Clarifications on the PWM signal

[electronicsLearner77]

I have few clarifications on the pwm signal very basic ones
a. If my input voltage is let us say 5V and the output contains resistor. Now i control this 5V signal using a pwm of 50% duty, then at the output do i measure 2.5V (5/2 V) or the RMS value? For sine wave it is the RMS values, so i assume the same for the pwm signal. Is it correct?
b. This is bit of confusing question, i am not sure if i can explain clearly. If i put a probe on the resistor and see the waveform do i see the pwm waveform or the RMS signal. I mean to say do i need to indirectly calculate how much voltage is applied on the resistor based on pwm or can i see it directly on the scope?

 

[Bluejets]

PWM is dc, on and off, not sine wave, no ac content no rms values etc. etc.
If you apply 50% duty from a 5v source then the average output is 2.5v.
The scope will show 5v pulses.

 

[Harald Kapp]

PWM is dc, on and off, not sine wave, no ac content no rms values etc. etc.

Pardon if I disagree.
On/off is correct, not a sine too. But:

• No AC content is wrong. A pulsed DC signal can be considered a pure DC superpositioned (added) by a square wave AC signal. In the op's example that would be 2.5 V (DC) + 2.5 V (AC square waveform).
• RMS values can be computed for any waveform. For non-sinusoidal waveforms it is generally just not as simple as dividing the amplitude by sqrt(2). The RMS of a pulse waveform is amplitude × sqrt( t_on / T).
  In the op's example that would be 5 V × sqrt(0.5) = 5 V × 1/sqrt(2) which is incidentally identical to the RMS of a sine with 5 V amplitude. But only for this particular duty cycle of 50 %.

 

[Harald Kapp]

If i put a probe on the resistor and see the waveform do i see the pwm waveform or the RMS signal.

An oscilloscope is meant to show voltage vs. time, i.e. the waveform. Any other values like mean, rms, frequency, ducty cycle etc. need to be calculated from the displayed waveform. Modern scopes can often do that for the user and display these computed values (and sometimes many more derived properties of the signal).

 

[ratstar]

A resistor as long as it isnt sharing - and + (a voltage divider) wont actually divide the volts, it drops the amps. If your measuring voltage your adding a loop to the signal and it gets a bit confusing (measuring volts means ur mixing the - and +) , if you measure the amps its half, and ur measuring just the + by itself.

PWM is like adding a resistor without having the resistor i think.

I could be wrong tho, im just a beginner too.

You can read current or volts with an oscilloscope, it depends on how you probe it in, just like a multimeter.

 

[bertus]

Hello,

Have a look at the wiki:
https://en.wikipedia.org/wiki/Pulse-width_modulation

Bertus

 

[Harald Kapp]

A resistor as long as it isnt sharing - and +

What does "sharing" mean here?
You really need to upgrade your manner of expressing electronic circuits in a way that is understood by others. As any profession there is a trade specific language. If you don't use it, you will be misunderstood at best.

 

[ratstar]

I mean by sharing, by you getting a point somewhere between + and - and then at the end, then you get a proportion of the volts.
Your right I do say things in strange ways. I think it could be a benefit that i'm different, cause you guys aren't necessarily easier to understand than what I say, since electronics is confusing to the beginner.

re-spec.

 

[electronicsLearner77]

From the above replies i am not sure which is the correct method, let me refer a text book and come back on clarifications.

 

[ratstar]

Just go with what they say, they own the forum, I'm saying the same thing I'm pretty sure, I'm just the additive element. =)

 

[electronicsLearner77]

Just go with what they say, they own the forum, I'm saying the same thing I'm pretty sure, I'm just the additive element. =)

S i am referring to them.

 

[bertus]

Hello,

Did you read the wiki page?
In there you will find this picture:



As it says, there is a percentage of the voltage given to the load.


Bertus

 

[electronicsLearner77]

Hello,

Did you read the wiki page?
In there you will find this picture:

View attachment 49733

As it says, there is a percentage of the voltage given to the load.
Bertus

Sorry not yet, i go through and clarify.

 

[ratstar]

I might be partially wrong with what I posted... because I forgot that a voltmeter is high impedance (it has a resistor?), so its not actually a voltage divider, because a voltage divider would be 0 resistance...

 

[Harald Kapp]

because a voltage divider would be 0 resistance...

Nonsense. With 0 Ω current would be infinite...

 

[ratstar]

Nonsense. With 0 Ω current would be infinite...

I mean on the output line/loop. (if you dont count the resistors meeting it.)

 

[bertus]

Hello,

Perhaps a simple circuit may show you how you can make a simple PWM circuit:

This circuit was designed by SgtWookie of the AAC.
The mosfet will switch the load on and off very fast.

Bertus

 

[ratstar]

You can get PWM out of AC, if you put a capacitor in the line, you get different amounts of on-off time if u change the capacitance, but the wave isnt a perfect square, its a discharge curve. (a quarter of a harmonic)

 

[electronicsLearner77]

Thank you for the circuit, i understand from the circuit that is the inverted logic, so when the pwm is high the voltage decreases and when it is low the voltage increases and i can see a linear increase and decrease of the voltage and is the average value. Few critical questions which i am having for a long time
a. When you apply a pwm signal for example a circuit below not a true circuit only representation

and i put a scope at the point as per post #4 i can only see the pwm signals, i cannot actually see the average voltage applied to the resistor, it is only based on calculations. Now my question is let say the pwm signal is something like below

And the applied input voltage is 5V the average voltage is 3.75V. Does this voltage apply at the end of PWM cycle as shown at the End of Region1 or during the Region1? As long as the PWM duty remains same the average voltage remains same and it does not vary again based on On time and Off time. And once the new duty is applied let us say 50% ON and 50% OFF the average voltage again changes. Am I correct?
b. The RMS value generally they refer to the dissipation equivalent of the DC. So, here in the case of PWM do we take the RMS value to calculate the dissipation? Or what is the reason for calculating the rms value of pwm signal, when some of the books refer?

 

[bertus]

Hello,

The following page will tell you more about the duty cycle:
https://www.quora.com/How-do-we-calculate-the-PWM-time-period

Bertus