your striking off on a tangent.

Your goal was to control current over a fixed resistor. That will generate a certain power that must be dissipated. You can choose to generate that heat in a continuous fashion, or you can pulse it in PWM fashion as twice the heat, half the time.

You've jumped back to the PWM thing when it was explained that the PWM was filtered to linearly run an amplifier.

Do you need PWM current or do you need linear current?

Sorry I guess I am not expressing myself clearly.

I don't need a particular form of current, I just need charges (Coulombs), and a specific quantity of them determined by experimental parameters. So that translates to Ampere*second. But I also need to use as little power as possible because there will be several of these reactors running 24/7 at considerable currents. It is part of the design considerations of this current regulator, so I don't deem it to be tangential.

What I was trying to get at is a point of confusion of mine. In my example, when I have a PWM signal at 50% duty, at 20V to generate an average of 1A, I consume 20*2*0.5=20Ws. But if I filter the PWM signal, then the actual voltage (not considering ripple) is going to be 10V, and will still give me the 1A, but now energy consumption becomes 10*1*1=10Ws. This does not make sense to me. How can the act of filtering have this effect on the energy consumption. So I must be thinking something in the wrong way, and that's why I am asking about it.

I hope the further explanation can illustrate what I am confused about.

(BTW, I am not passing current through a resistor, this is an electrolytic cell for organic compound degradation, so the load is purely resistive, but it is not a resistor "per se").

 

PWM does not control current or voltage. In your case, you're the one who's setting those values by whatever voltage you use to supply your load (I=E/R). Each period that the PWM output is high that formula will apply. PWM controls power expended vs time. The main goal and benefit of PWM over a linear approach is that nearly all power is dissipated in the load. A linear regulator will either control voltage or current to the load. Whatever energy isn't being delivered to the load is being dissipated (wasted) across the linear control element.

 

Sorry I guess I am not expressing myself clearly.

I don't need a particular form of current, I just need charges (Coulombs), and a specific quantity of them determined by experimental parameters. So that translates to Ampere*second. But I also need to use as little power as possible because there will be several of these reactors running 24/7 at considerable currents. It is part of the design considerations of this current regulator, so I don't deem it to be tangential.

What I was trying to get at is a point of confusion of mine. In my example, when I have a PWM signal at 50% duty, at 20V to generate an average of 1A, I consume 20*2*0.5=20Ws. But if I filter the PWM signal, then the actual voltage (not considering ripple) is going to be 10V, and will still give me the 1A, but now energy consumption becomes 10*1*1=10Ws. This does not make sense to me. How can the act of filtering have this effect on the energy consumption. So I must be thinking something in the wrong way, and that's why I am asking about it.

I hope the further explanation can illustrate what I am confused about.

(BTW, I am not passing current through a resistor, this is an electrolytic cell for organic compound degradation, so the load is purely resistive, but it is not a resistor "per se").

If you know then that 20v x 2a = 40watts, and that 10v x 1a = 10watts, why would you apply a 50% duty cycle of 40watts thinking that it would equal 10watts?

 

If you know then that 20v x 2a = 40watts, and that 10v x 1a = 10watts, why would you apply a 50% duty cycle of 40watts thinking that it would equal 10watts?

Becasue 50% duty gives me 1A from the 20V PWM, which is what I need and get from the 10V source in the example.

If I filter (integrate) the 20V/2A @ 50% duty PWM I get a signal of 10V which gives me 1A across the load. That means that the raw PWM signal is consuming 20Ws on average, while the filtered signal just 10Ws.

Ok, let me try to put it another way.

Say I need the system to have a current accross the load of 1A. The load resistance is 10 Ohm. Therefore I need V=IR=1*10=10V.

I get a 10V battery capable of delivering the current, and plug it in. The system will be consuming 10V*1A*1s=10W*sec.

Now,instead of the battery I get my 20V source with the PWM control circuit. My PWM pulses will be 20V and therefore I=V/R=20V/10 Ohm=2A high.
To get the 1A on average, I need to supply an average of 10V, that means a 50% duty cycle. So, for this case, the system is consuming 20V*2V*0.5s=20W*sec.

Furthermore, if I put a low pass filter at the output, the load will see the integrated signal, which would be 10V & 1A high. So now the consumption becomes the same as the battery: 10V*1A*1s=10Ws.

So either I am computing something the wrong way or the raw PWM consumes twice as much energy as the battery or the filtered output, while delivering the same amount of current. (I believe I must be computing something wrong, but I want to know what)

 

So the PWM version consumes twice as much energy for this example. Is this right, or am I missing something?

Here's an example of linear vs PWM. Check out the waste and tremendous power demands put on the pass transistor in the linear circuit.

 

So either I am computing something the wrong way or the raw PWM consumes twice as much energy as the battery or the filtered output, while delivering the same amount of current. (I believe I must be computing something wrong, but I want to know what)

Please read back over the replies because your ignoring some of them. This is why we're getting fuzzy thinking.

Edit: The fuzzy thinker was me! See post 51.

 

Here's an example of linear vs PWM. Check out the waste and tremendous power demands put on the pass transistor in the linear circuit.

Right, of course.

 

Please read back over the replies because your ignoring some of them. This is why we're getting fuzzy thinking.

I have read every single reply. I am not ignoring them at all.

If you can point out where my calculations are wrong or if they are right, it would help me a lot So far all replies I have read have not addressed the issue I am pointing out.

Perhaps if I ask differently:

For a 10 Ohm load, 10 volts are needed to deliver 1A. The energy consumption of a virtual 10V battery would be 10W*sec.

Question 1: What PWM duty cycle is needed to drive a 20V power supply in order to deliver an average 1A across the 10 Ohm load, and what are the V and I of the individual pulses?

Question 2: What is the energy consumption of the system at that duty cycle?

Question 3: If the output is filtered (integrated over time) what are the V and I seen by the load?

Question 4: What is the energy consumption of the filtered system?

My Answers:
1: 50%, 20V, 2A.
2: 20W*second
3: 10V, 1A
4: 10W*second

 

I thought it was explained in post 42. Is it the filtered/reintegrated values that you get that's sending you astray?

 

I thought it was explained in post 42. Is it the filtered/reintegrated values that you get that's sending you astray?

I understand post 42, it's basic PWM concepts. I just don't think what I am asking is being understood (judging from post 42's contents).

Of course it is me who sets the PWM, and in the example I am setting it to deliver 1A, and obtaining twice the energy consumption of what a battery would. Hence my whole series of inquiries.

 

Now I am thinking about power consumption.

Say I start with 20V, and a load resistance that gives me 2A.

If I need to deliver an average of 1A I would need an avg voltage of 10V. To do that with PWM I would have to use a 50% Duty cycle (half time ON, half time OFF).

If I power the load for 1 hour, that's a real ON-time of 0.5h, and my total energy consumption is

E = VIt = 20V*2A*0.5h = 20Wh

Here (bold) is your error. The average current, as defined, is 1 A; the average voltage is calculate as 10 V (i.e., 20V*0.5), so a perfect "averager" produces 10W.

Another way to look at it is:

Power = X*Y

If power is only on for 1/2 the time, then X is 50% and Y is 50%. (X/2)*(Y/2) = XY/4

Remember, the tracing you are showing is either Voltage vs. Time or Current vs. Time. Since they are in phase (in this example) only one is shown. In reality, there are two curves to contend with and both need to be averaged.

John

 

I understand post 42, it's basic PWM concepts. I just don't think what I am asking is being understood (judging from post 42's contents).

Of course it is me who sets the PWM, and in the example I am setting it to deliver 1A, and obtaining twice the energy consumption of what a battery would. Hence my whole series of inquiries.

In your 20V/10 Ohm PWM circuit the instantaneous current is 2A, P= 40W. If the duty cycle is 50% then the average current vs time = 1A, P= 20Wh. Because your supply voltage is 20V you will have to reduce the duty cycle to 25% to produce P=10Wh. If you're unhappy with this then lower the supply voltage.

Why fight this? You'll have full control over the duty cycle.

EDIT: See post 51 (by John) to see why my post is all wet!

 

Here (bold) is your error. The average current, as defined, is 1 A; the average voltage is calculate as 10 V (i.e., 20V*0.5), so a perfect "averager" produces 10W.

Another way to look at it is:

Power = X*Y

If power is only on for 1/2 the time, then X is 50% and Y is 50%. (X/2)*(Y/2) = XY/4


John

Ouch! A little voice kept telling me that I should not be treating the voltage as a constant while making I a variable. I should have listened to the little voice!

 

Here (bold) is your error. The average current, as defined, is 1 A; the average voltage is calculate as 10 V (i.e., 20V*0.5), so a perfect "averager" produces 10W.

Another way to look at it is:

Power = X*Y

If power is only on for 1/2 the time, then X is 50% and Y is 50%. (X/2)*(Y/2) = XY/4

Remember, the tracing you are showing is either Voltage vs. Time or Current vs. Time. Since they are in phase (in this example) only one is shown. In reality, there are two curves to contend with and both need to be averaged.

John

Oh I see. 20V half the time, and 2A half the time = 10W. I was just thinking of raw pulses 20V and 2A tall as a single waveform.

Thanks, that solves it.

 

Ouch! A little voice kept telling me that I should not be treating the voltage as a constant while making I a variable. I should have listened to the little voice!

See?, it is tricky.

(Actually I don't think it has to do with treating voltage as a constant, but realizing that both the current and the voltage are pulsed at 50%, which gives 0.5*0.5 = 0.25, so the PWM does decrease the power as a square factor)

 

See?, it is tricky.

(Actually I don't think it has to do with treating voltage as a constant

Yes it is, because that's what I did not factor in. I reduced 'I' by 50% but left 'E' constant at 20V!?!?!?! Considering a lifetime of doing this I'm very disappointed in myself. If I could find a hole I'd stick my head in it.

 

Yes it is, because that's what I did not factor in. I reduced 'I' by 50% but left 'E' constant at 20V!?!?!?! Considering a lifetime of doing this I'm very disappointed in myself. If I could find a hole I'd stick my head in it.

Nah, don't be so hard, I also missed it and I should have seen it right away.

I have to thank everyone for the help offered up to this point.

 

Just a follow up... The system for the 0-4A range is working fine. Again, thanks everyone.