Trjtdtk

Electoral considerations aside, what are potential benefits, for the US, of policy changes proposed by the tweet recognizing Golan annexation?

What are some good ways to treat frozen vegetables such that they behave like fresh vegetables when stir frying them?

Hero deduces identity of a killer

What exact color does ozone gas have?

What is the highest possible scrabble score for placing a single tile

Yosemite Fire Rings - What to Expect?

15% tax on $7.5k earnings. Is that right?

How do apertures which seem too large to physically fit work?

How to cover method return statement in Apex Class?

How does a computer interpret real numbers?

What is going on with 'gets(stdin)' on the site coderbyte?

When were female captains banned from Starfleet?

How do you respond to a colleague from another team when they're wrongly expecting that you'll help them?

How much character growth crosses the line into breaking the character

Why is the "ls" command showing permissions of files in a FAT32 partition?

What should you do when eye contact makes your subordinate uncomfortable?

Why is this estimator biased?

Mimic lecturing on blackboard, facing audience

Fear of getting stuck on one programming language / technology that is not used in my country

Non-trope happy ending?

PTIJ: Haman's bad computer

What should you do if you miss a job interview (deliberately)?

Can I still be respawned if I die by falling off the map?

Quoting Keynes in a lecture

Determine voltage drop over 10G resistors with cheap multimeter

How to measure Voltage & Current with a single multimeter, Simultaneously?Why does a multimeter put more voltage to measure a smaller resistance?Multimeter input impedance and its effect on the measurement of charged capacitor's voltage?How measure the voltage over a large resistance?LED Voltage Drop ConfusionCorrect way to choose resistors for loadCan I measure the relative output gain of a power amplifier with a multimeter?How to determine accuracy of multimeter?Is it really problem if 0,3 voltage more than required applied in digital multimeter?Measuring a small resistance, ~0.001 ohm

17

0

$begingroup$

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

share|improve this question

edited Mar 18 at 14:41

John Smith

asked Mar 18 at 13:42

John SmithJohn Smith

1146

$endgroup$

• 
  
  
  

  
  7
  

  
  
  
  
  
  

  
  $begingroup$
  Look up a 1993 ED article "What's All This Femtoampere Stuff, Anyhow?" by the late Robert Pease.
  $endgroup$
  – Spehro Pefhany
  Mar 18 at 14:00
  

  
  
  
  


• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  $begingroup$
  Why would you need such a circuit if I may ask? Any load attached will casue the same effect as the multimeter.
  $endgroup$
  – Huisman
  Mar 18 at 14:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  Mar 18 at 14:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 18 at 14:35
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  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♦
  Mar 18 at 14:57
  

  
  
  
  

 | 
show 2 more comments

17

0

$begingroup$

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

share|improve this question

edited Mar 18 at 14:41

John Smith

asked Mar 18 at 13:42

John SmithJohn Smith

1146

$endgroup$

• 
  
  
  

  
  7
  

  
  
  
  
  
  

  
  $begingroup$
  Look up a 1993 ED article "What's All This Femtoampere Stuff, Anyhow?" by the late Robert Pease.
  $endgroup$
  – Spehro Pefhany
  Mar 18 at 14:00
  

  
  
  
  


• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  $begingroup$
  Why would you need such a circuit if I may ask? Any load attached will casue the same effect as the multimeter.
  $endgroup$
  – Huisman
  Mar 18 at 14:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  Mar 18 at 14:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 18 at 14:35
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  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♦
  Mar 18 at 14:57
  

  
  
  
  

 | 
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

share|improve this question

edited Mar 18 at 14:41

John Smith

asked Mar 18 at 13:42

John SmithJohn Smith

1146

$endgroup$

share|improve this question

edited Mar 18 at 14:41

John Smith

asked Mar 18 at 13:42

John SmithJohn Smith

1146

share|improve this question

edited Mar 18 at 14:41

John Smith

asked Mar 18 at 13:42

John SmithJohn Smith

1146

asked Mar 18 at 13:42

John SmithJohn Smith

1146

asked Mar 18 at 13:42

John SmithJohn Smith

1146

3 Answers
3

active

oldest

votes

26

$begingroup$

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)

share|improve this answer

answered Mar 18 at 14:30

analogsystemsrfanalogsystemsrf

15.5k2822

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Clever answer, so +1. To calibrate, replacing R3 with another set of 10G resistors should allow the DVM to be set to zero.
  $endgroup$
  – Sparky256
  Mar 18 at 18:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Imbalance in impingment of external Efields, or Hfields, on the DVM leads, may be the residual error.
  $endgroup$
  – analogsystemsrf
  Mar 20 at 2:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  In this case it is wise to stay away from the meter wires, which should be as short as possible. A metal cage would be even better. At 10G ohm, you have a very sensitive mass detector. Any charged body close by will affect the readings.
  $endgroup$
  – Sparky256
  Mar 20 at 3:07
  

  
  
  
  

add a comment | 

2

$begingroup$

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

share|improve this answer

edited Mar 18 at 13:54

Dave Tweed♦

121k9152263

answered Mar 18 at 13:53

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
  Mar 18 at 15:08
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  CMOS op-amps with fempto-amp inputs are ideal for these type of devices.
  $endgroup$
  – Sparky256
  Mar 18 at 18:22
  

  
  
  
  

add a comment | 

0

$begingroup$

If you can get capacitors with zero leakage **, you can hang one across each resistor. Since you are working with DC, give the circuit a few weeks to stabilize, then measure the PEAK voltage. The capacitance required would be such that the time constant RC is several seconds, where R is the load resistance of your multimeter.

** I do not know of a source for such components. If you get one, you better not touch or breathe on it. :-)

share|improve this answer

answered yesterday

richard1941richard1941

34215

$endgroup$

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["\$", "\$"]]);
);
);
, "mathjax-editing");

StackExchange.ifUsing("editor", function ()
return StackExchange.using("schematics", function ()
StackExchange.schematics.init();
);
, "cicuitlab");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "135"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2felectronics.stackexchange.com%2fquestions%2f427794%2fdetermine-voltage-drop-over-10g-resistors-with-cheap-multimeter%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

26

$begingroup$

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)

share|improve this answer

answered Mar 18 at 14:30

analogsystemsrfanalogsystemsrf

15.5k2822

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Clever answer, so +1. To calibrate, replacing R3 with another set of 10G resistors should allow the DVM to be set to zero.
  $endgroup$
  – Sparky256
  Mar 18 at 18:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Imbalance in impingment of external Efields, or Hfields, on the DVM leads, may be the residual error.
  $endgroup$
  – analogsystemsrf
  Mar 20 at 2:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  In this case it is wise to stay away from the meter wires, which should be as short as possible. A metal cage would be even better. At 10G ohm, you have a very sensitive mass detector. Any charged body close by will affect the readings.
  $endgroup$
  – Sparky256
  Mar 20 at 3:07
  

  
  
  
  

add a comment | 

26

$begingroup$

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)

share|improve this answer

answered Mar 18 at 14:30

analogsystemsrfanalogsystemsrf

15.5k2822

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Clever answer, so +1. To calibrate, replacing R3 with another set of 10G resistors should allow the DVM to be set to zero.
  $endgroup$
  – Sparky256
  Mar 18 at 18:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Imbalance in impingment of external Efields, or Hfields, on the DVM leads, may be the residual error.
  $endgroup$
  – analogsystemsrf
  Mar 20 at 2:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  In this case it is wise to stay away from the meter wires, which should be as short as possible. A metal cage would be even better. At 10G ohm, you have a very sensitive mass detector. Any charged body close by will affect the readings.
  $endgroup$
  – Sparky256
  Mar 20 at 3:07
  

  
  
  
  

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)

share|improve this answer

answered Mar 18 at 14:30

analogsystemsrfanalogsystemsrf

15.5k2822

$endgroup$

share|improve this answer

answered Mar 18 at 14:30

analogsystemsrfanalogsystemsrf

15.5k2822

answered Mar 18 at 14:30

analogsystemsrfanalogsystemsrf

15.5k2822

answered Mar 18 at 14:30

analogsystemsrfanalogsystemsrf

15.5k2822

2

$begingroup$

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

share|improve this answer

edited Mar 18 at 13:54

Dave Tweed♦

121k9152263

answered Mar 18 at 13:53

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
  Mar 18 at 15:08
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  CMOS op-amps with fempto-amp inputs are ideal for these type of devices.
  $endgroup$
  – Sparky256
  Mar 18 at 18:22
  

  
  
  
  

add a comment | 

2

$begingroup$

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

share|improve this answer

edited Mar 18 at 13:54

Dave Tweed♦

121k9152263

answered Mar 18 at 13:53

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
  Mar 18 at 15:08
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  CMOS op-amps with fempto-amp inputs are ideal for these type of devices.
  $endgroup$
  – Sparky256
  Mar 18 at 18:22
  

  
  
  
  

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

share|improve this answer

edited Mar 18 at 13:54

Dave Tweed♦

121k9152263

answered Mar 18 at 13:53

Peter GreenPeter Green

11.9k11939

$endgroup$

share|improve this answer

edited Mar 18 at 13:54

Dave Tweed♦

121k9152263

answered Mar 18 at 13:53

Peter GreenPeter Green

11.9k11939

edited Mar 18 at 13:54

Dave Tweed♦

121k9152263

edited Mar 18 at 13:54

Dave Tweed♦

121k9152263

answered Mar 18 at 13:53

Peter GreenPeter Green

11.9k11939

answered Mar 18 at 13:53

Peter GreenPeter Green

11.9k11939

0

$begingroup$

If you can get capacitors with zero leakage **, you can hang one across each resistor. Since you are working with DC, give the circuit a few weeks to stabilize, then measure the PEAK voltage. The capacitance required would be such that the time constant RC is several seconds, where R is the load resistance of your multimeter.

** I do not know of a source for such components. If you get one, you better not touch or breathe on it. :-)

share|improve this answer

answered yesterday

richard1941richard1941

34215

$endgroup$

add a comment | 

0

$begingroup$

If you can get capacitors with zero leakage **, you can hang one across each resistor. Since you are working with DC, give the circuit a few weeks to stabilize, then measure the PEAK voltage. The capacitance required would be such that the time constant RC is several seconds, where R is the load resistance of your multimeter.

** I do not know of a source for such components. If you get one, you better not touch or breathe on it. :-)

share|improve this answer

answered yesterday

richard1941richard1941

34215

$endgroup$

add a comment | 

$begingroup$

If you can get capacitors with zero leakage **, you can hang one across each resistor. Since you are working with DC, give the circuit a few weeks to stabilize, then measure the PEAK voltage. The capacitance required would be such that the time constant RC is several seconds, where R is the load resistance of your multimeter.

** I do not know of a source for such components. If you get one, you better not touch or breathe on it. :-)

share|improve this answer

answered yesterday

richard1941richard1941

34215

$endgroup$

share|improve this answer

answered yesterday

richard1941richard1941

34215

answered yesterday

richard1941richard1941

34215

answered yesterday

richard1941richard1941

34215