DC motor control using PWM - is it really better?

Hello
OK, quick (I hope!) question:
PWM makes controlling speed of a DC motor via micro controller easier. I get that.... i think.
But in terms of benefits of PWM, I thought they were that PWM gave peek torque at low rpm, but as I've learnt in practice, this isn't true as the motor 'sees' simply an average (lower voltage), and hence using lower current to drive the motor at slow speeds - is that right?
So, why is PWM used in all sorts of DC drives from drills to robots as (presumably) you could have speed control via potentiometer controlled high gain transistor (either current BJT or voltage controlled FET I assume??) to control a DC motor while burning minimal resistive energy in the low current pot side.
So is the real reason for PWM popularity that a lot of devices these days use battery power and so need to conserve maximum energy, or is there an actual mechanical advantage, eg more torque at low speed or something I have not considered? Like I say, I've measured the torque of a 12v motor using PWM and half fixed voltage and they seemed the same to me.... I thought things like drills use PWM instead of a pot controlled transistor to get low speed torque but I think I may have got that wrong. Please confirm, someone?!?!?!

Many thanks if you can answer this helpfully!

M

Chopping the current so that it is either full on or full off is much more efficient than using a resistor or transistor to throttle the current.

If you have a motor that draws 6 amps at 12v (just for example) its effective impedance will be 12 / 6 = 2 ohms

If you want to use a resistor (or transistor) to reduce the current to 3 amps you need to increase the total resistance to 12 / 3 = 4 ohms so the "slow down" resistor will need to be 2 ohms. In effect you will have 6v across the resistor and 6v across the motor.

If you are passing 3 amps through a 2 ohm resistor the energy loss will be 6v * 2A = 12 watts. That would make a resistor very hot, and all the energy is just going to waste.

The maths here are over-simplified, but should illustrate the issue.

...R

Thanks, but why would you have to burn off current in a huge 2 ohm resister? Why not just use a transistor, FET or some amplifier control to control the motor current?? That way the 'burning off' of current is in the large resister in the base/gate of a transistor, not in a massive 2 ohm power resister...

It does not work this way. The transistor in linear region works like "variable resistor" and its resistance is controlled by base current/gate voltage (very roughly speaking). There is no way for transistor to reduce voltage other than burning excess power as heat.

I agree with @Smajdalf.

...R

Smajdalf:
It does not work this way. The transistor in linear region works like "variable resistor" and its resistance is controlled by base current/gate voltage (very roughly speaking). There is no way for transistor to reduce voltage other than burning excess power as heat.

But in the linear region it works as a current amplifier, so you use a variable resister (burning energy) in the small current (low power) base stage, and control a large motor current (with no losses) in the collector-emitter stage.

Please could you explain why this amplifier principle doesn't work for motors?

Thanks!

mechengncl:
Thanks, but why would you have to burn off current in a huge 2 ohm resister? Why not just use a transistor, FET or some amplifier control to control the motor current?? That way the 'burning off' of current is in the large resister in the base/gate of a transistor, not in a massive 2 ohm power resister...

No, the base/gate are only controlling what's happening. The burning off of heat is inside the transistor. And making the insides of a transistor very hot results in a dead transistor. So you need a large, heavy, expensive heat sink to keep the transistor reasonably cool.

If you have a 12V battery and you want your motor supplied with 5A at 6V then there's 60W coming out of the battery (12V@5A) and only 30W going into the motor. The other 30W is heating up the transistor.

It's like the difference between a linear voltage regulator and a switch-mode voltage regulator. I know which I prefer for most purposes.

Steve

mechengncl:
But in the linear region it works as a current amplifier, so you use a variable resister (burning energy) in the small current (low power) base stage, and control a large motor current (with no losses) in the collector-emitter stage.

Please could you explain why this amplifier principle doesn't work for motors?

Thanks!

Your concept of the transistor is incorrect. You need to find a website that explains how transistors work.

This Forum is for helping people write programs.

...R

slipstick:
No, the base/gate are only controlling what’s happening. The burning off of heat is inside the transistor. And making the insides of a transistor very hot results in a dead transistor. So you need a large, heavy, expensive heat sink to keep the transistor reasonably cool.

If you have a 12V battery and you want your motor supplied with 5A at 6V then there’s 60W coming out of the battery (12V@5A) and only 30W going into the motor. The other 30W is heating up the transistor.

It’s like the difference between a linear voltage regulator and a switch-mode voltage regulator. I know which I prefer for most purposes.

Steve

ok thanks, that makes more sense. I was assuming the CE current was being controlled by the base and not burning off power in the transistor itself… I can now move on!

mechengncl:
ok thanks, that makes more sense. I was assuming the CE current was being controlled by the base and not burning off power in the transistor itself... I can now move on!

The CE current is controlled by the base current and the power is burned off (dissipated) in the transistor itself.

By the way, the base is such an integral part of a transistor that I can't figure how you can think of the transistor as being separate.

...R

Robin2:
The CE current is controlled by the base current and the power is burned off (dissipated) in the transistor itself.

By the way, the base is such an integral part of a transistor that I can't figure how you can think of the transistor as being separate.

...R

Robin2:
Your concept of the transistor is incorrect. You need to find a website that explains how transistors work.

This Forum is for helping people write programs.

...R

Look, if it bothers you that I'm asking non-coding questions then just don't answer them.

As for why I don't understand the power dissipation in the CE part of a transistor compared to the Base, well, I walk in the shadow of your magnificence.

I have googled this topic before posting but most amplifier explanations don't really discuss the way a transistor uses power. As a switch though, from fully off (basically zero power wasted) to full on (close to zero power wasted), a hobbyist/newby may be expected to conclude that at 'half on' a similar behaviour may exist but where power is only used in the base supply resister(s). Apparently not, so I learn, I improve, and hopefully someone reading this will also be helped.