pulse width modulation

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Bhuvanesh123

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analogWrite(pin num,255)//100% duty cycle so it completely on
analogWrite(pin num,125)/50% dutycycle so its 50% of time on and 50 percent of time off)

my question is in the above image whatever may be the duty cycle it on (5v) and off(nearly o v)

then how do we getting half of that voltage i mean around 2.5 from the actual 5v(change in value between 0 to 255 only says on time of time )

.thank you in advance
 

The duty cycle is just the ratio between the on-time duration (5V here) and T (T = 1/frequency). Multiply the ratio with 100 to get the duty cycle in percent.

For getting a smooth DC output, you need a low pass filter (for example RC filter) to average the output waveform. In other words that low pass filter removes the AC components.

DC output voltage = {V1*(duty cycle) + V0*(100-(duty cycle)) }/100

V1 is voltage corresponding to "1" V0 is voltage corresponding to "0", Duty Cycle in percentage.
The formula just calculates the average value of the waveform.

when the device/load/etc has averaging by itself, the low pass filter can be omitted. This is done frequently in electric motor controllers. The mass of the moving parts of the motor are acting as the low pass filter. There may be some filtering to avoid electromagnetic interference to (for example) radio equipment.
 

thanks for your detail explaination sir
 

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