Questions: 1 2 3 4 5

PHYS 2420 Diode Solutions

1. A triangular wave is shown in the diagram below. It is purely AC and V_peak-to-peak = 20.0 Volts. The wave is used as the input for various rectifying circuits. Treat the diodes as ideal (V_B = 0).
   1. Sketch the output voltage for this wave for a half-wave rectifier. Determine V_DC.
   2. Sketch the output voltage for this wave for a centre-tapped full-wave rectifier where N_in:N_out is 5:2. Determine V_DC.
   3. Sketch the output voltage for this wave for a full-wave bridge rectifier where N_in:N_out is 1:3. Determine V_DC.

   

   1. For a half-wave rectifier, V(t) forward biases the diode only half the time. Since V_pp = 20 Volts, V_max in the diagram is 10 Volts. The output looks like

      

      The DC offset is the average of this signal over one period

      .

   2. Since the transformer is N_in:N_out = 5:2, the signal into the rectifier is reduced by 2/5. Furthermore the rectifier is centre-tapped so only half the signal is sent to each diode, but both positive and negative voltages contribute. The signal across each diode is

      

      The DC offset is the average of this signal over one period

      .

   3. Since the transformer is N_in:N_out = 1:3, the signal into the r ectifier is three time larges. In the bridge rectifier, both positive and negative voltages contribute. each The signal across each diode is

      

      The DC offset is the average of this signal over one period

      .

   

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2. The diagram below shows a simple diode circuit with a battery. In this particular case, the diode is said to be forward-biased. Treat the diode as ideal (i.e. V_B = 0). Sketch the voltage drop over the resistor for
   1. V = 0 ,
   2. V = ½V_s,
   3. V = V_s, and
   4. V = 2V_s.

   

   The potential over the resistor is the sum of V(t) + V_S minus the portion blocked by the diode. Remember that if the potential is not greater than V_BE, no current will get through the diode. We will solve each case by sketching the potential before the diode and then the sketch of the potential after the diode, that is everything in the first graph that is greater than V_BE.

   1. V = 0

      

   2. V = ½V

      

   3. V = V_s

      

      The diode is forward biased at all times, and there is no difference between the curve above and the voltage across the resistor.

   4. V = 2V_s

      

      The diode is forward biased at all times, and there is no difference between the curve above and the voltage across the resistor.

    

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3. The diagram below shows a reverse-biased diode circuit. Treat the diode as ideal. Sketch the voltage drop over the resistor for
   1. V = 0 ,
   2. V = ¼V_s ,
   3. V = ½V_s ,
   4. V = ¾V_s ,and
   5. V = V_s.

   

   The potential over the resistor is the sum of V(t) + V_S minus the portion blocked by the diode. Remember that if the potential is not greater than V_BE, now current will get through the diode. We will solve each case by sketching the potential before the diode and then the sketch of the potential after the diode, that is everything in the first graph that is greater than V_BE.

   1. V = 0

      

   2. V = ¼V

      

   3. V = ½V_s

      

   4. V = ¾V_s

      

   5. V = V_s

      

   

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4. An ac signal V = 10.0sin(1500t) is feed through a N_in:N_out = 1:2 bridge rectifier and then through a capacitor input filter. The diodes in the rectifier are silicon ( V_B = 0.7 V). The filter capacitor is C = 10 μF and the filter resistor is R = 2000 Ω .
   1. What is the peak voltage and frequency of the rectified wave?
   2. What is V_DC, V_{r(peak to peak)}, V_{r (rms)}, and the ripple factor r.
   1. The N_in:N_out = 1:2 transformer boosts the peak signal from 10.0 Volts to 20.0 Volts. The bridge rectifier rectifies the signal but decreases the peak signal to V_{peak (out)} = 20.0 Volts - 2V_B = 18.6 Volts. The signal going into the capacitor filter therefore looks like

   

   Note that the frequency of the rectified signal is f_rectified = 2f₀ = 3000/2π Hz = 477.5 Hz.

   2. The capacitor is charged up as the voltage increases, e.g. from 0 to ½T. Where the signal would normally decrease sharply from ½T to T, the energy from the capacitor is released slowly as the decay is approximately linear as shown in the diagram below. Note that the diodes in a bridge rectifier are arranged such that current cannot flow back from the capacitor through the transformer under normal operating conditions. Under abnormal conditions, such as if the voltage is high enough you can burn out the diodes and send current back. The result is a triangular or sawtooth wave with a DC offset. Note that the sawtooth occurs only if the discharge time R_LoadC is much larger than 1/f_rectified.

   

   Recall that the discharge of a capacitor is governed by the equation V_c(t) = V_peak e^-t/RC , where V_peak is the 18.6 Volts. The signal on the above graph would be

   V(t) = V_peake^-[t-¼T]/RC ¼T £ t £ ¾T,

   between ¼T and approximately ¾T, and repeating itself every ½T thereafter. Since the capacitor is discharging slowly, we can expand the exponential about t - ¼T = 0 to yield

   V(t) = V_peak{1 - [t-¼T]/RC + ½([t-¼T]/RC)² - …} .

   On the graph, we see that the capacitor is discharging for close to T/2 seconds or 1/2f_r seconds. Keeping only the first term in the expansion, the height of the curve at ¾T treating the curve from ½T to T as a straight line, the minimum value of the

   V_minimum = V_peak[1 - T/(2RC)] .

   The average value of a straight line depends only on its endpoints, so

   V_average = V_DC = ½[V_peak + V_min] = V_peak[1 - T/(4RC)]

   In terms of the rectified frequency f_rectirfied = 2/T, this becomes

   V_DC = V_out[1 - 1/(2f_rectifiedRC)] = 17.63 Volts.

   The peak value of the ripple voltage, V_{r (peak)}, is the maximum variation of the ripple about V_DC, or

   V_{r (peak to peak)} = 2(V_out - V_DC) = V_out/(f_rectifiedRC) = 1.948 Volts.

   Since the signal is so close to a sawtooth (as long as RC is big!), we can use the standard result that

   V_rms = V_{r (peak)}/Ö3 = ½V_{r (peak to peak)}/Ö3 = 0.562 Volts.

   The ripple factor, the ratio of the peak to peak value of the purely AC part of the signal to the DC offset,

   r	= Vr(peak to peak)/VDC
   	= [Vout/(frectifiedRC)]/[Vout[1 - 1/(2frectifiedRC)]]
   	= 2/(2frectifiedRC - 1) .

   Evaluating the above expression, we find r = 0.11 or 11%

   

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5. Repeat Question 4 for an inductor input filter. The inductor is L = 0.010 mH and has a resistance R_W = 75 Ω .

   The signal from the bridge rectifier is the same as in question 4. The filter circuit consists of

   

   The main difference from the pure capacitor filter is that the rectified signal passes through an inductor L before getting to the capacitor filter. The inductor is chosen so that its reactance is much greater than the reactance or the parallel combination of the capacitor and the resistor. As well, the DC resistance of the inductor must be small compared to R for best results. The differences are best understood by considering what happens to the DC signal separately from the AC signal.

   

   In the DC equivalent circuit, the inductor has a small resistance R_W. The capacitor is nearly fully charged and thus acts as an open. Then R_W and R are in effect in series, the voltage drop over the resistor is reduced from what it would be if the inductor were not there

   V_DC(out) = [R / (R_W + R)]V_DC(in)

   Where V_DC(in) is the average of the rectified signal from question 4. From previous work we know that the average of a rectified sine wave is V_average = 2V_peak/π. Note that we want the smallest R_W to get the largest DC output. In our case,

   V_DC = [2000 / (75+2000)](2)(18.6 Volts)/π = 11.41 Volts.

   For the AC portion of the signal, we see that having the inductor in the circuit reduces the voltage over the parallel branches. The impedance of the parallel arms is

   Z_|| = (1/R + jΩC)^-1 .

   The total impedance of the circuit is thus the sum of the impedance of the inductor plus the impedance of the parallel arms

   Z_total = Z_L + Z_|| = jΩL + Z_|| .

   Using Ohm's Law, the current produced by the signal would be

   I = V_in(ac)/Z_total .

   Thus the voltage drop over the inductor is

   V_L = IZ_L = V_in(ac)(Z_L/Z_total) ,

   while the drop over the parallel arms is

   V_|| = IZ_|| = V_in(ac)(Z_||/Z_total) .

   Thus the signal that charges up the capacitor is smaller by a factor Z_||/Z_total than the result in Question 4 where there was no inductor. The quantity Z_||/Z_total is complex. However, the phase shift of the signal is of little consequence. Only the magnitude of the signal is important. Using MAPLE to handle the algebra, that magnitude is

   .

   Note that if the reactance of the capacitor is much small than resistor R, the above expression reduces to

   |Z_||/Z_total| = 1/(Ω²LC - 1) = X_C/|X_L - X_C|,

   the value given in Floyd.

   The value of |Z_||/Z_total| for this problem is |Z_||/Z_total| = 0.01123 . Since the AC signal of Question 4 is scaled by this amount, the new value of V_{r(peak to peak)} with thus be

   V_{r(peak to peak)} = V_{r(pp) in} × |Z_||/Z_total| = (1.948 V)(0.01123) = 0.0219 Volts.

   Similarly the RMS value of the AC signal is scaled

   V_rms = V_{rms in} × |Z_||/Z_total| = (0.563 V)(0.01123) = 0.0063 Volts.

   The ripple factor will be

   r = V_{r(peak to peak)}/V_DC = 0.0219/11.41 = 0.0019 ,

   for a ripple factor of 0.19%. A big improvement over Question 4 although the DC component is substantially reduced.

   

Questions? mike.coombes@kwantlen.ca