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Many multimeters, when used to measure microamps, don't have an overly-high burden voltage when passing less than a milliamp, but the they can drop over a volt (and even blow a fuse) if a device has a momentary current draw of 100mA or more.

If one has a device under test which will usually draw a few microamps, but may occasionally draw half an amp or more, and if one is only interested in measuring the current during those times when it's below 2mA (e.g. if one were testing a lawn-sprinkler controller, which will occasionally need a lot of current to open or close the valve, but should otherwise only be drawing a few microamps most of the time) would there be any problem connecting a couple of reverse-parallel diodes in parallel with the inputs to the meter? I would expect there must be at least some type of diode that would pass only a very tiny amount of current at voltages below 0.1 volt, but which would nonetheless be able to handle an amp or more with roughly 0.7-volt drop. Would there be any problem with attaching such a pair of diodes in parallel with the meter? How much would doing so affect accuracy? What sort of diodes would be best?

PS--Alternatively, if one wants to get precision real-time current measurements at an adjustable regulated voltage, and supplying a higher input voltage wouldn't be a problem, are there any stock devices that would supply a regulated output voltage while also supplying a ground-referenced signal proportional to current? Assume as requirements: (1) device must supply 500mA with minimal drop, regardless of measurement scale; (2) current will always be flowing in one direction, (3) 100Khz bandwidth is desired to facilitate measuring the duration of high-current-draw periods; (4) some ripple on the output would be acceptable, to the extent that it does not interact with the input capacitance of the device under test so as to impede measurement accuracy while the current is within range, or within 100ms or so of it coming into range.

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    \$\begingroup\$ you should get a uCurrent from @eevblog... eevblog.com/projects/ucurrent or at least read about it :-) \$\endgroup\$ – vicatcu Oct 24 '11 at 16:26
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    \$\begingroup\$ Sounds interesting, though it doesn't sound like it solves the problem with the device having occasional peak currents which are much higher than those one wants to measure. On the 1,000uA scale (which would be the one I'd probably use most) the series resistance is 10 ohms, which would drop a full volt at 100mA. \$\endgroup\$ – supercat Oct 24 '11 at 17:16
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  • Back to back diodes act as protection as long as meter can survive about 0.7V across its terminals/

  • Back to back silicon diodes will not affect readings substantially below about 0.3V across meter terminals.

  • Use of a voltmeter on a 200 mV range and a series sense resistor will often allow destruction proof current metering plus superior burden voltages compared to typical good quality ammeters. (eg 1 mV/ma with 0.1 mA resolution or 0.1 mV/mA with 1 mA resolution).

  • Suggestion provided for superior test method.


Reverse parallel diodes do potentially work, but it depends somewhat on your meter characteristics. Having a "poor" meter spec may make diodes work better than having a good spec. As a guide (see below) a meter with protection diodes should not be used above about 300 mV dop across the metetr (if that) as as diode conduction is approached current rises exponentially.

In the article cited by Vicatu they mention figures of 1.8 mV/mA for a Fluke 87v meter. And they quote 70 uV/mA for their solution on the 300 mA range.

The term "burden voltage' is just jargon for saying that the meter has resistance, so that when current flows through it there is a voltage drop. Just Ohms law V = I x R.

1.8 mV/mA translates to 1.8 ohms meter resistance.
70 uV/mA translates to 0.07 ohm.

An acceptably good alternative can be had with multimeters which have a 200 mV range. In a 3.5 digit meter this gives 100 uV resolution in the LSD.

If you pass current through a 1 ohm resistor it drops 1 mV per mA. If you measure that voltage with a 200 mV meter range yu get a resolution of 0.1 mA and a maximum current of 200 mA. BUT if you then apply a high current you may fry the resistor BUT the meter, being on a volts range will survive.
If you use a 0.1 ohm resistor (= 0.1 mV/mA = 18 x better than the fluke meter) then at 1 mA you get 100 uV and the 200 mV range will resolve 1 mA. Note as aever that resolution and mV are not the same. This arrangement will now read up to 2 amps in 1 mA steps. Again, meter is safe against overload.

Using the 1 ohm resistor, you get 1 mV/mA so at say 2 mA you get 2 mV droo which is highly acceptable for testing almost anything. At that current you could often use 10 ohms for 10 mV / mA, 10 uA resolution (notionally), a drop of 100 mV at 10 mA and a full scale deflection of 20 mA on a 200 mV range. And you will not damage your meter when you attempt to draw an amp.

At 1 ohm resistance = 1 mV/mA to get 700 mV drop to activate a diode you need 700 mA. With 10 ohms series resistance you need 70 mA. With 0.1 ohm series resistance you need 7 amps. And in each case the meter will not be damaged with or without diodes.

Using the Fluke 87V at 1.8 mV/mA you need 700/1.8 =~ 390 mA to produce 700 mV for diodes to bite on. One would hope that a Fluke set 20 200 mA range with say 0.1 mA resolution (and maybe 0.01 mA) would survive a 2:1 overload.

The above article suggsts that other meters are far worse than the 87V for burden voltage. Which just means that they would be protected at even lower mAs of overload.

A problem is that a diode does not conduct sharply. The exponential knee will cause substantial current fklow at 500 mV and discernible flow at even 300 mV. So the effective mA range of a meter with diodes may be say half of 700 mV drop maximum.

I have a meter which is excellent for this.
It has a 10A range with auti ranging - so you get massively low shuntresistance combined with autoswitching from 10A to 200 mA ranges.The 200 mA range gives me a 100 uA resolution. Whether that is good enough is up to you.

You may b tempted to us Schottky diodes to reduce the forward voltage drop. Unfortunately these have very poor (_= high) reverse current specs and this gets worse or much worse as temperature goes up. So Schottky f=diodes will not usually be useful as protection diodes/

I have not seen the circuit of the micro-current adaptor but assume that it is essentially a suitable low offset voltage and low bias current amplifier - possibly just a single hi spec opamp. eg an 0.01 ohm resistor returns 10 uV per mA - amplifying the voltage across this with 100x gain gives 200 mA measurement on a 200 mV range with 0.1 mA resolution. Stepping upto a still acceptable 0.1 ohm sense resistor and still 100x gain gives 0.01 mA resolution on a 200 mV range.


Aha - they gave the circuit ...

enter image description here

Yes, just a single amplifier. The LMV321 is just used to set the 1/2 supply virtual ground point. The MAX4239ASA opamp datasheet here.

$2.47/1 at Digikey. A bargain. They say -

  • The MAX4238/MAX4239 are low-noise, low-drift, ultrahigh precision amplifiers that offer near-zero DC offset and drift through the use of patented autocorrelating zeroing techniques. This method constantly measures and compensates the input offset, eliminating drift over time and temperature and the effect of 1/f noise.

    Both devices feature rail-to-rail outputs, operate from a single 2.7V to 5.5V supply, and consume only 600µA. An activelow shutdown mode decreases supply current to 0.1µA.

    The MAX4238 is unity-gain stable with a gain-bandwidth product of 1MHz, whi le the decompensated MAX4239 is stable with AV ≥ 10V/V and a GBWP of 6.5MHz. The MAX4238/MAX4239 are available in 8-pin narrow SO, 6-pin TDFN and SOT23 package


Current through a forward biased PN junction at low Vf :

Current versus voltage across a PN junction can be approximated as exponentially increasing current with voltage (or logarithmically increasing voltage with current.) The underlying theory is "complex" (to put it mildly).
This website provides a superb explanation (as part of a full text book on "Princuples of semiconductor devices") in as much detail as you are liable to find anytwhere (and more than you may want) but in as accessible a form as such material can be.

A reasonably good summary is given by the chart below.
"n" is a diode ideality factor (often assumed to be 1) which varies between 1 & 2 in the different regions.

We are usually interested in the "high injectuon region" from about Vf = 0.5V to 0.8V. As a rough guide this indicates that at Vf = 0.3V current will be about I_0.6V/10,000 and at Vf = 0.1V current ~= I_0.6V/1,000,000 . ie you can expect minimal effects on readings at Vf = 0.3V and utterly minimal effects at Vf=0.1V at the sort of accuracies liable to be of interest.

enter image description here


I use the following method of providing a testing power supply with good success. I've shown the concept using an LM317 but I actually use a P Channel MOSFET, opamp and voltage reference in place of the LM317. The effect is much the same EXCEPT that the LM317 will add it's operating current to the measured current whereas my actual arrangement adds essentially nothing. I can draw up my actual arrangement and add it if there is any interest.

  • Provide an adjustable power supply

  • On the "upstream" side of the voltage regulated supply provide a series "current sense" resistor whose ** voltage drop** will be an indication of load current.

  • Measure voltage drop across the series resistor to determine current drain.

enter image description here - Provide a filter capacitor Ci at the input to the voltage regulator. Large rvalues will smooth the current meter response but improve the output stability under step load changes.

  • The current sense resistor can be eg 0.1 ohm or 1 ohm or 10 ohms or other.
    The value of this resistor is NOT reflected in the stability of the output under load variations BUT it does affect the required input voltage.

    Vin min = vout + Vregulator_dropout + Imax x Rsense.

Rsense = 1 ohm.
Vsense = 1mv per mA or V per amp (not surprisingly.) Power in R sense = I^2R = 1 Watt at 1 amp, so easily done.
200 mV meter will easily resolve 0.1 mA.

Rsense = 10 ohm.
Vsense = 10 mV/mA. Power in R sense = I^2R = 10 Watt at 1 amp, so design needed.
Power = 0.1 Watt at 100 mA. 1 mA = 10 V.
10 uA = 100 uV so this will resolve 10 uA on a 2 mv, 3.5 digit meter. I max measured on 200 mV range = 20 mA BUT meter will not be damaged by 100 mA or 1A overloading.

Use of an autoranging meter allows currents from say 10's of uV to say 1A to be measured.

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200 mV meter will easily resolve 0.1 mA.

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  • \$\begingroup\$ I don't think reverse-leakage would be a problem if it doesn't happen at a lower voltage than forward conduction. The bigger issue would be having the diodes not conduct any current while the meter is within range. The normal specs I see for diodes, however, don't specify how much current they'll leak at e.g. 100mV, and I have no idea how safely one may assume that e.g. forward current at 10mV will be one microamp or less, and forward current at 100mV will be ten microamps or less. \$\endgroup\$ – supercat Oct 24 '11 at 18:12
  • \$\begingroup\$ BTW, I think "burden voltage" includes not just meter resistance, but also any aspect of the meter that would cause a non-linear voltage drop. For example, a mechanical-movement meter used for measuring current on a 120VAC circuit might use a bridge rectifier on its input, which would add 1.4 volts of burden voltage beyond that imposed by meter resistance. \$\endgroup\$ – supercat Oct 24 '11 at 18:14
  • \$\begingroup\$ I've been thinking about some variations on that circuit, and wondering the best way to mirror the current as a voltage relative to the output ground, so as not to require a differential scope probe to view current draw relative to some other signal. \$\endgroup\$ – supercat Oct 25 '11 at 14:55
  • \$\begingroup\$ @supercat - Almost any differential amplifier will do it - probably including something as simple as the one opamp classic differential amplifier Connect inputs across RS and output is griund referenced, Bandwidth depends on implementation. \$\endgroup\$ – Russell McMahon Oct 25 '11 at 15:02
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In National AN-240, Bob Pease uses diodes in a similar fashion, to protect an op-amp input from big voltage swings:

enter image description here

But all the low-leakage planar diodes that I have found (the 1N457 is still in production) have measly current ratings like 200 mA.

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  • \$\begingroup\$ How much current will "ordinary" diodes pass at 100mV (would that be called "leakage" in the forward direction)? \$\endgroup\$ – supercat Oct 25 '11 at 14:41
  • \$\begingroup\$ (noticing the edit to Russel's answer) Mr. Russel makes it sound as though the 100mV current would be minimal; I wonder if there are any factors such as contamination that would cause such current to be significant on parts which otherwise pass QC. \$\endgroup\$ – supercat Oct 25 '11 at 14:57
  • \$\begingroup\$ I don't know; Schottky diodes are notorious for leakage current at high temperatures, but you are planning to use standard PN rectifiers. Ebers-Moll says the current will be minimal but rectifiers just aren't spec'ed at low currents like that. I think you should choose a pair of 1N400x and measure the forward and reverse leakage at 100mV. If it is acceptable (it probably will be) use that pair of diode without worry. \$\endgroup\$ – markrages Oct 25 '11 at 15:03
  • \$\begingroup\$ If a forward-biased diode conducts current when there's 100mV across it, is that called "leakage"? For standard PN rectifiers, it sounds like conduction at such voltages would be trivial. How about Schottky diodes? Would there be any particular difficulty with meter manufacturers including diodes for protection? It would seem that if one had a pair of diodes, one could get buy with a fairly high-current fuse or PTC, whose voltage drops at "normal" currents would be pretty minimal. \$\endgroup\$ – supercat Oct 26 '11 at 14:58
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I'll add the following to partially answer as an answer in its own right. This is an extended version of the bottom of my other answer. It's useful in reduced form there and addresses a subquestion here in its more detailed form.

A key point here is that measurement is that for the original question which was about avoiding meter damage and about not having high series resistance, here a voltmeter is used nd is innune to damage, and the series resistance is present but is not "seen" by the circuit under test as it is electronically isolated.


To the questions:

  • ... if one wants to get precision real-time current measurements at an adjustable regulated voltage, and supplying a higher input voltage wouldn't be a problem, are there any stock devices that would supply a regulated output voltage while also supplying a ground-referenced signal proportional to current?

    Assume as requirements:

    (1) device must supply 500mA with minimal drop, regardless of measurement scale;

    (2) current will always be flowing in one direction,

    (3) 100Khz bandwidth is desired to facilitate measuring the duration of high-current-draw periods

    (4) some ripple on the output would be acceptable, to the extent that it does not interact with the input capacitance of the device under test so as to impede measurement accuracy while the current is within range, or within 100ms or so of it coming into range.


Using the MOSFET and opamp method below, the above points are addressed as:

  • (1) Effectively zero output drop with any current. Response to load setp change depends on control loop.

    (2) Accepts unidirectional load current only - OK

    (3) Bandwidth depends on opamp / FET response.
    Capacitor Ci will need to be removed.

    (4) No fundamental reason for ripple. We can add some if you like ;-).

As shown the current signal is NOT ground referenced. It can be made so if desired by circuit rearrangement BUT having got this far, a simple 1:1 differential amp (or with gain as required) will translate Vrs down to ground.


I use the following method of providing a testing power supply with good success. I've shown the concept using an LM317 but I actually use a P Channel MOSFET, opamp and voltage reference in place of the LM317. The effect is much the same EXCEPT that the LM317 will add it's operating current to the measured current whereas my actual arrangement adds essentially nothing. I can draw up my actual arrangement and add it if there is any interest.

  • Provide an adjustable power supply

  • On the "upstream" side of the voltage regulated supply provide a series "current sense" resistor whose ** voltage drop** will be an indication of load current.

  • Measure voltage drop across the series resistor to determine current drain.

enter image description here - Provide a filter capacitor Ci at the input to the voltage regulator. Large rvalues will smooth the current meter response but improve the output stability under step load changes.

  • The current sense resistor can be eg 0.1 ohm or 1 ohm or 10 ohms or other.
    The value of this resistor is NOT reflected in the stability of the output under load variations BUT it does affect the required input voltage.

    Vin min = vout + Vregulator_dropout + Imax x Rsense.

Rsense = 1 ohm.
Vsense = 1mv per mA or V per amp (not surprisingly.) Power in R sense = I^2R = 1 Watt at 1 amp, so easily done.
200 mV meter will easily resolve 0.1 mA.

Rsense = 10 ohm.
Vsense = 10 mV/mA. Power in R sense = I^2R = 10 Watt at 1 amp, so design needed.
Power = 0.1 Watt at 100 mA. 1 mA = 10 V.
10 uA = 100 uV so this will resolve 10 uA on a 2 mv, 3.5 digit meter. I max measured on 200 mV range = 20 mA BUT meter will not be damaged by 100 mA or 1A overloading.

Use of an autoranging meter allows currents from say 10's of uV to say 1A to be measured.

.
200 mV meter will easily resolve 0.1 mA.

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