measure a voltage across of a dc motor

[mariomoskis]

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

http://imageshack.us/photo/my-images/401/dudaq.jpg/

 

[Harald Kapp]

It should be 3V*(duty cycle)
assuming duty cycle is defined as Ton/(Ton+Toff)

Are you shure the output of the 555 is 5 V? Since Vcc=6 V and the 555 sees a very small load, this voltage here could be 6 V, resulting in 3.6V at the voltage divider's tap. That (including some uncertainty in the measurement) could explain the 3.7 V measured by you.

Harald

 

[mariomoskis]

yes,it is sure that the 555 has 5V at the output,i measure it with a oscilloscope and 3V at the input,it is right,the problem is about the output

and why is 3*duty cyle and not 3*sqrt (duty cycle)? with the multimeter i am measuring effective or average voltage?
for example at the input where i don´t have the problem about the motor,with the multimeter i measure 2.12V for duty cyle 50%,so it is becasue V=3*sqrt0.5=2.12 not V=3*0.5=1.5V

and about the zener, when i put it at the output to fix the voltage in 3V and get that at the output the voltage won´t increase its value more than 3V, i am still measuring more than 3V with the multimeter?

 

[Harald Kapp]

If the multimeter is in DC mode, you measure average voltage. What is your definition of duty cycle? Same as mine? Sometimes it is defined as ton/(ton+toff), sometimes as ton/toff. With different results, of course.

A zener at the output will be only of limited help. If the amplifier puts aout more than th ezener voltage, the current through the zener will rise drastically. Unless you put a series resistor between the amplifier and the zener, but this will limit the current for the motor, too.

Harald

 

[mariomoskis]

duty cyle: ton/(ton+toff)

yes,i use a resistor in serie with the motor and i got low the current consumed for the motor,but the amplifier get hot,why?if the AO can has 1A at its output and the motor has its maximum current consumed in 0.3A

 

[Harald Kapp]

I'm a bit at a loss.
Have you looked at the voltage waveform across the motor? And at the current waveform (measure voltaeg across the series resistor)? Any anomalies? Have you tried another motor?

To isolate the cause for the behaviour of the circuit, replace the motor by an equivalent resistor. Now you don't have the inductance and can look at a purely resistive behaviour. What are the results (current, voltage, average volate wit respect to ducty cycle)?

Harald

 

[mariomoskis]

yes,i saw the waveform across the motor: 3Vhigh state 1V low state,so it is correct i think

i didn´t try another motor(i don´t have another one)

the current consumed for the motor is 0.35A if i don´t use a resistor series with the motor,and 0.29A if i use a power resistor 1k serie with motor, so i must use resistor because the maximum current consumed for the motor can be 0.3A max

with the resistor in series with the motor and without resistor i think that i had the same behaviour without a resistor,i don´t remember now.i saw the next:

the increase of voltage across the motor isn´´t linear with the increase of duty cycle,because i saw that when i increase duty cyle from 50% to 75% the motor increase its speed more than when i increase the duty cycle from 75% to 100%
why is that? should it happens?

about replace the motor by an equivalent resistor i must wait to do it until monday

 

[(*steve*)]

I very much doubt that resistor is 1K. If it had 0.29A running through it, that means your power supply is around 300 volts.

It is far more likely to be 1 ohm.

Rather than adding a resistor, you just limit the duty cycle to the point where the average current is 0.3A. So (for example) you limit the maximum duty cycle to 80% if the average current at 80% duty cycle is 0.3A.

However that is probably wrong anyway, because I imagine your motor is currently running without a load? When it has a load placed on it, it will draw more current.

Is this the same motor we've seen before in some other thread? Why are you starting another thread?

I can't find the specs on the motor. Can you post them? (again, possibly)

 

[mariomoskis]

i start a new thread because in the another case i am using a transistor,so i though that they are diferent things

specs of the motor are:
http://imageshack.us/photo/my-images/401/motordc.png/

i am not using a load now,but later i will use a gearbox,so the current will be increased a little,i checked about 0.04A higher

yes,with the resistor i limit the duy cycle too,
but for example with the inductance of the motor ,the current which i am reducing with the resistor , will be increase later about the inductance,so i think that it should work.
do u know another way to reduce that current consumed for the motor without a resistor?? i tried with a zener diode at the output of the AO,but it didn´t work.

 

[duke37]

There are no details of the type of motor, if it is a motor with a commutator, it will generate a voltage when it is turning which can upset your measurements. Get the PWM circuit working into a resistor and then substitiute the motor.

There seems to be a negative spike in the waveforms, a diode across the motor (not normally conducting) would clip this spike.

The motor uses about 0.3A unloaded, it uses over 1A when run at maximum efficiency so why are you worried about 0.3A?

 

[mariomoskis]

i don´t have load,so the current must be lower than 0.3A

later i will use a gearbox,is it considered as a load?when i use the gearbox the current consumed is increased,but no so much 0.04A more or less

i don´t understand what u mean about this:
''There seems to be a negative spike in the waveforms, a diode across the motor (not normally conducting) would clip this spike.''

 

[Harald Kapp]

The gearbox alone will need little additional power, as you have noticed. However, the final goal of motor and gearbox surely is to move some load. THIS will increase the power consumption of the motor.

The negative spike duke refers to is visible in the oscilloscope image to the right of your circuit. One assumes that this scope image belongs to this circuit. If so, what duke means is that a suitable diode in reverse orientation across the motor should limit these spikes to ~-0.7 V.

Harald

 

[(*steve*)]

The motor uses about 0.3A unloaded, it uses over 1A when run at maximum efficiency so why are you worried about 0.3A?

100% true. And we're not even talking about the stall current (which you should really spec your circuit for according to Murphy's law)

 

[mariomoskis]

i am worried because i think:
if with no load the maximum current is 0.3A and i have a value higher than it I=0.35A so i am out of range when i don´t have load,and i guess that even if the motor can have 1A at maximum efficiency,i am not in that case.

which is the different between maximum efficiency and no load,and load?gearbox is like load or no load?my problem is that i don´t understand the range of currents.

 

[duke37]

If the motor were 100% efficient, it would take no current with no load. In practise, there is resistance in bearings, air, copper etc so power is needed to run the moter even when there is no power coming out. As load (torque) is increased, the motor slows and the current rises to provide the output power.

The efficiency is zero with no load by definition. As the load is increased the power wasted in the motor becomes a smaller proportion of the input power and the efficiency rises. If the motor is stalled, the power output is zero so the efficiency is again zero. There is a maximum efficiency at a certain speed, quoted in the data sheet.

The gearbox should provide very little load to the motor if the friction is low. Putting a load on the gearbox output will increase the load on the motor and so increase the motor current. I would suggest you do not go much above 1 or 1.5A.

If the unloaded current is a little high, it may be due to slightly stiff bearings or brushes rubbing a little too hard , do not worry about it.

 

[mariomoskis]

ok,so if i understand well:

when i don´t have load,it isn´t a problem to have 0.35A?even if the data sheet says that the current is 0.3 max no load

with the gearbox, it is like load but it will increase the current only a little as i checked when i tried it in the practice,so nothing to be worried

so the thing that i need is with this circuit that i posted at the begining,to get that the average voltage across the motor won´t be higher than 3V when the duty is 100% or little less, but i can´t get it because the current consumed for the motor is increased to 0.35A,so that was why i put the resistor in series with the motor, and i don´t find another way to do it.

and one question:
when duty cycle is 50% i should get average voltage 1.5V more or less across the motor, or isn´t it linear?do u know what i mean?