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Diode as a ripple source.

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electricity86

New Member
Hello there.
I read about power supplies and it was said there that its not good that the time the diode remains ON will be long since the diode causes ripples in the output voltage.

The PSU I read about was half wave rectifier, and full wave rectifier, that use a filter capacitor in the output.

Why does the diode cause ripples to the output voltage?
 

Boncuk

New Member
Without diode you have 100% ripple - pure AC.
 

Hero999

Banned
The diode does not cause ripple.

Ripple is a result of the AC waveform going through 0V every half cycle.
 

electricity86

New Member
Here is what they said:

"The slope of this line increases as the current increases,
bringing point c lower. Consequently the diode
conduction time (c-d) increases, increasing ripple.".



So the ripple here is the c-d part of the graph?
How do you condier the b-c line, if its not part of the ripple?

This c-d line is crossing the zero every complete cycle, not half cycle, is it still considered to be a ripple?
 

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MikeMl

Well-Known Member
Most Helpful Member
Hello there.

Why does the diode cause ripples to the output voltage?
The diode doesn't cause the ripple. The ripple results from the fact that the filter capacitor gets a momentary recharge every 16.6msec (or 20msec if 50Hz), and then it is up to the capacitor to supply the load current until it gets another recharge during the next AC peak when the diode conducts to refill the capacitor. The larger the capacitor, the less the ripple....

Look at the attached sim. Note the peak diode current as compared to the load resistor current and its duty cycle.

Exercise for student: How big would the filter capacitor have to be to reduce the load ripple voltage to < 0.5V? Why would you rather use a full-wave rectifier?
 

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crutschow

Well-Known Member
Most Helpful Member
So the ripple here is the c-d part of the graph?
How do you condier the b-c line, if its not part of the ripple?

This c-d line is crossing the zero every complete cycle, not half cycle, is it still considered to be a ripple?
The ripple is simply the peak-to-peak difference between the two voltage levels. It includes c-d and b-c.
 

Boncuk

New Member
Here is what they said:

"The slope of this line increases as the current increases,
bringing point c lower. Consequently the diode
conduction time (c-d) increases, increasing ripple.".
The diode conduction time is constant, depending on mains frequency.

Think logic! What is changing in the circuit? - the diode, mains frequency or the load current?

If the diode remains the same and also does mains frequency and the load current changes it is consequently the load increasing ripple.

The rectified AC charges the smoothing capacitor which has to deliver the supply DC. If AC changes polarity the cap is not charged during that period, hence the voltage across the cap decreases, depending on how fast it is discharged. The highter the load current the faster will be cap be discharged.

It takes 20ms at 50Hz and 16 2/3 Hz at 60Hz to start a new charge cycle for the cap. Using a full wave rectifier both times are half of that value meaning less ripple with the same load current.

Boncuk

P.S. c - d is contained in vertical (Y-axis) of the graph displaying voltage. Time is contained in the horizontal (X-axis, time) of it, displaying voltage change.
 
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