Eletronics Lab Report - Rectifier Circuit 104e5m

EGR220

Than & Bhavin

Introduction and Objectives One of the most applications of diodes are rectifier circuits, which converts AC to DC. In fact, diodes rectifier forms the fundamental building block of the DC power supplies for electronic devices [1]. Therefore, in this lab, we were instructed to measure and analyze behavior of rectifier circuits, both halfwave and full wave (bridge) rectifier and peak rectifier. The primary objectives of this lab are: 1. To analyze and understand the nature of halfwave rectifier 2. To understand about the nature of bridge rectifier and peak rectifier 3. To be able to approximate ripple voltage by using ideal diode model 4. To be able to understand the effect of capacitance on peak rectifier circuits

Equipments and Components used In this lab, the equipments and components we used are:- Power Diodes: 1N4004 (x5); Transformer [12.6V, 1A]; Resistors: 1KΩ @ 1W (x1), 2KΩ @ 1W (x1), 100Ω @ 3W, 200Ω@ 2W (x1), 510Ω@ 1W (x1); Capacitor: 470μF @ 50V (x5) a breadboard, a waveform generator, ±20V power supply, a multimeter, an Oscilloscope to capture the I-V curve, wires and cords.

Lab #3

Figure 1 First, we checked the secondary voltage of the transformer, and we found that Vpp is 20V, and the average voltage is 12.7V and the current is about 1 mA. We, then, estimated the max current and peak inverse voltage across the diode. PIV (Peak Inverse Voltage) was actually the peak value of the secondary voltage of the transformer, which is 10V. The maximum current was calculated from V0/R which was 9.3 mA, where V0 ≈ Vp – VD and VD = 0.7 V (We assumed that rD << RL)

In the prelab, we calculated the average voltage of V0 by using the following formula: Avg V0 = (1/π)Vp – VD / 2

(1)

where VD = 0.7 V (Ideal Model). We got 2.83V for average V0. Average power can be calculated by Avg P = (Avg V0 )2 / RL

(2)

Therefore, average power dissipated by RL was 8.03 mW. With Oscilloscope, we captured the screen image of both Vs (Secondary Voltage of Tranformer) and V0.

Procedures Procedure 1: Analyzing the Nature of of Half-

Figure 2: 2 Waveforms of Vs (10Vpp) and Vo (1KΏ)

Wave Rectifier Circuits

From the screen image, we found that there was an offset between output voltage and Vs. By moving cursor, we measured the gap between two wave forms,

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EGR220

Than & Bhavin

and found that it was almost 0.7V. Then we measure the average output voltage with multimeter, and found that the average output voltage 2.78V. When compared to the calculated value, which was 2.83V, we found that there was error of 0.05, which might be the result of cumulative errors built up from approximation used in ideal model.

Lab #3

We built the circuit as in figure 4, and captured the screen image of output voltage and source voltage.

Then, we doubled RL analyze the effect of load resistance on the circuit. From our prediction from prelab, we assumed that the increment on load resistance would not affect the output voltage, as long as diode resistance is far much smaller than load resistance. We captured the screen image, and then analyzed and compared the result with our prediction. We found that the amount of change in output voltage was negligible.

Figure 5: 2 Waveforms of Vs (10Vpp) and Vo (1KΏ and 470 uF) In the prelab, we calculated the ripple voltage Vr by using the following formula: Vr = Vp / fCR

(3)

We got Vr = 0.35V. The output voltage was calculated from Avg V0 = Vp – 0.5Vr

Figure 3: 2 Waveforms of Vs (10Vpp) and Vo (2KΏ)

Procedure 2: Analyzing the Nature of of HalfWave Peak Rectifier Circuits

(4)

We got avg V0 = 9.82 V. We measured both output voltage and ripple voltage by using multimeter, we found that avg V0 = 9.77V and Avg Vr = 0.19V, and that the measurement results were not very far from the prelab calculation. Then, we replaced RL with 100Ώ, 200Ώ and 510Ώ to analyze the effect of load resistance on the circuit. Since load current IL is V0 /RL, when frequency and capacitance is kept constant at 470 uF, Vr is directly proportional to load current (Assume Vp ≈ V0). Vr = IL / fC

(5)

In prelab, we calculated the value of IL and Vr and plotted the data on the I-V graph.

Figure 4

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EGR220

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Ripple Voltage

Lab #3

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Figure 7: Ripple Voltage Vs. Capacitance

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Figure 5: Ripple Voltage Vs. Load Current

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Figure 8: Measured Ripple Voltage Vs. Capacitance

Current

Although ripple voltage we measured was average value, the graph showed the inverse proportionality. Figure 6: Measured Ripple Voltage Vs. Load Current We also plotted the measure load current and average ripple voltage with multimeter, and the curve showed that it is pretty much linear.

Procedure 3: Analyzing the Nature of of FullWave Bridge Peak Rectifier Circuits

Again, we kept constant for load resistance at 100Ώ, and changed the capacitance, and analyzed the ripple voltage. We predicted that the ripple voltage would be inversely proportional to the capacitance.

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EGR220

Than & Bhavin

Figure 8 In bridge configuration, neither side of terminal of the transformer is grounded so that during each half-cyle (positive or negative), the current flows from D1 through RL to D2 or from D4 through RL to D3.

Lab #3

The non-linearity nature of diodes are very useful in building rectifier circuits. Although we used the ideal model of diode for Half-Wave Peak rectifier and Full-Wave Peak rectifier, we found that the approximations were pretty close to observations as the supply voltage was far much greater compared to the voltage of diode. However, if the supply voltage is much close to that of diode, the ideal model of diode will not work and we need to use more elaborate model such as constant voltage drop or piece-wise linear.

References [1] Sedra, Adel S., and Smith. Kenneth C. “Microelectronics Circuits”. 5th. New York: Oxford University Press, 2004.

Figure 9: Ripple Voltage of Bridge Peak Rectifier We built the circuit as in figure 8, and captured the screen image of average output voltage and figured out the ripple voltage. In the prelab, we calculated the ripple voltage Vr by using the following formula: Vr = Vp / 2fCR V0 = Vp – 0.5Vr

(6) (4)

We got Vr = 0.175V, and V0 = 9.9V. From our measurement, we got Vr = 0.17V, and V0 = 10V. When compared the ripple voltage of bridge rectifier with that of half-wave rectifier, we found that bridge's ripple voltage was half of the half-wave's ripple voltage.

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