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The example, worked out
Given
I want
A ripple after a rectifier
Load current
1 A
Peak voltage
15.6 V
Ripple allowed
1 V
Mains
50 Hz
Rectifier
Full wave (a bridge)
Working
two refills per cycle
the load drinks I for a whole period, the voltage may drop ΔV
with 10 mF
Results
Capacitance10 mF
From the shelf (E6, above)10 mF
Trough14.6 V
Ripple frequency100 Hz
Voltage rating25 V
Ripple current, roughly2.5 A
The formulas
C = I / (f ΔV)the load current, the ripple frequency, the dip allowed
f = 2 × mains (full wave), mains (half wave)two humps per cycle, or one
C = 2 P t / (V1² − V2²)hold-up: energy for a time
The usual traps
Full-wave ripple is at 100 or 120 Hz, not 50 or 60: half the capacitor.
The ripple current through the capacitor is two or three times the load current: check its rating.
Rate the capacitor a quarter above the peak, and the peak is √2 × the RMS.
Questions
How big must the capacitor after a rectifier be?
C = I / (f ΔV): the load current, the ripple frequency, the dip allowed. f = 2 × mains (full wave), mains (half wave): two humps per cycle, or one. C = 2 P t / (V1² − V2²): hold-up: energy for a time.
What does it give for typical values?
10 mF for 1 V of ripple at 1 A, full wave on 50 Hz. Capacitance: 10 mF; From the shelf (E6, above): 10 mF.
What are the usual mistakes?
Full-wave ripple is at 100 or 120 Hz, not 50 or 60: half the capacitor. The ripple current through the capacitor is two or three times the load current: check its rating. Rate the capacitor a quarter above the peak, and the peak is √2 × the RMS.
In the app
Your circuit, drawn live
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The part you can buy
Results rounded to standard values (E12, E24, E96) and power ratings.
The working, shown
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Open the calculatorOpens in the app on 1 December, with Circuit Notebook 1.0