Does PWM power supply divide the current drawn of a non-inductive Load along with the divided apply voltage?

[mannok]

Let say I have non-inductive Load and it is rated as follow:

• 0.2A @ 1V applied
  2A @ 5V applied

Which is powered by a DC battery and is rated as follow:

• 5V output
  1A max current supply

I am going to divide the supply voltage by switching it with PWM, say 20% duty cycle and as a result we can simulate a 1V voltage is applying onto the Load. In this case, may I know how much current will the Load draw? As it is a non-inductive Load, does it mean that it would (tend to) draw 2A current for every high state in PWM (because high state is 5V)? However, the battery can provide only 1A at maximum, so it may not work properly.

OR... Should I just treat it as 1V applied and so that the current drawn by the Load should be only 0.2A?

[Burngate]

Being told the load is non-inductive isn't much help, unfortunately, in part because

0.2A @ 1V applied
2A @ 5V applied

Tells us it's very non-linear - not resistive (is it something like an LED?)

What it would do with a supply switched by PWM isn't something I could guess at.
However, you could perhaps take that pwm-switched-supply and filter it (inductors and capacitors spring to mind) to produce a fairly smooth 1v that it would be happy with.

[drgeoff]

Let say I have non-inductive Load and it is rated as follow:

• 0.2A @ 1V applied
  2A @ 5V applied

Which is powered by a DC battery and is rated as follow:

• 5V output
  1A max current supply

I am going to divide the supply voltage by switching it with PWM, say 20% duty cycle and as a result we can simulate a 1V voltage is applying onto the Load. In this case, may I know how much current will the Load draw? As it is a non-inductive Load, does it mean that it would (tend to) draw 2A current for every high state in PWM (because high state is 5V)? However, the battery can provide only 1A at maximum, so it may not work properly.

OR... Should I just treat it as 1V applied and so that the current drawn by the Load should be only 0.2A?

If everything was exactly as per the info you have given about the PWM and load then in theory for 20% of the time you would be applying 5 volts and 80% of the time 0 volts. The load would take 2A for 20% of the time and 0A for 80%. The average current would be 2 * 20% which is 0.4A.

However, for 20% of the time the load will be trying to take 2A from the battery which you say has a 1A maximum. Therefore from the info given a definitive answer as to what will happen in practice is not possible.

[mannok]

Let say I have non-inductive Load and it is rated as follow:

• 0.2A @ 1V applied
  2A @ 5V applied

Which is powered by a DC battery and is rated as follow:

• 5V output
  1A max current supply

I am going to divide the supply voltage by switching it with PWM, say 20% duty cycle and as a result we can simulate a 1V voltage is applying onto the Load. In this case, may I know how much current will the Load draw? As it is a non-inductive Load, does it mean that it would (tend to) draw 2A current for every high state in PWM (because high state is 5V)? However, the battery can provide only 1A at maximum, so it may not work properly.

OR... Should I just treat it as 1V applied and so that the current drawn by the Load should be only 0.2A?

If everything is exactly as per the info you have given then for 20% of the time you will be applying 5 volts and 80% of the time 0 volts. The load will take 2A for 20% of the time and 0A for 80%. The average current will be 2 * 20% which is 0.4A.

But the Battery can supple only 1A at max, which mean it may not work properly?

[drgeoff]

But the Battery can supple only 1A at max, which mean it may not work properly?

I was updating my post while you wrote that. See aboove.

[mannok]

Being told the load is non-inductive isn't much help, unfortunately, in part because

0.2A @ 1V applied
2A @ 5V applied

Tells us it's very non-linear - not resistive (is it something like an LED?)

What it would do with a supply switched by PWM isn't something I could guess at.
However, you could perhaps take that pwm-switched-supply and filter it (inductors and capacitors spring to mind) to produce a fairly smooth 1v that it would be happy with.

@Burngate thank you for helping.

In reality I will add a capacitor for the Load, but for this question I hope that I can get some basic concept.

What if I change the question a little bit, say the Load is very resistive. 1A@1V, 5A@5V, etc. Then what will the answer be?

The reason I ask this is because I want to get myself clear to what situation should I add a capacitor? Or for most of the cases we can just treat the PWM switches power supply as its average voltage, since everything even wire may have some inductance... just like a capacitor

[mannok]

Let say I have non-inductive Load and it is rated as follow:

• 0.2A @ 1V applied
  2A @ 5V applied

Which is powered by a DC battery and is rated as follow:

• 5V output
  1A max current supply

I am going to divide the supply voltage by switching it with PWM, say 20% duty cycle and as a result we can simulate a 1V voltage is applying onto the Load. In this case, may I know how much current will the Load draw? As it is a non-inductive Load, does it mean that it would (tend to) draw 2A current for every high state in PWM (because high state is 5V)? However, the battery can provide only 1A at maximum, so it may not work properly.

OR... Should I just treat it as 1V applied and so that the current drawn by the Load should be only 0.2A?

If everything was exactly as per the info you have given about the PWM and load then in theory for 20% of the time you would be applying 5 volts and 80% of the time 0 volts. The load would take 2A for 20% of the time and 0A for 80%. The average current would be 2 * 20% which is 0.4A.

However, for 20% of the time the load will be trying to take 2A from the battery which you say has a 1A maximum. Therefore from the info given a definitive answer as to what will happen in practice is not possible.

@drgeoff thank you.

So in this case I can add a capacitor to filter the switched power supply in order to filter the current drawn as well?

Because I thought PWM frequency is always so high comparing to the inductance of any components in the circuit like wire, air, etc. and so that we can always omit the high and low state of PWM and treat the power supply as its average voltage at anytime.

[Burngate]

Because I thought PWM frequency is always so high comparing to the inductance of any components in the circuit like wire, air, etc. and so that we can always omit the high and low state of PWM and treat the power supply as its average voltage at anytime.

What you're describing is, in essence, a switch-mode power regulator.

Although PWM frequencies can be quite high, they don't have to be. That's one reason why you see inductors built in to switch-mode supplies.

If you put a capacitor directly across the PWM output, 20% of the time it's trying to pour current into what it sees a a short circuit to raise its voltage, and the other 80% it's trying to drag it back down to ground.

Putting an inductor in series gives everything a chance - but you'll need an idea of the frequency involved, and how smooth you want the output voltage to be, to work out the sizes of your inductors and capacitors.

So in this case I can add a capacitor to filter the switched power supply in order to filter the current drawn as well?

The load will take whatever current it wants. Smoothing the voltage you feed to it will help it to want smooth current, but like horses, we know nothing about your load, so our advice may be as good as my advice about the Kentucky Derby.

[mannok]

Because I thought PWM frequency is always so high comparing to the inductance of any components in the circuit like wire, air, etc. and so that we can always omit the high and low state of PWM and treat the power supply as its average voltage at anytime.

What you're describing is, in essence, a switch-mode power regulator.

Although PWM frequencies can be quite high, they don't have to be. That's one reason why you see inductors built in to switch-mode supplies.

If you put a capacitor directly across the PWM output, 20% of the time it's trying to pour current into what it sees a a short circuit to raise its voltage, and the other 80% it's trying to drag it back down to ground.

Putting an inductor in series gives everything a chance - but you'll need an idea of the frequency involved, and how smooth you want the output voltage to be, to work out the sizes of your inductors and capacitors.

So in this case I can add a capacitor to filter the switched power supply in order to filter the current drawn as well?

The load will take whatever current it wants. Smoothing the voltage you feed to it will help it to want smooth current, but like horses, we know nothing about your load, so our advice may be as good as my advice about the Kentucky Derby.

@Burngate, may I know how do I determine what situation/load should I add a capacitor? Some loads like motor they may have inductance and I may not need a capacitor as the current drawn has already been smoothed by the load itself. However, some loads like sensor I must use capacitor to smooth the switch-mode power input...

[Burngate]

@Burngate, may I know how do I determine what situation/load should I add a capacitor? Some loads like motor they may have inductance and I may not need a capacitor as the current drawn has already been smoothed by the load itself. However, some loads like sensor I must use capacitor to smooth the switch-mode power input...

Hmm. It's complicated - enough so that I can't really answer that in a straight-forward way. I'm not a teacher.

The PWM output is rectangular - high and low with nothing in between, with some undefined source impedance.
A purely resistive load will see exactly that.
A purely inductive load will integrate that into a rising/falling current.
A purely capacitive load will form some sort of low-pass filter with the source impedance, so what you get will depend on the parameters of that source.

Putting just an inductance in series with the source will hopefully swamp any nasties in the source impedance, so your load will just see an inductive source.
But if it decides to reduce the current it wants to take, the inductance will raise the output voltage to maintain the current already passing through.

Putting a capacitor across the output after the inductor provides somewhere for the excess current to go when you don't want it, and somewhere to get current from when you want extra.

That probably doesn't make a lot of sense - as I said, I'm not a teacher. Sorry.

You mention motors as having inductance.
But, as well as the inductance of the motor winding, there's the mass of the rotor that acts like a capacitance, in that if you remove the electrical power the rotor supplies power back as it slows down.
So it's more complex than a simple inductance.

[drgeoff]

A motor must have a resistive component as well. it couldn't do any work without consuming power and pure inductances and capacitances don't consume power. That resistive part will depend on how much power , ie it varies with the load on the motor.

[mannok]

Thank you both of you. Let me do more research on this topic.