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Disclaimer:
This circuit is based on normal algebra, where to increase the height of y axis, a constant is added
y=sinx → y=sinx + c, where c ε Z
Given:
V(rms) =10V
Average voltage=13V
Load=Led, V(led) =2V, I(led) =10mA
Diode voltage drop, Vd=0.7V
Frequency=50HZ
Output calculation:
The given voltage is RMS, convert it into peak voltage:
V(peak) =√2*V(rms) =√2*10V=14.14≈14.2V
The capacitor voltage=V(peak) -Vd=14.2-0.7
=13.5V(since the capacitor voltage matches the Average Voltage, let's just continue, the if condition is given at end of the description)
To provide stiff clamping
RC>100T, where T=time of signal
(Continued on load calculation, but originally, since the current and voltage are taken as purpose, this calculation is done before the load calculation)
Clamped Voltage, V(clamp) =V(peak new)
=V(peak old) + V(capacitor)
=14.2+13.5=27.7V
So, Vo(max) =27.7V.
Load calculation:
Load resistor, R=(Vo(max)-V(led)) /I(led)
R=(27.7-2) /10mA
R=25.7/10mA=2.57kΩ
Now, for Time calculation, Where
T=1/f=1/50Hz=0.02s
C≥100T/R=100(0.02) /2.57=778μF≈800μF
Warning:
If the capacitor voltage ≠ The average voltage, then use a resistance series to diode or before the load part.
If this happens, calculate impedance, then resistance and reactance.
Else, use series Diodes or a voltage divider circuit(if the frequency is very low and the capacitance is greater than millifarads only)
The small variations in output is due to impedance
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