, the power lost to heat is reduced to e.g. 60 µA * 60 kV = 3.6 W, which is only 0.036 W per resistor.
answered 19 hours ago
jpa
1,294611
So in this case, would I want to set up my connections to be measuring voltage or current? I initially designed the circuit from my post to reduce the 60 kV DC and then read a small voltage (around 150 V).
– Super Nerds Team
18 hours ago
@SuperNerdsTeam For measuring current. That way all the measurement current goes to good use, instead of being wasted as heat in voltage divider.
– jpa
8 hours ago
add a comment |Â
up vote
3
down vote
A small change would be to use the multimeter in current measuring range, instead of voltage measuring.
This way you can reduce the measurement current to microamperes, and it would also be much less sensitive to the impedance of the multimeter. And the voltage on the wires going to multimeter will be less than a volt.
It is still a good idea to leave R2 in place to limit the voltage in case the meter is not connected. For µA level measurement current, your value of 1 Mohm is suitable to limit the voltage to quite safe ~60 volts.
And like other answers already mentioned, you will have to construct R1 from multiple resistors as a physically long chain, to avoid arcing. For example, 100x of 10Mohm resistors in series would give 1 µA = 1 kV response for the meter.
The advantage is that in current measurement mode, the multimeter's impedance is very low (at most a few kilo-ohms) compared to R2. This way all measurement current goes to good use in the meter, instead of being wasted as heat. Also the exact impedance matters less, because in any case it is much smaller than R2, whereas in voltage mode the impedance is several megaohms and is thus similar to R2 value.
By reducing the measurement current to microamperes, the power lost to heat is reduced to e.g. 60 µA * 60 kV = 3.6 W, which is only 0.036 W per resistor.
answered 19 hours ago
jpa
1,294611
So in this case, would I want to set up my connections to be measuring voltage or current? I initially designed the circuit from my post to reduce the 60 kV DC and then read a small voltage (around 150 V).
– Super Nerds Team
18 hours ago
@SuperNerdsTeam For measuring current. That way all the measurement current goes to good use, instead of being wasted as heat in voltage divider.
– jpa
8 hours ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
A small change would be to use the multimeter in current measuring range, instead of voltage measuring.
This way you can reduce the measurement current to microamperes, and it would also be much less sensitive to the impedance of the multimeter. And the voltage on the wires going to multimeter will be less than a volt.
It is still a good idea to leave R2 in place to limit the voltage in case the meter is not connected. For µA level measurement current, your value of 1 Mohm is suitable to limit the voltage to quite safe ~60 volts.
And like other answers already mentioned, you will have to construct R1 from multiple resistors as a physically long chain, to avoid arcing. For example, 100x of 10Mohm resistors in series would give 1 µA = 1 kV response for the meter.
The advantage is that in current measurement mode, the multimeter's impedance is very low (at most a few kilo-ohms) compared to R2. This way all measurement current goes to good use in the meter, instead of being wasted as heat. Also the exact impedance matters less, because in any case it is much smaller than R2, whereas in voltage mode the impedance is several megaohms and is thus similar to R2 value.
By reducing the measurement current to microamperes, the power lost to heat is reduced to e.g. 60 µA * 60 kV = 3.6 W, which is only 0.036 W per resistor.
answered 19 hours ago
jpa
1,294611
A small change would be to use the multimeter in current measuring range, instead of voltage measuring.
This way you can reduce the measurement current to microamperes, and it would also be much less sensitive to the impedance of the multimeter. And the voltage on the wires going to multimeter will be less than a volt.
It is still a good idea to leave R2 in place to limit the voltage in case the meter is not connected. For µA level measurement current, your value of 1 Mohm is suitable to limit the voltage to quite safe ~60 volts.
And like other answers already mentioned, you will have to construct R1 from multiple resistors as a physically long chain, to avoid arcing. For example, 100x of 10Mohm resistors in series would give 1 µA = 1 kV response for the meter.
The advantage is that in current measurement mode, the multimeter's impedance is very low (at most a few kilo-ohms) compared to R2. This way all measurement current goes to good use in the meter, instead of being wasted as heat. Also the exact impedance matters less, because in any case it is much smaller than R2, whereas in voltage mode the impedance is several megaohms and is thus similar to R2 value.
By reducing the measurement current to microamperes, the power lost to heat is reduced to e.g. 60 µA * 60 kV = 3.6 W, which is only 0.036 W per resistor.
answered 19 hours ago
jpa
1,294611
edited 8 hours ago
answered 19 hours ago
jpa
1,294611
answered 19 hours ago
jpa
1,294611
answered 19 hours ago
jpa
1,294611
1,294611
So in this case, would I want to set up my connections to be measuring voltage or current? I initially designed the circuit from my post to reduce the 60 kV DC and then read a small voltage (around 150 V).
– Super Nerds Team
18 hours ago
@SuperNerdsTeam For measuring current. That way all the measurement current goes to good use, instead of being wasted as heat in voltage divider.
– jpa
8 hours ago
add a comment |Â
So in this case, would I want to set up my connections to be measuring voltage or current? I initially designed the circuit from my post to reduce the 60 kV DC and then read a small voltage (around 150 V).
– Super Nerds Team
18 hours ago
@SuperNerdsTeam For measuring current. That way all the measurement current goes to good use, instead of being wasted as heat in voltage divider.
– jpa
8 hours ago
So in this case, would I want to set up my connections to be measuring voltage or current? I initially designed the circuit from my post to reduce the 60 kV DC and then read a small voltage (around 150 V).
– Super Nerds Team
18 hours ago
So in this case, would I want to set up my connections to be measuring voltage or current? I initially designed the circuit from my post to reduce the 60 kV DC and then read a small voltage (around 150 V).
– Super Nerds Team
18 hours ago
@SuperNerdsTeam For measuring current. That way all the measurement current goes to good use, instead of being wasted as heat in voltage divider.
– jpa
8 hours ago
@SuperNerdsTeam For measuring current. That way all the measurement current goes to good use, instead of being wasted as heat in voltage divider.
– jpa
8 hours ago
add a comment |Â
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Maybe look into a capacitive divider. I know they've been mentioned here when discussing how to measure high DC voltages.
– JRE
21 hours ago
7
This is extremely dangerous. You have to make sure that all of your connections will not leak current at these high voltages. Don't plan to use a PCB. Please, please find a way to buy a good bit of commercial test equipment.
– Elliot Alderson
20 hours ago
2
Any high voltage like this will also create tiny quantities of static charge (as part of a gradient from one end to the other of a galvanic path) that immediately set themselves up in order to cause the current to flow. Those charges can be independently measured. That's another approach to consider. How often are the measurements? What kind of resolution do you require? Is there a separate accuracy requirement?
– jonk
20 hours ago
What is (1) the output impedance and (2) output capacitance of this -60 kV 10 mA DC PS? A schematic would help.
– Mike Waters
18 hours ago
Even wire is not easily arranged at high voltage; instead of 'a multimeter', consider using an electrostatic meter (a moving-vane capacitor, or even a piezoelectric crystal or bimorph) that draws no current.
– Whit3rd
7 hours ago