problems to control a dc motor with a pwm signal

[mariomoskis]

my circuit is this, but now i have 2 power resistor in series with the motor too
View attachment 62704
i just try with the gearbox and the current consumed by the motor is higher about 0.5A,so it is higher than the limit for no load 0.3A

will be it a problem for my motor which had:
0.3A no load
1.05A max efficiency

or this current could be until 1.05A with the gerabox?

 

[mariomoskis]

i think taht it won´t be a problem but i want to ask before up the supply voltage to 5V to the AO
and another question i forgot to do:
this higher current consumed by my motor,can affect to the voltage at the output of the AO,or it will keep being the same?

 

[alec_t]

now i have 2 power resistor in series with the motor

Not recommended.

current consumed by the motor is higher about 0.5A

Average current probably a lot more than that if that's the reading of a typical hobbyist DMM. Peak (stalled) current may be >> 1.5A (we don't know, as the datasheet doesn't give enough info).

will be it a problem for my motor

Not for the motor. The problem will be the opamp over-heating.

or this current could be until 1.05A with the gerabox?

I would guess much more, peak.

i think taht it won´t be a problem but i want to ask before up the supply voltage to 5V to the AO

That's your view. Mine is different .

this higher current consumed by my motor,can affect to the voltage at the output of the AO,or it will keep being the same?

The opamp has a finite (small) output impedance so one will affect the other.

As the opamp is rated at 1A maximum and your motor can clearly draw more than 1.05A, perhaps up to 2A or more at switch-on, the opamp is likely to run hot and fail quickly.

BTW, the opamp in your diagram has no ground connection, and components R and C do nothing useful.

 

[mariomoskis]

yes,the circuit has some mistakes but i have it well conected

so do you know some a power operational amplifier with more than 1A at its output? to solve this problem?
and is there a way to how with current is really at the output of the amplifier, because with the multimeter i never see more than 1A as you said.

thanks

 

[audioguru]

Your circuit has too many voltage drops. The input to the power opamp is from a 555 with a 6V supply driving a voltage divider.
The output high of the 555 is +4.7V. The voltage divider reduces the input to the power opamp to only 2.82V. The output high voltage of the power opamp is shown as about 3.65V less than its supply voltage when its supply voltage is much higher than you have so it might not have ANY voltage for the motor.

Who says the power opamp works with a supply as low as 6V? The datasheet shows 30V and 24V.
Oh, your power opamp has a negative 6V supply but has no positive supply. Then maybe its total supply is 12V?

 

[alec_t]

so do you know some a power operational amplifier with more than 1A at its output?

No, I've never needed/used one. I'd use the 555 or existing opamp to drive a power MOSFET, as already recommended.

is there a way to how with current is really at the output of the amplifier

You could use a current shunt (say 0.1Ω) in series with the motor and use a O'scope to monitor the voltage dropped across that. Alternatively you would need a very expensive DMM.

 

[mariomoskis]

when you said current shunt,is it like a multimeter?
ot it is another diffrent thing?which is the different?

 

[alec_t]

A current shunt is just a resistor, with an accurately known value. The value is low, so that it doesn't affect significantly the current you are trying to measure.

 

[colin55]

You could use a current shunt (say 0.1Ω) in series with the motor

A current shunt is not a series resistor. It is a PARALLEL RESISTOR.

You are talking about a current-limit resistor but more-accurately a CURRENT-DETECT resistor, by measuring the voltage-drop across a low value resistor that will not affect the current-flow in a circuit.

Why don't you use a simple transistor such as BD679 and save all the hassles.

 

[alec_t]

A current shunt is not a series resistor. It is a PARALLEL RESISTOR.

Agreed. In normal use it is used that way and is sold as such. However, for the OP's purpose he could buy one under that name and use it in series with the motor as a current-sense resistor.

Why don't you use a simple transistor such as BD679 and save all the hassles.

Alternative solutions have already been suggested to the OP, but he is determined to go his own way

 

[mariomoskis]

because i have to develop both circuits,first with the amplfier and later with the transistor. it isn´t my choice,it is about my teacher jeje

and about the transistor i did with a BD135(which can have 1.5A in its colector) but i saw the next thing:
i need at least to supply the 555 with 5V,so if i use this supply voltage and duty cycle 60% i should see 3V across my motor,yes?

but when the duty cycle is 60% and supply 3V, the motor consumes its maximum current 0.3A, but i think that it should happen with 5V of supply because 0.6*5=3V??

 

[alec_t]

because i have to develop both circuits,first with the amplfier and later with the transistor. it isn´t my choice,it is about my teacher jeje

It's a pity you didn't tell us that right at the beginning .

i need at least to supply the 555 with 5V,so if i use this supply voltage and duty cycle 60% i should see 3V across my motor,yes?

Not necessarily. As has previously been explained to you a motor is not simply a resistor which 'averages' the applied voltage. Re-read all the previous posts.

when the duty cycle is 60% and supply 3V, the motor consumes its maximum current 0.3A,

Where did you get that from? Re-read the datasheet.

 

[mariomoskis]

OK,so the only way to get this with the transistor is give a supply of 3V to my motor

so what can i do about the 555,how can i give 5V to it,if i only have one power supply?

 

[ericgibbs]

OK,so the only way to get this with the transistor is give a supply of 3V to my motor

so what can i do about the 555,how can i give 5V to it,if i only have one power supply?

hi,
Look at the CMOS version of the 555, it will work with a 3V supply.

 

[mariomoskis]

hi,
Look at the CMOS version of the 555, it will work with a 3V supply.

ok,i will ask if we have that

Re-read all the previous posts.
Where did you get that from? Re-read the datasheet.

it was which i saw when i conected the circuit.

ok,i understand about it now,so i need 3V supply becasue the motor is not a simple resistive load.

and how can i explain to my teacher that if i have supply 5V,and even if i have a zener diode at the emitter of 2V, the motor will have more than 3V across it with duty cycle 100%??

 

[mariomoskis]

i have ask about the motor and my tacher said me that the motor reads averages voltages from 1.5 to 3V, so if it is true,i should can do about supply the motor with 5v and have a duty cycle of 60%,yes?

 

[alec_t]

my tacher said me that the motor reads averages voltages from 1.5 to 3V

Ask your teacher if he/she can explain the relationship between applied voltage, duty-cycle, motor inductance and motor-generated back-emf and then provide an equation linking all of those factors; and preferably look at the actual voltage variation on an oscilloscope.

 

[colin55]

My tacher doesn't know what he is talking about either.
How can a motor read? It hasn't got any eyes. Only a needle can read because it has an eye.
The operation of a motor is much more complex than any equation can provide.
You need to deliver a voltage, load the motor and read the current.
There is no other way.
If there were, I would be a millionaire.

 

[mariomoskis]

View attachment 62747

then.why if i try this circuit and i check the voltage across the motor with a multimeter i see 2.7V with duty cycle 100% and 1.7V with duty cycle 50% ? it looks work well not?

 

[audioguru]

The duty cycle of the pulses are 50% but the current in the motor keeps going between the pulses because it becomes a generator (its back EMF). Then a meter reads the voltage too high.

We don't know your meter. If it is digital then the frequency of the pulses might mess up its reading.
If it is analog with a needle then the pulses might quickly move the needle high but the needle might drop slowly between pulses so it reads too high.

1.7V/2.7V is 63%, not 50%.