Split current over Wire and Multimeter?

Thread Starter

Daan Lageschaar

Joined Jul 8, 2019
36
Hi,

so I got this multimeater with a built in oscilloscope (or the other way around, depends on how you look at it)
I want to put it in series with my amp and speakers but i'm afraid that it will supply more current than the multimeter can handle.
Would putting the multimeter in parallel with a seperate wire lessen the current through my multimeter?

That's how it should work, right?

Thanks in advance! :D
-Daan
circuit.PNG
 

ebeowulf17

Joined Aug 12, 2014
3,274
How much current can your meter handle? How much current do you expect your amp to deliver?

Could you just turn down the volume to a very low level, start measuring, and slowly increase volume to see how much current the speaker draws at various volume levels? Maybe you don't need any limiting at all.
 

Thread Starter

Daan Lageschaar

Joined Jul 8, 2019
36
How much current can your meter handle? How much current do you expect your amp to deliver?

Could you just turn down the volume to a very low level, start measuring, and slowly increase volume to see how much current the speaker draws at various volume levels? Maybe you don't need any limiting at all.
Well that's the weird thing, I can't find it for measuring AC volts. I thought it was 20 amps but that's for measuring amps

I've attached the manual as a file
 

Attachments

ebeowulf17

Joined Aug 12, 2014
3,274
Well that's the weird thing, I can't find it for measuring AC volts. I thought it was 20 amps but that's for measuring amps

I've attached the manual as a file
If it's volts that you want to measure, then you shouldn't be putting the meter in series with the load in the first place. In that case you'd want the meter in parallel with the load.
 

Reloadron

Joined Jan 15, 2015
5,646
Per your drawing in post #1 what you label as "wire" will in fact have a small, very small, voltage drop across it. There will be a very small or better put very low resistance. The voltage drop across that low resistance will be proportional to the current. Your meter would be set to measure a very low AC voltage. One problem is unless the signal is constant amplitude and frequency the power to the speaker is constantly changing and any meter used not only must have enough sensitivity but also have enough frequency range to cover the frequencies sent to the speaker.

Measuring current your meter does have a 20 Amp range so it has an internal shunt. Note how the 20 Amp range is all by itself. If you were to look inside you would find a straight piece of wire just like what you drew. The meter measures the voltage drop across the known resistance of that shunt wire. With a known shunt resistance and measured voltage it's only a matter of doing the math which the meter does. The meter would be placed in series with the speaker. Problem now is you need to know the True RMS current and voltage at any point in time of the signal to derive the actual power.

Ron
 

Thread Starter

Daan Lageschaar

Joined Jul 8, 2019
36
If it's volts that you want to measure, then you shouldn't be putting the meter in series with the load in the first place. In that case you'd want the meter in parallel with the load.
Yeah of course lol
Don't really know what I was thinking xD

But will that multimeter be able to measure the voltage if it only gets about 0.0000191 amps?
 
Last edited:

bertus

Joined Apr 5, 2008
20,766
Hello,

You can switch between AC and DC with the button next to the power button.
From page 5 of the posted manual:

AC-DC setting.png

Bertus
 

ebeowulf17

Joined Aug 12, 2014
3,274
Yeah of course lol
Don't really know what I was thinking xD

But will that multimeter be able to measure the voltage if it only gets about 0.0000191 amps?
Sure, why not? The meter operates totally differently in voltage measurement mode than in current measurement mode. When set to read voltage, the meter is configured to draw as little current as it can from the circuit without sacrificing too much accuracy. Typically this means using an input impedance of 10Meg.

Anyway, where did you get that number? Are you expecting some undisclosed portion of your circuit to limit the current available to the meter, or were you trying to calculate current through the meter based on its input impedance specs?

If it's the former, then you might possibly have problems. If you're working with really high impedance circuits, at some point the impedance of the meter can start affecting the measurement too much. But all indications until this last post would seem to indicate that you don't have anything dramatically limiting current in your circuit.

So, assuming that 0.0000191A figure isn't limited by your circuit, only by meter impedance, that's plenty of signal. Multiply that by 10Meg and you've got 191V to measure. Easy, peasy!
 

Thread Starter

Daan Lageschaar

Joined Jul 8, 2019
36
Sure, why not? The meter operates totally differently in voltage measurement mode than in current measurement mode. When set to read voltage, the meter is configured to draw as little current as it can from the circuit without sacrificing too much accuracy. Typically this means using an input impedance of 10Meg.

Anyway, where did you get that number? Are you expecting some undisclosed portion of your circuit to limit the current available to the meter, or were you trying to calculate current through the meter based on its input impedance specs?

If it's the former, then you might possibly have problems. If you're working with really high impedance circuits, at some point the impedance of the meter can start affecting the measurement too much. But all indications until this last post would seem to indicate that you don't have anything dramatically limiting current in your circuit.

So, assuming that 0.0000191A figure isn't limited by your circuit, only by meter impedance, that's plenty of signal. Multiply that by 10Meg and you've got 191V to measure. Easy, peasy!
Since I wanna measure my amplifiers output voltage, I have to put the Multimeter in series with my speaker (which is 4 ohms, 'speaker' B is the multimeter with a resistance of 10MOhms).
As you can see in the file I attached it would get about 0.00095... watts. I assumed my amp would output about 50 volts (might be more, might be less I don't really know, that's one of the reasons i'm measuring it)
0.00095.../50 ≈ 0.0000191A
So that's how I got to that

impedance.PNG

But I think I know enough now,
Thanks for clearing things up for me, really like this forum when it comes to how many replies you get :D
-Daan

PS: the website I used in the picture is https://www.speakerimpedance.co.uk/?act=three_series_parallel&page=calculator
 

ebeowulf17

Joined Aug 12, 2014
3,274
Since I wanna measure my amplifiers output voltage, I have to put the Multimeter in series with my speaker (which is 4 ohms, 'speaker' B is the multimeter with a resistance of 10MOhms).
As you can see in the file I attached it would get about 0.00095... watts. I assumed my amp would output about 50 volts (might be more, might be less I don't really know, that's one of the reasons i'm measuring it)
0.00095.../50 ≈ 0.0000191A
So that's how I got to that

View attachment 200176

But I think I know enough now,
Thanks for clearing things up for me, really like this forum when it comes to how many replies you get :D
-Daan

PS: the website I used in the picture is https://www.speakerimpedance.co.uk/?act=three_series_parallel&page=calculator
Yeah, no problem! A few more comments:

If "speaker B" in the drawing represents your meter, then you're not putting it in series with the other speaker, you're putting it in parallel with the other speaker. You've drawn it correctly - you're just using the wrong terminology to describe it.

Your online calculator has just given you an insight into the inner workings of meters. You can see that, even with 10Meg input impedance, adding the meter to the circuit changes the behavior of the rest of the circuit. If course, the change is tiny, much smaller than you could measure in real life, so it can be ignored. But if you play with the numbers, you'll see that if you used the same meter, but you were measuring voltage across a 1 Meg load, or if you kept the load the same but only had a meter with 10k or 100k impedance, suddenly the meter would have a much larger effect on circuit performance, and the very act of measuring anything would be disruptive and provide inaccurate readings. This is why meter input impedances are usually kept fairly high.
 

Thread Starter

Daan Lageschaar

Joined Jul 8, 2019
36
Yeah, no problem! A few more comments:

If "speaker B" in the drawing represents your meter, then you're not putting it in series with the other speaker, you're putting it in parallel with the other speaker. You've drawn it correctly - you're just using the wrong terminology to describe it.

Your online calculator has just given you an insight into the inner workings of meters. You can see that, even with 10Meg input impedance, adding the meter to the circuit changes the behavior of the rest of the circuit. If course, the change is tiny, much smaller than you could measure in real life, so it can be ignored. But if you play with the numbers, you'll see that if you used the same meter, but you were measuring voltage across a 1 Meg load, or if you kept the load the same but only had a meter with 10k or 100k impedance, suddenly the meter would have a much larger effect on circuit performance, and the very act of measuring anything would be disruptive and provide inaccurate readings. This is why meter input impedances are usually kept fairly high.
Erm, yeah, parallel, that's what I meant to say but I was really tired yesterday so I mixed some things up xD

Thanks for the extra explanation btw :)
 
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