1 Why Not Use Batteries? (10 pts) HW #4: Power Supply Work this problem in symbols, then clearly state the values of any parameters you need before plugging in to get final numbers. 1.1 How much current would be drawn, on average, by a 4-digit display? Each LED segment of the display draws current when emitting light. Given the amount of current you know the LEDs need (check your previous lab results if you cant remember) A typical current per LED is 10 to 20 ma. There are four displays with seven segments, so 28 LEDs. Half of the segments will be lit at any one time, so on average there are 14 LEDs lit at all times. I avg = 1 2 N LEDI LED = 1 2 (28)(15 10 3 ) = 210mA 1.2 Now consider supplying your clock using AA batteries. How many will you need and in what arrangement? What will be the voltage and current capacity of this combination? Think carefully about which quantities add and which do not. Hint: Two AA batteries should have the capability to supply twice the energy of one AA battery. You can use this to check your answer. We require 5 V and AA batteries provide 1.5 V. So we need four batteries in series such that the voltages add. That will give us 6 V when the batteries are fresh, declining somewhat as they are used. According to the Wikipedia article on AA batteries, a typical charge capacity is 400-900 mah. This will not add - the same current will travel through each battery. Using the hint, if the voltages add and the charge stays constant, then the energy which is voltage times charge increases linearly with number of batteries. If the current also increased, the energy supplied would increase like the square of the number of batteries, which makes no sense. 1.3 Estimate how long a 4 digit clock would run before depleting your batteries. The time to discharge the batteries is Q avg (400 to 900mA h) = = 2 to 4.5 hours I avg 210 ma. In other words, not through a single night. So this would make a lousy alarm clock. One could stack up more batteries in parallel to provide more charge, but it would take 6 sets (24 batteries) to even last 12 hours worst case. This is why we need a power supply. Grading: (3 pts) calculate the average current (2 pts) number of the battery needed and arrangement (2 pts) calculate the voltage and current capacity (3 pts) estimate the running time Version 2.3, September 25, 2014 Page 1
2 RL Circuit Charging (20 pts) 2.1 Assuming that there is no current through the inductor initially, find the current through and voltage on the inductor as a function of time > 0 assuming the switch is closed at t=0. Plot these two functions. At what time does the inductor current reach 90% of its final value? R 1 = 1kΩ V 1 = 10V L 1 = 1mH Solution: I(t = ) = 1V 1kΩ = 10mA τ L = L R = 1mH 1kΩ = 1µs I(t) = I (1 e t τ L ) = 0.01(1 e t 1 10 6 )[A] V (t) = L di dt = I L t e 1 10 6 = 0.9I [V ] τ L 0.9I = I (1 e t τ L ) t = τ L ln(0.1) = 2.6µs Note that the voltage across the inductor starts at 10 V (the source voltage) because no current is flowing, then decays to zero as inductor evolves towards a short. Plots of the two functions are below. (a) I L vs t (b) V L vs t Figure 1 Version 2.3, September 25, 2014 Page 2
Grading: (2 pts) calculate the time constant (4 pts) solve for I(t) (4 pts) solve for V(t) (4 pts) plot the function of I(t) (4 pts) plot the function of V(t) (2 pts) calculate the time that the current reaches 90% of its final value 3 Basic Transformer Operation (10 pts) 3.1 If a transformer delivers 500 ma at 24 V peak from its secondary side when driven with a sinusoidal voltage source with peak voltage of 170 V on its primary side, how many turns are on the secondary if the primary has 3000 turns? What is the primary current? Grading: (2 pts) write N/V relation for the transformer (3 pts) solve for Ns (2 pts) write N/I relation for the transformer (3 pts) solve for Ip Version 2.3, September 25, 2014 Page 3
4 Tapped Transformer Operation (10 pts) A transformer is wound with three taps on both the primary and secondary as shown. The number of turns between each tap, as shown is black to white = 100, white to red = 200, blue to green = 400 and green to orange = 200. A 100 V peak sinusoid is connected between the black and white wires and a 2000 Ohm resistor is connected between green and orange. 4.1 What is the peak current by the source? 4.2 What is the peak voltage between the blue and green wires? 4.3 What is the peak voltage between the white and red wires? Grading (1 pts) write N/V relation for the transformer (3 pts) solve for the peak current delivered by the source (3 pts )solve for the peak voltage between the blue and green wires (3 pts) solve for the peak voltage between the white and red wires Version 2.3, September 25, 2014 Page 4
5 A Load Seen Through a Transformer In the circuit below, the peak current in the secondary circuit, I1 = 100 ma, N1 = 50, and N2 = 40. 5.1 First find the RMS current and voltage on the secondary side and the average power dissipated in the resistor. 5.2 Find the RMS current through and voltage on the primary coil. The find the average power delivered by the source and compare to the average power dissipated in the resistor. Find each quantity in symbols, plugging in numbers only in the final step. Note that the expression for power delivered by the primary is exactly that for power dissipated by the resistor. Grading (2 pts) find the RMS current on the secondary side (2 pts) find the RMS voltage on the secondary side (2 pts) solve for the average power dissipated on the resistor (3 pts) find the RMS current on the primary coil (3 pts) find the RMS voltage on the primary coil (3 pts) solve for the average power delivered by the source Version 2.3, September 25, 2014 Page 5
6 Effeciency of a Bridge Rectifier Consider a full-wave bridge rectifier driven by a 120 V RMS function generator oscillating at 60 Hz (that is, identical to US line voltage). Assume the diodes are silicon with a forward bias voltage drop of 0.7 volts. 6.1 What is the peak voltage of the rectified waveform delivered to the load? First calculate the peak voltage of the source, then subtract the diode forward bias of the two diodes. 6.2 What is the peak current of the rectified waveform delivered to the load? The load current is related to the load voltage by Ohms law. 6.3 How much average power is delivered to the load? The average load power is given by the load RMS voltage and the normal power expression. 6.4 Defining efficiency of the rectifier as power delivered to the load over power expended by the source, what is the efficiency of the rectifier? The source current is equal to the load current, so find the source RMS current and multiply by the source RMS voltage to the find the source power. Version 2.3, September 25, 2014 Page 6
It is instructive to work the entire problem symbolically: From this we see that (since the source and load currents are equal) that the power loss is just due to the voltage drop of the two diodes relative to the peak voltage of the source. Since the 0.7 V drop is small in comparison to 120 V, the diode rectifier is efficient. However, if we were rectifying a 5 V RMS waveform, the efficiency would drop to 80%. Grading: (4 pts) solve for peak voltage of the rectified waveform delivered to the load (4 pts) solve for peak current of the rectified waveform delivered to the load (5 pts) solve for the average power delivered to the load (7 pts) solve for the efficacy of the rectifier Version 2.3, September 25, 2014 Page 7
7 Putting them Together (15 pts) A 120 V RMS, 60 Hz source is connected to pins 1 and 2 of the transformer T1. The primary winding of this transformer has 1000 turns and the secondary winding has 20 turns. The diode rectifier is made from silicon diodes with a forward bias voltage drop of 0.7 volts. 7.1 Calculate the peak voltage and plot the voltage for t = 0 to 50 ms for the input voltage across pins 1 and 2. 7.2 Calculate the peak voltage and plot the voltage for t = 0 to 50 ms for the transformer secondary voltage across pins 3 and 4 Version 2.3, September 25, 2014 Page 8
7.3 Calculate the peak voltage and plot the voltage for t = 0 to 50 ms for and the rectifier output voltage relative to ground. Grading (3 pts) calculate the peak voltage across pins 1 and 2 (2 pts) plot the voltage for t = 0 to 50 (3 pts) calculate the peak voltage across pins 3 and 4 (2 pts) plot the voltage for t = 0 to 50 (3 pts) calculate the rectifier output voltage relative to ground (2 pts) plot the voltage for t = 0 to 50 Version 2.3, September 25, 2014 Page 9