I'm a newbie in electronics, so before delving into the high amp side of electronics, I found it better to get some advice first.

I need to measure current consumption of two 14V DC motors in parallel under several scenarios (no load, full load, max speed, etc). The motors are of the type found in R/C monster trucks (traxxas e-maxx).

My DMM only goes up to 10 AMP, and on the traxxas forums, I've heard that these motors may go as high as 20AMP. I don't want to burn my DMM (although I believe there is a fuse in there to protect it against overcurrent).

In first place, I will install a 10AMP fuse (could only find a 250V one, is there a problem with that?) between my multimeter and the motor, but it is not going to be enough, as it will allow to measure a few situations only.

Is there any other way to measure high currents?


Considering that I'm a newbie (but I have enough skills to program a PIC, and I even know how to distinguish a resistor from a capacitor :wink: ), how easy is to build a logger with a PIC that will store peak amperes up to 30AMP over time? Is there an IC that does that?


Cheers

Padu

 

use known low resistance in series with motors, measure voltage drop and apply Ohms Law.
that's exactly how your meter does it.
you can also ask for samples or buy chips from Allegro.
I got few ACS750 but there are ones that cover lower current range...

 

Humnn, I'm not sure how to measure it though. I know Ohms law, but how to apply it in this case? (I told you I were a newbie)

Let's say I connect a 10ohms resistor in series with the motors (I know that somewhere I'll need to calculate power ratings in order to choose the resistor, but seems pretty obvious a 1/4W won't do it).

Then I measure current between going through the resistor on a certain scenario (let's say motor has no load, minimum speed) and current is 1AMP. Am I correct in assuming that at this point that 1AMP is going through the motor too? What if the resistor wasn't there, what would be the current going through the motor?

Now scenario 2: same resistor but now motor has a big load on its output shaft and I'm driving it to full speed. How do I calculate the current flow now?


Thanks and sorry if my questions are too basic.

 

you would NOT/CANNOT measure 20A with a 10R resistor

why?
1) the power dissipated would be 20*20*10 4kW !!!!!!!
2) the voltage developed would be 20*10 = 200V !!!!!

you would use a resistor with an ohmage of about 0.1R or less
.You would put the resistor in the return path of the machine windings and use a differential amplifier configured OPAMP to measure the voltage across that resistor. That voltage is proportional to the current flowing through it (the constant of proportionality is the resistor value).

simple as that.

as to the

whether the resistor is there or not, the current flowing through the machine is 1A.

If hte load on the machine increases then the current drawn must increase, whether the sense resistor is there or not the current will still flow.


I think you need to take a step back and go over the basics. It is one thing jumping into PIC electroncis with next to no infomation, but when dealing with power the basics need to be such you dont even think abt them any more

Mr Kirchoff and MR Ohm should be 2nd nature to you

__________________
Nothing is impossible.
Once a problem is realised, the rest is just details

 

Ok, I've been reading a few articles online about series circuits, let me see if I got it straight:


Back to the my previous scenario - 10 ohms in series with the motor, battery has 14V:

First step: measure voltage drop 'X' and verbose known parameters
Let's suppose voltage drop is 4V (I'm away from my multimeter, so I have no way to come up with real values), then using ohms law we can calculate current across the resistor right?
We should also know that if voltage drop is 4V, then the motor should have the complementar 10V drop across it correct?


Now, it is my understanding that in a series circuit, the current is the same across all the components, if that's true, then

From there, we can calculate all the other parameters, total resistance = 35ohms and motor resistance = 25ohm.

If I remove the resistor, I assume the voltage drop across the motor will be 14V, but resistance will still be 25ohms, therefore current will be 0.56 amp. Am I correct?

What if I want to repeat the process with the maximum speed/load situation, how do I make sure I won't exceed the 10AMP limit on my DMM?

 

Wow, that's right! When you say 0.1R, is that the same as 0.1Ohm?

ok, that's over my head, I'll do some research on that. thanks for the pointer though



I know, as you may have realized, I'm a software engineer trying to get into electronics... You're right, for PIC programming I had very few problems so far, but I clearly need more knowledge in order to continue...

 

I quickly scanned the responses and didn't see "shunt" - which is more or less a low value resistor that can be used to extend the range of a meter. Most of the current flows thru the shunt - some thru the meter. I've made them from copper wire. What you need is wire big enough that it won't get red hot and melt under the current draw but yet be small enough so that a modest (a foot or two) length provides the value that you need.

__________________
stevez

 

that is what I have described above.
The 0.1R resistor acts a a shunt resistor for the 100k input resistoance to a diff-amp

__________________
Nothing is impossible.
Once a problem is realised, the rest is just details

 

Yeah, but he really shouldn't need a diff amp for DC measurements. All 2 wire, ungrounded voltmeters are differential.
For 20 amps and 14V, 0.1R is probably too high. 0.01R would cause much less degradation in current, torque, and RPM.

__________________
Ron

 

I just ordered an eval board of that Allegro current sensing device. The one I ordered senses up to 75amp. Seems pretty simple, only requires one additional bypass capacitor and has analog output (vcc/2 = 0amp).

I can use it now to measure current consumption and I can use it later on my robot design as an additional stall sensor.

 

If these are R/C model motors (as used in Tamiya R/C car models?), then the stall current is likely to be far greater than 20 amps - a LOT depends on your power source, which 'may' be the limiting factor?. NiCd batteries have very low internal resistances, and can provide enough current to destroy themselves - which is what makes them fairly dangerous.

__________________
PIC programmer software, and PIC Tutorials at:
http://www.winpicprog.co.uk

 

I agree, the OP said he was using a DMM, why the extra amp stages? Even 2 wire grounded voltmeters are differential. The fact that it is DC doesnt remove the requirement to reject the common mode voltage, so isnt a diffamp needed?

The dmm will average and the shape of the current waveform may cause inaccuracies (I'm not so sure its a clean DC amps) Also would it be better to pick a sense R such that the full scale current reading will match the full scale DMM reading? If the DMM has a 2V, 200mV scale Rsense = .1 or .01 respectively is good. If it has 1V, 100mV range then maybe R = .005 on the 100mV range is best?

 

I think I'm starting to make sense.. and it is much simpler than I thought.... yes, when you guys started talking about differential things and opamps, I think I got lost.

So, if I have a .1 ohms in series with the motor and measure 2V across it, it means that the motor is drawing 20Amps, right?

I figured that .1 is still to high, because power dissipation at 20Amps will still be 20A*2V = 40W correct?

 

Well, I was assuming a handheld DMM, which don't know nuthin' 'bout no common mode voltage. Even an AC-powered DMM can probably handle 14V common mode - which only exists if you put the sense resistor on the high side.

The difference between RMS and average current isn't too significant (for a motor, anyway) unless the ripple amplitude is greater than, say, 40% p-p. (I'm too lazy to calculate that. I simulated it and came up with about a 2% max difference).
Regarding the sense resistor value, I was just thinking that 2V drop out of 14V is too much. What you say makes sense, so long as the reading stays within range.

__________________
Ron

 

Well, I was assuming a handheld DMM, which don't know nuthin' 'bout no common mode voltage. Even an AC-powered DMM can probably handle 14V common mode - which only exists if you put the sense resistor on the high side.

The difference between RMS and average current isn't too significant (for a motor, anyway) unless the ripple amplitude is greater than, say, 40% p-p. (I'm too lazy to calculate that. I simulated it and came up with about a 2% max difference).
Regarding the sense resistor value, I was just thinking that 2V drop out of 14V is too much. What you say makes sense, so long as the reading stays within range.

__________________
Ron