problems to control a dc motor with a pwm signal

[mariomoskis]

yes,i refer about average current(which is really across my motor)
so ithink that i am right,yes?

 

[mariomoskis]

You can put a resistor in series with the motor, but I wouldn't recommend it because the motor needs a high current to kick-start it.

i try about a power resistor 1ohm series with the motor,i got to reduce the current consumed by the motor, but the power operational amplifier get hot,why is this happening?

 

[Reloadron]

yes,i refer about average current(which is really across my motor)
so ithink that i am right,yes?

No, if you want to PWM a 3.0 VDC motor you use 3 volt pulses. You do not use for example 6 volt pulses with a 50% duty cycle and call it 3 volts.

Ron

 

[mariomoskis]

but i only can hace supply 5V,because i am supplying another circuits with the same supply power 5V,and it is the minimum

can i fix 2V at the emittor with a diode o diode zener?to get always 3V across the motor?
maybe 3 diodes 1n4004?

 

[Reloadron]

Then as Alec mentioned and also mentioned it was not a recommended practice would be to place a resistor in series with your motor of you can try a few diodes as you mention. The main problem also as Alec covered is that when the motor starts, especially when under a load it may not start correctly. The 1N4XXX series are 1 amp diodes, you can try some. The real problem is that you have a 5 volt circuit and a 3 volt motor you are trying to PWM.

Ron

 

[PICMICRO]

I am getting confused here.
Why won't applying 6V at 50% duty_cycle create same average current in the motor as applying 3V continuous ?
Let me analyze it.
Lets say a motor is spinning at its no_load rated speed (w) an say, its Back Emf (generator action) is Eb = 2.7V. (for this given speed)
Suppose its DC armature resistance is 1ohm.
Suppose we are applying 3V continuous to this motor.
Then, Average current = continuous current = (3-2.7)/1 = 0.3A.
Since, the motor is spinning at its no_load rated speed, this 0.3A current is just sufficient to provide enough torque to over_come friction, and maintain constant speed.
Now, switch the supply to 6V at 50% to this motor (spinning at the same speed)
Then, during the On_time, the current through it is I = (6-2.7)/1 = 3.3A.
During off time, current is 0.
So, average current is 3.3A/2 = 1.65.
So, clearly the average current is lot more than the previous case, and the motor will fly to very high speed. (A little maths revealed it must be twice the rated speed, but that is assuming the friction/windage torque remaining same, which they aren't.)
If you really have to apply 6V then, the duty cycle needs to be 9.09% (for this case), such that the average current would be = 3.3 * 0.0909 = 0.3A.

So, can I conclude that -" You can apply 6V to a 3V motor, but the duty-cycle needs to be ridiculously lower and should be judiciously calculated. It won't follow a linear pattern like that in a resister." ?

 

[alec_t]

Why won't applying 6V at 50% duty_cycle create same average current in the motor as applying 3V continuous ?

Because a motor is not a simple resistive load. It has resistance and inductance. Also, when rotating, it acts as a speed-dependent voltage generator opposing the applied voltage.

 

[PICMICRO]

Why don't you people suggest the OP to use lower duty-cycle instead of suggesting to change the power supply ?

@alec_t Please, see my previous post again. I have answered my own question. . So, it wasn't really a question, but just thinking_aloud.
There is an unanswered question though, if you would like to answer.
Thanks anyways.

 

[audioguru]

Why doesn't the OP use a 5V motor to match his 5V power supply?

 

[mariomoskis]

so,the problem is that i need supply voltage 5V because i just have one supply power and i have another circuits which have supply voltage 5V and it can´t be less

the motor must be that,(1.5-3V),i don´t have load!

and i have two transistors 2n2222a or BD135,so is there a way to get 3V across my motor with duty cycle 100%??

and altec_t one question for you or the other members of this forum,i don´t know if you will read this:
when i tried it when the power operational amplifier

before try with another transistor i saw that maybe i can do it with a power opeartional amplifier like follow voltage(3V) with 1A output,like this:
View attachment 62416

but i have this question:
if the supply voltage for the amplifier must be 5V,
why when i have supply voltage=6V with duty cycle 50%,the motor is already consumes the 0.3A maximum that it can? and i see too that the voltage at the output of the amplifier is higher than at the input,but it shouldn´t happen?

and i see too that if i increase the supply voltage for the amplifer to more than 6V,the motor is gonna comsume more current,why is this happen?

i did your suggest,i put power resistor in series with the motor to reduce its current consumed,i got it but which is the higher value that i can use??? and why when i put 1ohm resistor the operational get little hot???

 

[PICMICRO]

Why doesn't the OP use a 5V motor to match his 5V power supply?

Sorry, I seem to be missing something here.
If its possible to use the OP's original setup by just triming down the Potentiometer a little more (to achieve lower duty-cylce), why are you insisting on getting another motor/power supply ?
I am often poor at arguing with people without them considering my words harsh, so please bear with me. Its not me, but my poor expertise at the language.
Also I am not sure if my analysis at post #46 is correct/complete.
Hoping to learn something.
Thanks.

 

[alec_t]

@PICMICRO

Also I am not sure if my analysis at post #46 is correct/complete

I would agree with most of it, except :
"Then, during the On_time, the current through it is I = (6-2.7)/1 = 3.3A." Because of the winding inductance the current doesn't instantly reach that value but instead ramps up to it. Not knowing the inductance value or exact PWM rate we can't say what the average 'on' current is.

So, can I conclude that -" You can apply 6V to a 3V motor, but the duty-cycle needs to be ridiculously lower and should be judiciously calculated.

In view of the above I'm not sure about 'ridiculously lower'.

It won't follow a linear pattern like that in a resister." ?

I agree.

@mariomoskis

i did your suggest,i put power resistor in series with the motor

But I also said I didn't recommend it.

but which is the higher value that i can use?

I can't recommend any value.

why when i put 1ohm resistor the operational get little hot?

I don't know. Possibly the switching time of the opamp is affected by load conditions. Anyone else know?

 

[mariomoskis]

ok,but i saw that is the only way to get reduce that current,so i think that i will try to do it
how could i know until which value of resitor can i use in series with my motor?

later i am gonna put a gearbox in my motor,so current will be 1.05A as we saw in the datasheet of the motor? the power operational amplifier have 1A at its output
will be it enough?

and about the pwm with transistor what can i do to get this, and i will use a gearbox too later:

so,the problem is that i need supply voltage 5V because i just have one supply power and i have another circuits which have supply voltage 5V and it can´t be less

the motor must be that,(1.5-3V),i don´t have load!

and i have two transistors 2n2222a or BD135,so is there a way to get 3V across my motor with duty cycle 100%??

is there a solution to do this or i am trying to do one impossible thing?

 

[PICMICRO]

@alec_t
I agree with both of your objection.
Since the switching frequency is quite low (some hundreds of Hz) and the motor is small (I guess L/R quite small), I thought the current rise time would be quite fast w.r.t the switching time period. But, yes, as you suggested, we need more details.
Yes, the choice of the word 'ridiculously lower' is ridiculous.
Thanks.
@mariomoskis
There is two way to control the current into your motor. (which is what really matters)
i) You use Supply Voltage much greater than the motor ratings.
i.a) Then you need to use PWM at very low duty_cycle.
i.b) You need the transistor with much higher peak current rating than the average current that flows through motor.( Ipeak ~= I_avg / K where K is the duty_cycle ).

ii) You use Supply Voltage matching the motor ratings
ii.a)Then you will be using PWM mostly near 100% duty_cylce.The dutycylce will only be lowered, when you want the motor to spin at lower speed than the rated speed.
ii.b)You need transistor with peak current rating no_more than the Nominal Motor current ratings.

And, here you seem to choose option i) but yet refuse to do i.b).
Either fully follow i) or ii). You can't mismatch.

 

[mariomoskis]

@alec_t
There is two way to control the current into your motor. (which is what really matters)
i) You use Supply Voltage much greater than the motor ratings.
i.a) Then you need to use PWM at very low duty_cycle.
i.b) You need the transistor with much higher peak current rating than the average current that flows through motor.( Ipeak ~= I_avg / K where K is the duty_cycle ).

then if this is true and i always need a supply voltage 3V to get 100%duty cyle,(not 5V,even if i put a diode at the emittor of 2V)
my teacher was mistaken about this topic

when you speak about peak current you refered when i see in the datasheet 0.3A no load and 1.05 at max efficient yes?

 

[alec_t]

i am gonna put a gearbox in my motor,so current will be 1.05A as we saw in the datasheet of the motor?

1.05A is stated in the datasheet to be the current at maximum motor efficiency. That is not necessarily the same thing as the maximum motor current in all circumstances, so for reliability you need a circuit to cope with a possibly higher current than 1.05A. Certainly adding a gearbox will increase the average current above what you are experiencing now, because of the added friction of the gearbox. The peak current will occur when the motor is stalled.

 

[mariomoskis]

that current with the gearbox could be higher than 1A?

and about the resistor in series with the motor i see this for the case of the operational amplifier:

generating the pwm signl with the 555,the AO get little hot as i said,but i put a amperer to see the current at the output of the amplifier and it is 0.3A,so why is this happening?
and generating the pwm with a module of Compactrio ni9401(digital output) instead of the 555, the AO don´t get hot and it is working well

why i have this two differents situations??

 

[alec_t]

that current with the gearbox could be higher than 1A?

Yes.

i put a amperer to see the current at the output of the amplifier and it is 0.3A,

A digital meter (if that's what you used) set on the current range will not give an accurate measurement of a variable pulsating current.

and generating the pwm with a module of Compactrio ni9401(digital output) instead of the 555, the AO don´t get hot and it is working well

Good.

why i have this two differents situations??

I don't know your circuit layout so can't say definitely. But if you have this set up on a breadboard then there is lots of stray capacitance which could affect pulse rise and fall times and hence power dissipation.

 

[mariomoskis]

there is a way to don´t have that stray capacitance?

i put a capacitor beside the power supply,but it has the same behaviour

 

[alec_t]

there is a way to don´t have that stray capacitance?

Yes. At least, it can be reduced (not entirely eliminated) by using a correctly designed pcb which minimises conductor lengths.

i put a capacitor beside the power supply,but it has the same behaviour

That is good practice; but it won't solve any stray capacitance (or stray inductance) problem.