A full-wave rectifier circuit is driven by a sinusoidal input voltage with Vrms=15V and frequency 50Hz. If the load resistance is 100Ω, what is the ripple voltage with a filter capacitance of 1.5mF? (Assume the diodes to be ideal, with Von=0V.)
T = 1/F = 1/(2*50) = 0.01 s. = Period.
R*C = 100 * 1.5*10^(-6) = 1.50*10^-4 s.
T/RC = 0.01/1.5*10^-4 = 66.7.
Your RC time constant is much too short to be used as a filter at 50 Hz!
The T/RC ratio should be less than 1; it is 66.7!
Let C = 500uF.
T/RC = 0.01/(100*5*10^-4) = 0.2.
Vo min = Vomax/e^(t/RC) = 10.6/e^0.2 = 8.68 Volts.
Vr = (Vomax-Vomin)/2 = (10.6-8.68)/2 = 0.96 Volts, peak. = Ripple voltage.
Vo = (V0max+Vomin)/2 = 10.6+8.68)/2 = 9.64 Volts, D.C.
when getting the period, why did you multiple by 2?
To find the ripple voltage in a full-wave rectifier circuit, we need to determine the output voltage and then calculate the difference between the maximum and minimum output voltages.
1. Calculate the maximum voltage (Vmax) at the output of the rectifier. This can be done by multiplying the root mean square (RMS) voltage with a factor of √2:
Vmax = Vrms * √2
Vmax = 15V * √2
Vmax ≈ 21.21V
2. Calculate the minimum voltage (Vmin) at the output of the rectifier. In this case, since the diodes are ideal with a voltage drop (Von) of 0V, there will be no voltage drop across the diodes:
Vmin = 0V
3. Calculate the ripple voltage (Vripple) as the difference between Vmax and Vmin:
Vripple = Vmax - Vmin
Vripple = 21.21V - 0V
Vripple ≈ 21.21V
4. Since the calculation above gives the peak-to-peak ripple voltage, we can calculate the average ripple voltage (Vripple_avg) by dividing the peak-to-peak ripple voltage by 2:
Vripple_avg = Vripple / 2
Vripple_avg = 21.21V / 2
Vripple_avg ≈ 10.6V
Therefore, the ripple voltage with a filter capacitance of 1.5mF is approximately 10.6V.
To find the ripple voltage in a full-wave rectifier circuit, we need to first calculate the peak-to-peak voltage of the output waveform, and then divide it by 2 to find the ripple voltage.
Step 1: Calculate the peak voltage (Vpeak) of the input waveform.
The peak voltage can be calculated using the formula:
Vpeak = √2 * Vrms
where Vrms is the root mean square voltage of the sinusoidal input.
In this case, Vrms = 15V, so using the formula:
Vpeak = √2 * 15V
= 21.21V (approx.)
Step 2: Calculate the peak voltage (Vp) of the output waveform.
Since the diodes are ideal with Von (forward voltage drop) being 0V, the peak voltage of the output waveform will be equal to the peak voltage of the input waveform.
Vp = Vpeak = 21.21V
Step 3: Calculate the time constant (τ) of the filter circuit.
The time constant (τ) is given by the formula:
τ = R * C
where R is the load resistance and C is the filter capacitance.
In this case, R = 100Ω and C = 1.5mF. However, for calculation purposes, we need to convert the capacitance from millifarads (mF) to farads (F).
1 mF = 0.001 F
So, C = 1.5 * 0.001 F
= 0.0015 F
Now, we can calculate the time constant:
τ = 100Ω * 0.0015 F
= 0.15 s (seconds)
Step 4: Calculate the ripple voltage (Vr).
The ripple voltage can be calculated using the formula:
Vr = Vp / (2 * π * f * C)
where f is the frequency of the input waveform and π is a constant (approximately 3.14159).
In this case, f = 50Hz and C = 0.0015 F. So, substituting the values:
Vr = 21.21V / (2 * π * 50Hz * 0.0015 F)
= 21.21V / (3.14159 * 2 * 50Hz * 0.0015 F)
= 0.090 V (approx.)
Therefore, the ripple voltage with a filter capacitance of 1.5mF is approximately 0.090V.