# Measuring Current via voltage measurement

Discussion in 'Electronics Homework Help' started by Stoneww, Oct 20, 2019.

1. ### Stoneww

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Apr 18, 2017
Hey I am doing a project using the TCRT5000 which is a phototranistor with an IR LED. I need to measure the current through the transistor. We get told to add a resistor and measure the voltage accross it using a voltmeter (and then I= V/R) instead of measuring current directly using a ammeter.

So I am curious, why it is better to use the voltage approach over the direct current measurement?

Also how would I determine what the optimal value of that resistor should be?

Thank you,

A lost, confused student

Last edited: Oct 20, 2019
2. ### Harald KappModeratorModerator

9,878
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Nov 17, 2011
Who said this is better? It is just another method. If you look at the current measuring circuit of a digital multimeter, it i sexactly what happens there: curtrent passing through a shint resistor develops a voltage across the resistor. Said voltage is then measured.
Optimal in respect to which parameter? Optimization always requires at least one or more parameters that need to come to a target value as close as possible. You could optimize e.g. for
- cost
- size
- power dissipation (ususally minimize)
- measurement accuracy

Thee parameters more often than not influence one another in contradictory ways. Take for example accuracy. At a first glance it seems logical to use a resistor with a very accurate value (say 0.1 % tolerance or better) to get a precise measurement result. But: high precision resistors are more expensive than less precise ones. So you'll have to trade cost vs. accuracy. Have alook at your voltmeter. How accurate is it? If your meter has an accuracy of say 5 %, it makes no sense to use a 0.1 % resistor as the resistor's accuracy will be completely masked by the "inacuracy" of the meter.
Stating it on a high level, you need an optimization function which will allow you to find the best fitting parameters for an optimal solution. There's lots of fancy math behind this.
More simply: state which property (or parameter) is most important for you, then adjust the other parameters to optimite this property.

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3. ### Frankchie

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5
Nov 14, 2017
Many times the optimal resistor is one that has the least affect on circuit operation. So without extensive analysis, I would choose the lowest value possible that still produces a voltage large enough to be measured by your voltmeter.

Of course, if you have some idea of what the circuit current range might be and the characteristics of your voltmeter, that helps to pick a starting value for the resistor. In any case some "trial and error" may be necessary. That is, start with a very low value and work your way up if your voltmeter doesn't give a satisfactory reading. IOW, if you are using an analog voltmeter and the needle barely moves, that would not be satisfactory.

Hopefully as you add/increase this resistor you have a way to determine if your circuit is being adversely affected.

If my answer is not clear, let me know and I'll try to clarify.
Frank

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4. ### Kabelsalat

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22
Jul 5, 2011
It gives you control over the total resistance. You're also free to meassure using other equipment than an amperemeter, such as an osciloscope. Also an amperemeter will cut the connection if you decide to measure something else on the table, but with an serie resistor you can choose to measure whenever you want, and also in the meantime use the multimeter for something else.

It depend on the circuit, and if power loss is an important factor.

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5. ### Hunter64

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12
Nov 20, 2018
Check the TCRT5000 datasheet and look for the maximum collector-current.

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6. ### Frankchie

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5
Nov 14, 2017
Hunter64, maybe I missed something, but it sounds like this is a working circuit and the goal is to measure the collector current. Therefore, adding the lowest possible resistor would be desirable to have the minimum impact to the collector current.

Frank

7. ### Harald KappModeratorModerator

9,878
2,092
Nov 17, 2011
I think you did.
The phototransistor in the TCRT5000 requires a resistor eithe rin the collector path or the emitter path to operate. See e.g. here for an explanation. You can directly use this resistor and measure the voltage drop.
Then show us the circuit diagram. Lacking information it may as well be us who miss something?

8. ### Frankchie

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5
Nov 14, 2017
The problem stated a requirement to "add" a resistor and measure the voltage across it.

Also the Op asked "what would be the optimum value of the resistor"..

I think I provided an appropriate answer that stayed in context with the Op's post.

9. ### Harald KappModeratorModerator

9,878
2,092
Nov 17, 2011
That is indeed what the op stated, but as there should already be one, the optimum resistor to add is 0 Ω and use the existing resistor.

Your introductory sentence in post #3 is no doubt absolutely correct. But lacking information about the circuit my answer was intended to view the problem from another perspective.
Unfortunately @Stoneww hasn't replied to this thread.

10. ### Frankchie

91
5
Nov 14, 2017
Harald,
Yup, the optimum resistor is zero. Therefore, outside the classroom using an existing resistor would be the preferred solution.
Good point.
Frank

11. ### bertusModerator

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179
Nov 8, 2019
Hello,

Did you read the datasheet of the device?
The pages 2,3 and 4 show a lot of information about the behaviour of the device.
How much current is used for the emmitor (led) ?
What is the supply voltage for the transistor ?

Bertus

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12. ### davennModerator

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1,850
Sep 5, 2009

the OP never came back