Switches (and components in general) are not ideal.

WIth PWM only, there is no energy storage (flywheel) so there is constant switching of X high energy peaks with resistive losses to the PWM duty cycle for X voltage and current output. With inductive energy (storing energy with lower losses in a magnetic field) storage, I can design the circuit to be very efficient with lower peak energy per duty cycle switch pulses, letting the stored (inductor flywheel) energy flow for X voltage and current output.
View attachment 320385
A current level example.

I understand that . I get the whole energy storage concept.

Can you give me your thoughts on my above question. I want to make sure i even understand PWM and a FET Correctly.


So lets say i make a simple voltage divider circuit Where Vin = 10V and R1 = 10 ohms. R2 is an IDEAL IGBT driven by a PWM generator
Say 50% Duty Cycle

The gate turns off and on such that the switch opens and closes rapidly.
The current is 1A when closed and 0 amps when open.

With a 50% duty cycle should the average current not be .5A? and therefore the average power equal to 2.5 Watts????

Or am i so screwed up i dont even understand that much?

 

With a 50% duty cycle should the average current not be .5A? and therefore the average power equal to 2.5 Watts????

No!!!!
Yes, the average current is 0.5A but the voltage is 10V, so the power is 10V * 0.5A =5W
Another way to look at it is the power dissipated in the resistor is 1/2(10² / 10) = 5W

The error (and I have made this one myself) is that the power dissipated in a resistor is the square of the current, so you can't use the PWM signal average current value to calculate resistor power.

 

You are correct about a PWM controller’s ability to deliver half power to a device with near 100% efficiency IF the device operates at the same voltage as the input voltage.

But that is not what a buck converter does. A buck converter actually reduces the voltage to the load. PWM does not do that. That is the difference. A buck converter acts like series resistor or voltage regulator, but does not suffer the power loss in the resistor or linear mode transistor. The voltage drop occurs in the inductor, which, unlike a resistor, can store the wasted energy, then give it back later (when the switch is off.)

Filtered PWM without an inductor can also reduce the voltage, but not without the loss.

 

No!!!!
Yes, the average current is 0.5A but the voltage is 10V, so the power is 10V * 0.5A =5W
Another way to look at it is the power dissipated in the resistor is 1/2(10² / 10) = 5W

The error (and I have made this one myself) is that the power dissipated in a resistor is the square of the current, so you can't use the PWM signal average current value to calculate resistor power.

Ok I guess I’m mentally challenged
Since when is i^2r not valid?

if the average current is .5 then it’s .5^2 x 10 ohms for 2.5 watts

if you Do it with p = Iv the complete voltage drop across the resistor can’t be 10v

why is i^2 r not valid in this equation
That is just crazy to me. I know you are correct because I see 5 watts in the sim

so you can't use the PWM signal average current value to calculate resistor power. (how is this even possibly true although i know you are correct but dont understand why)

 

You are correct about a PWM controller’s ability to deliver half power to a device with near 100% efficiency IF the device operates at the same voltage as the input voltage.

But that is not what a buck converter does. A buck converter actually reduces the voltage to the load. PWM does not do that. That is the difference. A buck converter acts like series resistor or voltage regulator, but does not suffer the power loss in the resistor or linear mode transistor. The voltage drop occurs in the inductor, which, unlike a resistor, can store the wasted energy, then give it back later (when the switch is off.)

Filtered PWM without an inductor can also reduce the voltage, but not without the loss.

tracking with you, thanks for the continued replies

 

No!!!!
Yes, the average current is 0.5A but the voltage is 10V, so the power is 10V * 0.5A =5W
Another way to look at it is the power dissipated in the resistor is 1/2(10² / 10) = 5W

The error (and I have made this one myself) is that the power dissipated in a resistor is the square of the current, so you can't use the PWM signal average current value to calculate resistor power.

The assumption i dont understand here is why we are assume a voltage drop of 10V across R1 for 10*.5 = 5W
If our current is represented as the average = .5A then why is there thus also not an average voltage drop across the IGBT also?
Why are we assuming that the whole 10V gets dropped across the resistor. Its like well..... sometimes we consider the average and sometimes we dont and sometimes some equations work and some times they dont......

 

For power dissipated in a PWM signal, you cannot simply use the current squared.

 

No!!!!
Yes, the average current is 0.5A but the voltage is 10V, so the power is 10V * 0.5A =5W
Another way to look at it is the power dissipated in the resistor is 1/2(10² / 10) = 5W

The error (and I have made this one myself) is that the power dissipated in a resistor is the square of the current, so you can't use the PWM signal average current value to calculate resistor power.

So i found this video....
Is this the reason behind this? At the end of the little 6 min video she explains why the two are not equal.
Is that what your getting at?

https://www.bing.com/videos/rivervi...F5A6634F4DBC8D6E50CFF5A6634F4DBC8D6&FORM=VIRE

 

Alright..... well... lessons have been learned here.

I will spare you all the details on how my brain finally started working.

But this video was the key.

@ minute 5.48 she explains how the <i^2> not equal to <I>^2 Which is the underlying basis for my screw up....
OK! Can we get back to buck converters now!

Thanks for hanging in there with me gents

 

Talk about over-thinking an isue!

I usually use software tools to get voltage conversion circuits working, but my simplified model is:

A duty cycle is the ratio of input voltage to output voltage, a low loss low pass filter (the inductor and the filter capacitor averages tge voltage.

The diode is considered part of the switch that established the duty cycle.

The above gets to a good starting point for a real dedign. Rely on modern tools like SPICE or manufacturer supplied tools if you like, but I have done well moving to hardware from there.