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Unless you have an expensive digital multimeter it will probably give wildly inaccurate measurements of pulse signals. An analogue meter, on the othe hand, will tend to 'average' any pulse voltage applied.when i try to check voltages with a multimeter, as i have a pulse signal, all voltages which i check in each part of the circuit must be calculate like this
It can and does happen, because the motor is not a simple resistive load. Its winding has inductance (read up about that if you're unfamiliar with the term).i see too that the voltage at the output of the amplifier is higher than at the input,but it shouldn´t happen?
Whatever the duty cycle is the motor maximum current can be > 1A, even with a 3V supply. Read the motor spec again.when i have supply voltage=6V with duty cycle 50%,the motor is already consumes the 0.3A maximum that it can?
Ohm's Law.if i increase the supply voltage for the amplifer to more than 6V,the motor is gonna comsume more current,why is this happen?
Whatever the duty cycle is the motor maximum current can be > 1A, even with a 3V supply. Read the motor spec again.
Ohm's Law.
Yes. But A0 is not a rail-to-rail output type of opamp. At 1A current the datasheet for A0 says the output voltage is typically never below 1.3V (when the opamp supply is 24V....its not stated what it would be if the supply was only 5V).but if i am fixing the output of the AO to 3V because the input is 3V too, and it is working as a follow voltage,
if i increase the supply voltage, the voltage at the output should be fixing to 3V ,not?
The opamp bandwidth is specified as 1MHz, so the problem may lie in stray external (to the IC) circuit capacitance reducing the A0 gain at high frequencies.but when the frequency is 10kHz the voltage at the input and output is exactly the same maybe because it isn´t working well at this frequency?
That sounds as though you've found your own solutioni tried to fix the voltages at the input and output to 2V instead of 3V and it looked better about the motor current consumed. Do you know some solution to solve this problem about the current consumed by the motor?
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.is there a way to reduce the current which the motor consumes or i will be always limited to do it with less voltage at the ouput and input?
Because this IC has been designed for driving motors and can handle more current than the transistor you were using.and about this kind of circuit, why this works for control pwm motors too ? which is the different bewteen this and when iused the transistor?
The average motor current should increase as you increase the duty cycle, more or less proportionally provided the motor winding doesn't saturate (at which point the current is limited more by the motor resistance than its inductance.should the motor consumes almost the same current when i change the duty cycle from 80% to 50%
You would need a tachometer.how could i know the speed of the motor in each moment to know if this is working well?
I think the 1N4001 rectifier diode is too slow for PWM. It should be a much faster diode like maybe a 1N5817 Schottky 1A diode.
I get them from cheap old solar garden lights.
The specifications for your motor clearly spell out with a 3.0 volt constant the motor speed will be about 14,800 RPM and the motor will draw about .300 amp. That is for the motor free running and doing no work.
Let's forget the motor for a moment. Draw a series circuit with a battery and resistor. The battery voltage is 3.0 volts and the resistor is 10 ohms. Therefore the circuit current is .300 amp. Now what happens if we increase the battery voltage to 5 volts? What happens to the circuit current?
Why do you insist on trying to run a 3 volt motor at voltages above 3 volts?
Ron
but the duty cycle which i am using is 60%,it will give me the 3V no more
for example if i try a diode zener o 3 or 4 normal diodes at the emittor i can 100%