# Multimeter questions

#### ApacheKid

Joined Jan 12, 2015
385
Hi,

I studied electronics and even wrote articles for electronics magazines in the UK in late 70s early 80s, but I never worked in the field professionally (ended up working in software).

So I'm rusty as hell and have some questions, how things have changed. I've just started setting up a new workshop at home and I'm kitting out piece by piece as I go shelving, cabinets, scopes, sig gens etc etc, will be repairing some old 1930s tube radios as well as putting some hew stuff together, I also dabble with STM32 devices from time to time.

The last multimeter I worked with for actual bench electronics was an Avo 8 - so you'll appreciate I have some catching up to do.

OK so I do have a small utility multimeter, a small portable inexpensive thing - Southwire - 10040N - I bought a few years ago as a utility for doing very basic stuff around the house, I fully appreciate this has severe limitations for serious electronic circuit development.

I was playing around last night and needed to use this meter to measure the DC current through an Infred LED, the (very simple) circuit was powered by a reasonable bench PSU but it's current readout is crude, very low resolution.

With the meter on the mA range it was say 1.987 mA, but when I switched the meter to the μA range it read like 1,654 μA. (and yes, it does have a brand new battery in it).

So I have no idea which of those readings is closer to reality, and I was doing some rudimentary calculations too, so having uncertainty about the current obviously impacts what I'm trying to do.

Anyway I realize I need a better meter and I like Siglent a lot (I have their 4 channel 200 MHz scope and one of their signal gens) so I looked at some of these.

The SDM3045 and the more expensive SDM3055 are in the right price range and clearly blow that utility meter out of the water.

But something else came up as I was fiddling around, I wanted to be able to measure the current through one part of a circuit and the voltage in some other part of the circuit, so I wondered if in this day and age there are multi channel bench meters, like we have multi channel scopes and so on.

To my surprise I'm not seeing any, it seems they don't exist (except as very expensive serious professional units - like 7 grand).

Is this really true? if so I wonder why? surely its easy to make a multi channel bench meter, we have 4 channel scopes after all.

It seems the only way people do this is to have two meters - very costly.

Thought?

#### SamR

Joined Mar 19, 2019
3,406

#### ApacheKid

Joined Jan 12, 2015
385
Or look at the used market. I picked up a nice Fluke 8050A bench meter for ~\$40 with a 200uA - 2000mA range.
Man, the Fluke 8050A does seem pretty decent, certainly for amateur fiddling around work.

Do they need calibrating? if so what do you do when that time comes?

#### ApacheKid

Joined Jan 12, 2015
385
Well I just bought two 45 dollar Fluke 8050A units from same vendor, ex-college units, apparently tested out OK.

#### wayneh

Joined Sep 9, 2010
17,152
With the meter on the mA range it was say 1.987 mA, but when I switched the meter to the μA range it read like 1,654 μA. (and yes, it does have a brand new battery in it).

So I have no idea which of those readings is closer to reality...
What makes you doubt either one? For one thing, I'd say they're basically the same. But the meter is part of the circuit and so when you change the setting, there's no reason to expect identical results.

#### ApacheKid

Joined Jan 12, 2015
385
What makes you doubt either one? For one thing, I'd say they're basically the same. But the meter is part of the circuit and so when you change the setting, there's no reason to expect identical results.
Are you being serious here?

#### MrChips

Joined Oct 2, 2009
23,519
Are you being serious here?
I believe he is. Every ammeter has a burden resistor across which the voltage is measured. When you switched meter range you switched the burden resistor and hence you changed the circuit.

#### ApacheKid

Joined Jan 12, 2015
385
I believe he is. Every ammeter has a burden resistor across which the voltage is measured. When you switched meter range you switched the burden resistor and hence you changed the circuit.
Yes I am aware of that but these are digital devices, so they could compensate the displayed value by adjusting for the expected deviation arising from that resistance.

Perhaps more sophisticated instruments do that...

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#### KeithWalker

Joined Jul 10, 2017
1,780
Are you being serious here?
When you measure DC current, you are putting the resistance of the test leads and meter shunt in series with your circuit. The shunt will be a different resistance on each current range. They will affect the amount of current flowing. You also need to look at the accuracy and resolution of the meter in both of the ranges you used. You will most likely find that the difference between the two readings is within the specification.
For hobby electronics you will find that you rarely need to make very accurate measurements. Mostly you make them to verify that a signal or level is there, or to compare it to another reading. Digital panel meter modules are available for ridiculously low prices at AliExpress. They are accurate enough for everyday measurements. Some even display volts and amps at the same time.They are fixed range but are quite useful for some measurements. I have three of them built into my protoboard variable power supplies for easy, quick setup.

#### SamR

Joined Mar 19, 2019
3,406
Do they need calibrating? if so what do you do when that time comes?
When I found the manual for it online I was quite pleased to find it had the entire calibration procedure in it.

#### KeithWalker

Joined Jul 10, 2017
1,780
Yes I am aware of that but these are digital devices, so they could compensate the displayed value by adjusting for the expected deviation arising from that resistance.

Perhaps more sophisticated instruments do that...
The meter can not compensate for the resistance of the shunt because the effect on the reading will depend on the amount of resistance in the circuit being measured. Better meters have lower shunt resistance so that the effect is less. When you make current measurements, use short, heavy gauge test leads to keep the resistance low..

#### MrChips

Joined Oct 2, 2009
23,519
I believe he is. Every ammeter has a burden resistor across which the voltage is measured. When you switched meter range you switched the burden resistor and hence you changed the circuit.
No. It is not just about the meter. You also have to take into the consideration the resistance of your circuit.

As an example, suppose your circuit resistance is 1Ω.
Suppose in one range the burden resistor is 0.1Ω and in the other range the resistance is 0.01Ω.
Can you see that 1.1Ω and 1.01Ω would indicate two different currents, differing by 9%?

#### MrChips

Joined Oct 2, 2009
23,519
Yes I am aware of that but these are digital devices, so they could compensate the displayed value by adjusting for the expected deviation arising from that resistance.

Perhaps more sophisticated instruments do that...
No. It is not just about the meter. You also have to take into the consideration the resistance of your circuit.

As an example, suppose your circuit resistance is 1Ω.
Suppose in one range the burden resistor is 0.1Ω and in the other range the resistance is 0.01Ω.
Can you see that 1.1Ω and 1.01Ω would indicate two different currents, differing by 9%?

#### ApacheKid

Joined Jan 12, 2015
385
Thanks.

If the burden resistor was electronically variable, its value being controlled by the unit's processor, then we could - in principle - compensate.

We can derive a set of simultaneous equations that express the circuit voltage and circuit resistance in terms of two sets of voltage measurements and solve for these, the ratio of circuit voltage to circuit resistance is then the true current, the current that would flow if the meter had zero resistance.

I don't know what actual values are used for shunt resistors in the devices but so long as we can set it to two values and measure the voltage across it for each value we can do the arithmetic and calculate the actual current, the current that would flow if the meter was a short circuit.

So we'd actually measure two successive voltages Va and Vb corresponding to the two shunt resistor values Ra and Rb, one could have two resistors in parallel and electronically disconnect one, or this could be an option, one presses a key to perform that second measurement and refine the displayed value.

I'm rusty so doing the algebra is taking longer than it should, but Rx (the resistance of the circuit as if meter were not in the way) looks to be:

Rx = RaRb (Vb + Va) / (RbVa + RaVb)

A similar exercise will give us Vx the voltage across the circuit, once we have these we can calculate Ix.

Of course doing that (altering the resistance dynamically) is interrupting the circuit behavior and that may not be desirable, but in cases where this is not a concern it could give us very accurate current readings, eliminating the effect of the shunt resistor entirely.

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#### MrChips

Joined Oct 2, 2009
23,519
Yes and no.
In order to make the compensation you need to take two measurements.
The meter does not know that you are making the same measurement on two different ranges.

#### ApacheKid

Joined Jan 12, 2015
385
Yes and no.
In order to make the compensation you need to take two measurements.
The meter does not know that you are making the same measurement on two different ranges.
Yes, but it would do this automatically not on different ranges but on any given range it would take two measurements each with a different value of shunt (e.g 0.1 and 0.05 ohms) you'd just see the displayed value but under the hood it would have taken two measurements perhaps 0.5 seconds apart of whatever.

#### MrChips

Joined Oct 2, 2009
23,519
You are really trying to push your case.
When you insert a different shunt resistor into the circuit you are inadvertently altering the behaviour of the circuit.
How do you know that the circuit resistance is constant and therefore obeys Ohm's Law?
Just for example, suppose you want to measure the current flowing through a diode?