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measure a voltage across of a dc motor

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mariomoskis

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hello,i am trying to measure a square wave(pwm) with amplitude 3V, so when i try to measure it with a multimeter across my motor,should i watch the next:

Vmultimeter=3*sqrt(duty cyle)

is it correct? because sometimes i am watching more than 3V on my multimeter across the motor about 3.7V but i think that it shouldn´t be higher than 3V.
i think that it can be caused about the inductance of the motor but i try to put a zener diode of 3V at the output of the AO but the problem is still happening.

somebody could help me with that? the Vcc=6V and the multimeter is set between the output of the power operational amplifier and ground

thanks

duda.JPG
 

the inductance can cause above voltage readings.
you should be aware of the technicalities whentrying to measure a pwm waveform with a digital multimeter .
you should know the duty cycle.
or if pwm , no of cycles in pwm variation.

a better method is to use a scope.

you can remove the motor and measure with meter for checking.
 

ok,i have check the measure at the input of the AO and i have 1.5V with duty cycle 50%,and 3V with duty cycle 100%, so with the multimeter i am measuring avegare or effectuve voltage?

i think that the formulas are: Vaverage=Vmax*(duty cycle) ,is it correct? and what about the Veffective? which is the different between both voltages(average and effective)?

with a osciloscope i have the problem that i can´t see exactly the wave,because the motor causes distorsion on the wave

and one question more, i tried to set a power resistance in serie with the motor, and i solve the problem a little about this voltage at the output. but i saw that the operational amplifier get hot,why is this happening? because the motor has its maximum current consumed in 0.3A and the AO can has 1A at its output!
 

you are measuring average value.
Vavg=Vmax*dutycycle is correct.
try to have two diodes at the o/p of AO . in reverse direction to supply and ground.
heating may reduce.
 

but maybe it could reduce the voltage at the output? and i don´t want it .

you refered this:


AO------(-2*DIODE+)-----------motor----Rpower-------

__________________ --------(+diode-)-----------------
 

[

no. not that diode in series with the o/p.
it is:

ao---->diode-anode---diode-cathode------>+ve supply.
ao---->diode-cathode---diode-anode------>ground.
 

sorry but i donñt understand what you mean

you said a diode(-+) between the output of the AO and the parallel conexion of the motor with the diode?
 

in addition to your existing connection , connect two freewheeling diodes at the output AO.
one to +ve and another to ground.
 

ah ok,i understand what you mean, i will try it and i will say you the results,but i think that the diodes maybe can get so hot about the current which will be across them

i have 2 diodes 1n4004,are them correct to use in this case?
 

you can use 1n4004. and post your observation.
 

it didn´t work,when i set the diodes as you said the current got even high

if the data sheet of the motor says that the maximum current no load is 0.3A is there a problem is that current for my case is 0.4A?? even if the data sheet says too that at maximum efficiency that current can be 1A

the motor works between 1.5 and 3V,with duty cycle 50% should i have 1.5V or it can be higher (about 1.8V)
because this relation doesn´t have to be linear i think,should it?
 
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if the motor is rated for 3v , give 3v supply with reqd duty cycle.
can you post full circuit diagram for finding the cause of heating in the sugested setup .

and observe , the output waveform with diodes. are they square?
is the final amplifier is designed for such square waveshapes?
 

i put a power resistor 0.5ohm at the output of the AO to try check the current here
it is difficult to see because the waves are distorsionated about the inductance of the motor,but more or less i saw a different of voltage between the output and the wire parallel of 0.4V,with a power resistor of 0.5ohm,so tyhe current is about 0.8A,no load

so maybe when i use the gearbox this current will be higher,and my AO only can have 1A at its output,so i am sure that i am going to have a higher current, then this is why the AO is getting hot
 

just connect a resistor of say, 10 ohms or 8,2 ohms in the o/p removing motor.
see the waveform and heating. without your current measuring resistor , see the waveform.

can you give the data of yhe amp AO. for what purpose it was meant for ?
 


is TCA032DP1,the supply voltage for my case is +6V
and the data sheet says that the minimum is 5V or +-2.5V,and the output of the AO has 3V because of the voltage divisor

with a 10ohm resistor instead of the motor, and a power resistor to check the current at the output,the voltage is 0.2,so the current is 0.2/0.5=0.4A, but when i checked it with the motor instead of the 10ohm resistor this current was 0.8A
 

so the motor gets 3v always which is not correct.
try to have the two tca032 digram given in datasheet.
 

the voltage at the output will be a little higher at the output because of the inductance of the motor

and i don´t understand the diagran that u said
 

you can first substitute the resistor(10 or 8.2 ohms/2w to 3w) and observe whether o/p goes fro 3v to 0v for a 50 % duty cycle i/p.
 

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