$begingroup$
I have two 10G resistors connected in series with a 3V battery. I want to determine the voltage drop across one of them, which of course is 1.5V. When I use my multimeter to check the voltage drop, it reads ~3mV, which I believe is because it has a 10M impedance so the circuit is really one 10G resistor in series with (a 10G resistor and a 10M resistor in parallel), so the voltage drop when the multimeter is part of the circuit is 2.99 mV.
simulate this circuit – Schematic created using CircuitLab
How can I measure the voltage drop? Is there something I can build so that that I can adapt the multimeter impedance to be high enough that it won't affect the circuit as much?
multimeter voltage-measurement
asked 2 hours ago
John SmithJohn Smith
895
$endgroup$
|
show 2 more comments
$begingroup$
I have two 10G resistors connected in series with a 3V battery. I want to determine the voltage drop across one of them, which of course is 1.5V. When I use my multimeter to check the voltage drop, it reads ~3mV, which I believe is because it has a 10M impedance so the circuit is really one 10G resistor in series with (a 10G resistor and a 10M resistor in parallel), so the voltage drop when the multimeter is part of the circuit is 2.99 mV.
simulate this circuit – Schematic created using CircuitLab
How can I measure the voltage drop? Is there something I can build so that that I can adapt the multimeter impedance to be high enough that it won't affect the circuit as much?
multimeter voltage-measurement
asked 2 hours ago
John SmithJohn Smith
895
$endgroup$
2
$begingroup$
Look up a 1993 ED article "What's All This Femtoampere Stuff, Anyhow?" by the late Robert Pease.
$endgroup$
– Spehro Pefhany
1 hour ago
3
$begingroup$
Why would you need such a circuit if I may ask? Any load attached will casue the same effect as the multimeter.
$endgroup$
– Huisman
1 hour ago
$begingroup$
@Huisman for trying to build ammeters that can go to very low current. I want very low current sources, then to try to measure them. If I'm dividing the voltage down first before passing through a 10G resistor (or higher) it's especially helpful to be able to measure that the actual voltage drop is what I expect it to be.
$endgroup$
– John Smith
1 hour ago
$begingroup$
Could you please draw a schematic (by pressing Schematic button in editor in your original post) where the ammeter is located? I think you'd better divide the voltage by e.g. 10k pot and connect its branch with a 10G resistor to the ammeter.
$endgroup$
– Huisman
1 hour ago
$begingroup$
As drawn, it looks like the divider will be passing 150pA. There are definitely more things that can sneak up on you at that point to mess with your measurement. Maybe you don't need femto amp precision, but seeing what they do to ensure fA accuracy is probably a good step.
$endgroup$
– W5VO♦
59 mins ago
|
show 2 more comments
$begingroup$
I have two 10G resistors connected in series with a 3V battery. I want to determine the voltage drop across one of them, which of course is 1.5V. When I use my multimeter to check the voltage drop, it reads ~3mV, which I believe is because it has a 10M impedance so the circuit is really one 10G resistor in series with (a 10G resistor and a 10M resistor in parallel), so the voltage drop when the multimeter is part of the circuit is 2.99 mV.
simulate this circuit – Schematic created using CircuitLab
How can I measure the voltage drop? Is there something I can build so that that I can adapt the multimeter impedance to be high enough that it won't affect the circuit as much?
multimeter voltage-measurement
asked 2 hours ago
John SmithJohn Smith
895
$endgroup$
I have two 10G resistors connected in series with a 3V battery. I want to determine the voltage drop across one of them, which of course is 1.5V. When I use my multimeter to check the voltage drop, it reads ~3mV, which I believe is because it has a 10M impedance so the circuit is really one 10G resistor in series with (a 10G resistor and a 10M resistor in parallel), so the voltage drop when the multimeter is part of the circuit is 2.99 mV.
simulate this circuit – Schematic created using CircuitLab
How can I measure the voltage drop? Is there something I can build so that that I can adapt the multimeter impedance to be high enough that it won't affect the circuit as much?
multimeter voltage-measurement
multimeter voltage-measurement
asked 2 hours ago
John SmithJohn Smith
895
asked 2 hours ago
John SmithJohn Smith
895
edited 1 hour ago
John Smith
asked 2 hours ago
John SmithJohn Smith
895
asked 2 hours ago
John SmithJohn Smith
895
asked 2 hours ago
John SmithJohn Smith
895
895
2
$begingroup$
Look up a 1993 ED article "What's All This Femtoampere Stuff, Anyhow?" by the late Robert Pease.
$endgroup$
– Spehro Pefhany
1 hour ago
3
$begingroup$
Why would you need such a circuit if I may ask? Any load attached will casue the same effect as the multimeter.
$endgroup$
– Huisman
1 hour ago
$begingroup$
@Huisman for trying to build ammeters that can go to very low current. I want very low current sources, then to try to measure them. If I'm dividing the voltage down first before passing through a 10G resistor (or higher) it's especially helpful to be able to measure that the actual voltage drop is what I expect it to be.
$endgroup$
– John Smith
1 hour ago
$begingroup$
Could you please draw a schematic (by pressing Schematic button in editor in your original post) where the ammeter is located? I think you'd better divide the voltage by e.g. 10k pot and connect its branch with a 10G resistor to the ammeter.
$endgroup$
– Huisman
1 hour ago
$begingroup$
As drawn, it looks like the divider will be passing 150pA. There are definitely more things that can sneak up on you at that point to mess with your measurement. Maybe you don't need femto amp precision, but seeing what they do to ensure fA accuracy is probably a good step.
$endgroup$
– W5VO♦
59 mins ago
|
show 2 more comments
2
$begingroup$
Look up a 1993 ED article "What's All This Femtoampere Stuff, Anyhow?" by the late Robert Pease.
$endgroup$
– Spehro Pefhany
1 hour ago
3
$begingroup$
Why would you need such a circuit if I may ask? Any load attached will casue the same effect as the multimeter.
$endgroup$
– Huisman
1 hour ago
$begingroup$
@Huisman for trying to build ammeters that can go to very low current. I want very low current sources, then to try to measure them. If I'm dividing the voltage down first before passing through a 10G resistor (or higher) it's especially helpful to be able to measure that the actual voltage drop is what I expect it to be.
$endgroup$
– John Smith
1 hour ago
$begingroup$
Could you please draw a schematic (by pressing Schematic button in editor in your original post) where the ammeter is located? I think you'd better divide the voltage by e.g. 10k pot and connect its branch with a 10G resistor to the ammeter.
$endgroup$
– Huisman
1 hour ago
$begingroup$
As drawn, it looks like the divider will be passing 150pA. There are definitely more things that can sneak up on you at that point to mess with your measurement. Maybe you don't need femto amp precision, but seeing what they do to ensure fA accuracy is probably a good step.
$endgroup$
– W5VO♦
59 mins ago
2
2
$begingroup$
Look up a 1993 ED article "What's All This Femtoampere Stuff, Anyhow?" by the late Robert Pease.
$endgroup$
– Spehro Pefhany
1 hour ago
$begingroup$
Look up a 1993 ED article "What's All This Femtoampere Stuff, Anyhow?" by the late Robert Pease.
$endgroup$
– Spehro Pefhany
1 hour ago
3
3
$begingroup$
Why would you need such a circuit if I may ask? Any load attached will casue the same effect as the multimeter.
$endgroup$
– Huisman
1 hour ago
$begingroup$
Why would you need such a circuit if I may ask? Any load attached will casue the same effect as the multimeter.
$endgroup$
– Huisman
1 hour ago
$begingroup$
@Huisman for trying to build ammeters that can go to very low current. I want very low current sources, then to try to measure them. If I'm dividing the voltage down first before passing through a 10G resistor (or higher) it's especially helpful to be able to measure that the actual voltage drop is what I expect it to be.
$endgroup$
– John Smith
1 hour ago
$begingroup$
@Huisman for trying to build ammeters that can go to very low current. I want very low current sources, then to try to measure them. If I'm dividing the voltage down first before passing through a 10G resistor (or higher) it's especially helpful to be able to measure that the actual voltage drop is what I expect it to be.
$endgroup$
– John Smith
1 hour ago
$begingroup$
Could you please draw a schematic (by pressing Schematic button in editor in your original post) where the ammeter is located? I think you'd better divide the voltage by e.g. 10k pot and connect its branch with a 10G resistor to the ammeter.
$endgroup$
– Huisman
1 hour ago
$begingroup$
Could you please draw a schematic (by pressing Schematic button in editor in your original post) where the ammeter is located? I think you'd better divide the voltage by e.g. 10k pot and connect its branch with a 10G resistor to the ammeter.
$endgroup$
– Huisman
1 hour ago
$begingroup$
As drawn, it looks like the divider will be passing 150pA. There are definitely more things that can sneak up on you at that point to mess with your measurement. Maybe you don't need femto amp precision, but seeing what they do to ensure fA accuracy is probably a good step.
$endgroup$
– W5VO♦
59 mins ago
$begingroup$
As drawn, it looks like the divider will be passing 150pA. There are definitely more things that can sneak up on you at that point to mess with your measurement. Maybe you don't need femto amp precision, but seeing what they do to ensure fA accuracy is probably a good step.
$endgroup$
– W5VO♦
59 mins ago
|
show 2 more comments
3 Answers
3
active
oldest
votes
$begingroup$
Do what the ancients did ==== use a Wheatstone bridge. Like this
simulate this circuit – Schematic created using CircuitLab
Rotate the 10,000 ohm potentiometer for ZERO reading.
Then measure the pot voltage (and compensate for the DVM loading)
answered 1 hour ago
analogsystemsrfanalogsystemsrf
15.4k2721
$endgroup$
add a comment |
$begingroup$
sure, a voltage follower built with a FET op-amp that has extremely low input bias current.
https://www.mouser.co.uk/Semiconductors/Amplifier-ICs/Operational-Amplifiers-Op-Amps/_/N-4h00g?Rl=4h00gZgjdhpmZ1yvbz5oZ1yve6dbSGT
edited 2 hours ago
Dave Tweed♦
121k9151260
answered 2 hours ago
Peter GreenPeter Green
11.9k11939
$endgroup$
$begingroup$
Is it enough to use a single low input bias current follower (i.e. only use it for one of the multimeter probes with the 2nd directly probing the circuit under test) or would I need 2?
$endgroup$
– John Smith
48 mins ago
add a comment |
$begingroup$
If you want a tunable low current source, I suggest using something like the following circuit.
The potmeter 'adjusts' the voltage and so the current through the DIY ammeter.
In this circuit, the voltmeter hardly influences the voltage measurement and the DIY ammeter hardly influences the current measurement.
simulate this circuit – Schematic created using CircuitLab
answered 12 mins ago
HuismanHuisman
791111
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Do what the ancients did ==== use a Wheatstone bridge. Like this
simulate this circuit – Schematic created using CircuitLab
Rotate the 10,000 ohm potentiometer for ZERO reading.
Then measure the pot voltage (and compensate for the DVM loading)
answered 1 hour ago
analogsystemsrfanalogsystemsrf
15.4k2721
$endgroup$
add a comment |
$begingroup$
Do what the ancients did ==== use a Wheatstone bridge. Like this
simulate this circuit – Schematic created using CircuitLab
Rotate the 10,000 ohm potentiometer for ZERO reading.
Then measure the pot voltage (and compensate for the DVM loading)
answered 1 hour ago
analogsystemsrfanalogsystemsrf
15.4k2721
$endgroup$
add a comment |
$begingroup$
Do what the ancients did ==== use a Wheatstone bridge. Like this
simulate this circuit – Schematic created using CircuitLab
Rotate the 10,000 ohm potentiometer for ZERO reading.
Then measure the pot voltage (and compensate for the DVM loading)
answered 1 hour ago
analogsystemsrfanalogsystemsrf
15.4k2721
$endgroup$
Do what the ancients did ==== use a Wheatstone bridge. Like this
simulate this circuit – Schematic created using CircuitLab
Rotate the 10,000 ohm potentiometer for ZERO reading.
Then measure the pot voltage (and compensate for the DVM loading)
answered 1 hour ago
analogsystemsrfanalogsystemsrf
15.4k2721
answered 1 hour ago
analogsystemsrfanalogsystemsrf
15.4k2721
answered 1 hour ago
analogsystemsrfanalogsystemsrf
15.4k2721
answered 1 hour ago
analogsystemsrfanalogsystemsrf
15.4k2721
15.4k2721
add a comment |
add a comment |
$begingroup$
sure, a voltage follower built with a FET op-amp that has extremely low input bias current.
https://www.mouser.co.uk/Semiconductors/Amplifier-ICs/Operational-Amplifiers-Op-Amps/_/N-4h00g?Rl=4h00gZgjdhpmZ1yvbz5oZ1yve6dbSGT
edited 2 hours ago
Dave Tweed♦
121k9151260
answered 2 hours ago
Peter GreenPeter Green
11.9k11939
$endgroup$
$begingroup$
Is it enough to use a single low input bias current follower (i.e. only use it for one of the multimeter probes with the 2nd directly probing the circuit under test) or would I need 2?
$endgroup$
– John Smith
48 mins ago
add a comment |
$begingroup$
sure, a voltage follower built with a FET op-amp that has extremely low input bias current.
https://www.mouser.co.uk/Semiconductors/Amplifier-ICs/Operational-Amplifiers-Op-Amps/_/N-4h00g?Rl=4h00gZgjdhpmZ1yvbz5oZ1yve6dbSGT
edited 2 hours ago
Dave Tweed♦
121k9151260
answered 2 hours ago
Peter GreenPeter Green
11.9k11939
$endgroup$
$begingroup$
Is it enough to use a single low input bias current follower (i.e. only use it for one of the multimeter probes with the 2nd directly probing the circuit under test) or would I need 2?
$endgroup$
– John Smith
48 mins ago
add a comment |
$begingroup$
sure, a voltage follower built with a FET op-amp that has extremely low input bias current.
https://www.mouser.co.uk/Semiconductors/Amplifier-ICs/Operational-Amplifiers-Op-Amps/_/N-4h00g?Rl=4h00gZgjdhpmZ1yvbz5oZ1yve6dbSGT
edited 2 hours ago
Dave Tweed♦
121k9151260
answered 2 hours ago
Peter GreenPeter Green
11.9k11939
$endgroup$
sure, a voltage follower built with a FET op-amp that has extremely low input bias current.
https://www.mouser.co.uk/Semiconductors/Amplifier-ICs/Operational-Amplifiers-Op-Amps/_/N-4h00g?Rl=4h00gZgjdhpmZ1yvbz5oZ1yve6dbSGT
edited 2 hours ago
Dave Tweed♦
121k9151260
answered 2 hours ago
Peter GreenPeter Green
11.9k11939
edited 2 hours ago
Dave Tweed♦
121k9151260
edited 2 hours ago
Dave Tweed♦
121k9151260
edited 2 hours ago
Dave Tweed♦
121k9151260
121k9151260
answered 2 hours ago
Peter GreenPeter Green
11.9k11939
answered 2 hours ago
Peter GreenPeter Green
11.9k11939
answered 2 hours ago
Peter GreenPeter Green
11.9k11939
11.9k11939
$begingroup$
Is it enough to use a single low input bias current follower (i.e. only use it for one of the multimeter probes with the 2nd directly probing the circuit under test) or would I need 2?
$endgroup$
– John Smith
48 mins ago
add a comment |
$begingroup$
Is it enough to use a single low input bias current follower (i.e. only use it for one of the multimeter probes with the 2nd directly probing the circuit under test) or would I need 2?
$endgroup$
– John Smith
48 mins ago
$begingroup$
Is it enough to use a single low input bias current follower (i.e. only use it for one of the multimeter probes with the 2nd directly probing the circuit under test) or would I need 2?
$endgroup$
– John Smith
48 mins ago
$begingroup$
Is it enough to use a single low input bias current follower (i.e. only use it for one of the multimeter probes with the 2nd directly probing the circuit under test) or would I need 2?
$endgroup$
– John Smith
48 mins ago
add a comment |
$begingroup$
If you want a tunable low current source, I suggest using something like the following circuit.
The potmeter 'adjusts' the voltage and so the current through the DIY ammeter.
In this circuit, the voltmeter hardly influences the voltage measurement and the DIY ammeter hardly influences the current measurement.
simulate this circuit – Schematic created using CircuitLab
answered 12 mins ago
HuismanHuisman
791111
$endgroup$
add a comment |
$begingroup$
If you want a tunable low current source, I suggest using something like the following circuit.
The potmeter 'adjusts' the voltage and so the current through the DIY ammeter.
In this circuit, the voltmeter hardly influences the voltage measurement and the DIY ammeter hardly influences the current measurement.
simulate this circuit – Schematic created using CircuitLab
answered 12 mins ago
HuismanHuisman
791111
$endgroup$
add a comment |
$begingroup$
If you want a tunable low current source, I suggest using something like the following circuit.
The potmeter 'adjusts' the voltage and so the current through the DIY ammeter.
In this circuit, the voltmeter hardly influences the voltage measurement and the DIY ammeter hardly influences the current measurement.
simulate this circuit – Schematic created using CircuitLab
answered 12 mins ago
HuismanHuisman
791111
$endgroup$
If you want a tunable low current source, I suggest using something like the following circuit.
The potmeter 'adjusts' the voltage and so the current through the DIY ammeter.
In this circuit, the voltmeter hardly influences the voltage measurement and the DIY ammeter hardly influences the current measurement.
simulate this circuit – Schematic created using CircuitLab
answered 12 mins ago
HuismanHuisman
791111
answered 12 mins ago
HuismanHuisman
791111
answered 12 mins ago
HuismanHuisman
791111
answered 12 mins ago
HuismanHuisman
791111
791111
add a comment |
add a comment |
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2
$begingroup$
Look up a 1993 ED article "What's All This Femtoampere Stuff, Anyhow?" by the late Robert Pease.
$endgroup$
– Spehro Pefhany
1 hour ago
3
$begingroup$
Why would you need such a circuit if I may ask? Any load attached will casue the same effect as the multimeter.
$endgroup$
– Huisman
1 hour ago
$begingroup$
@Huisman for trying to build ammeters that can go to very low current. I want very low current sources, then to try to measure them. If I'm dividing the voltage down first before passing through a 10G resistor (or higher) it's especially helpful to be able to measure that the actual voltage drop is what I expect it to be.
$endgroup$
– John Smith
1 hour ago
$begingroup$
Could you please draw a schematic (by pressing Schematic button in editor in your original post) where the ammeter is located? I think you'd better divide the voltage by e.g. 10k pot and connect its branch with a 10G resistor to the ammeter.
$endgroup$
– Huisman
1 hour ago
$begingroup$
As drawn, it looks like the divider will be passing 150pA. There are definitely more things that can sneak up on you at that point to mess with your measurement. Maybe you don't need femto amp precision, but seeing what they do to ensure fA accuracy is probably a good step.
$endgroup$
– W5VO♦
59 mins ago