Electronics 1 Lab (CME 2410)

Transcription

1 Electronics 1 Lab (CME 241) School of Informatics & Computg German Jordanian University Laboratory Experiment (4) -Diode Applications & Power Supply 1. Objective: - To detere the load fluence on the output voltage of a -diode stabilized power supply. - To detere the put voltage fluence on the output voltage of a -diode stabilized power supply 2. Theory: Almost every electronic circuit needs a stabilized power supply, which provides a constant voltage dependent of put voltage changes delivered from the rectifier or the battery as well as dependent of a changg load. In all of these circuits the -diode is the core of this circuit. In low power supplies stabilizg is simply done by a current limitg resistor 1 followed by the -diode V1 (see Fig. 5.1). To meet higher demands the current limitg resistor is replaced by an electronic current source and to feed higher loads there is the need of a controlled current amplifier. The -diode is still needed as a voltage reference element. Current source as well as the current amplifier makes the power supply circuit lookg more sophisticated. Simple -diode stabilization: The diagram of the basic circuit is shown Fig. 5.1 with 1 as the current limitg resistor, as the load and V1 as -diode. 1 ectifier U V1 U o Fig. 5.1 Power supply with -diode stabilization 1/7

2 As the -diode is always used the reverse biased mode it is more convenient to display the U/I-characteristic the form I = f( U) stead the forward biased mode IF = f( UF) as we do with normal diodes. (We prefer the first quadrant to handle with positive numbers.) Fig. 5.2 shows the U/I-characteristic of a -diode. The dividual data sheet will specify the values for the -voltage U which is defed at I = I (for general purpose -diodes I = 5 ma) and the peak current I = I. Instead of I the data sheet might provide the imum power P = I V the -diode can tolerate. In between the limits I and I the -diode will work as a stabilizg element. I /ma 1 I I U /V U Fig. 5.2 Example for the U/I-characteristic of a -diode ( U = 6 V; I = 1 ma) How to choose the current limitg resistor 1: The current limitg resistor 1 has to be chosen a way that under operatg conditions it ensures the -diode will work between the limits: I > I (otherwise there is no regulation any more) and I < I (otherwise we could have the diode break-down). The operation conditions are: imum put voltage: imum put voltage: U, U, imum load: it should be the open circuit: I =, U imum load (imum load current): I =. If the condition I < I < I is fulfilled the output voltage U is nearly constant and equal to U. The 2 limits I and I result 2 values for the current limitg resistor 1 and 1. Only one value between these limits has to be chosen (see below). 2/7

3 In Fig. 5.3 the graphical solution is given to detere the current limitg resistor 1 under different conditions. I The put voltage U /ma followed by the current 1 I limitg resistor 1 are 1 regarded as a voltage source. 8 The U/I-characteristic of a source is given by a straight le detered by the open 6 loop voltage U = U (I =, without the load) and the 4 resistance 1 (voltage drop at I +I 1 dependg on I): 1 2 U = U 1 I. The meetg pot between I U /V the straight le (voltage source) and the given nonlear U/I-characteristic is the U U U operatg pot of the Fig. 5.3 Graphic solution for the current limitg resistor -diode. As we have two different operation conditions for the put voltage ( U and U ) we will get two different straight les (see Fig. 5.3) with two different values for the current limitg resistor 1. (a) 1 : For imum load (open loop) all the current will flow across the -diode. The actual - diode current might exceed the peak current I when the imum put voltage U is applied. Under this condition the current limitg resistor 1 is allowed gettg lager but never smaller. We are gog to detere the imum value 1. To detere the numerical value you can use the triangle Usg the values out of Fig ΔU U = ΔI I U 1 13 V 6 V 1 ma = 7Ω (b) 1 : For full load most of the current will flow across the load and the -diode is on the verge of gettg not enough current I. This will get worse for the imum put voltage U. The current limitg resistor 1 is allowed gettg smaller but never larger. We are gog to detere the imum value 1. To fd the graph of the put voltage source workg with this value 1 we need to regard the -diode parallel to the full load as the total load for the source. This is simply done by addg up the imum -diode current I and the load current I at the 3/7

4 -voltage U. The straight le between U (on the U -axis) and I + I (at U ) deteres the imum value 1. To detere the numerical value you can use the triangle Usg the values out of Fig = ΔU U U = = Δ I I + I 1 V 6 V 5 ma + 25 ma = 133Ω (c) Choose the best value for 1: The circuit can only be realized with one value for 1 between the limits: 1 < 1 < 1. Out of the Fig. 5.3 it can be seen that small changes Δ U result a shift of the operation pot and hence a change of Δ U =Δ U, the stabilized output voltage. To get the smallest change output voltage Δ U while acceptg big changes put voltage Δ U we defe a smoothg factor SF which should be as high as possible. Usg the voltage divider rule we get SF Δ U 1+ r ΔU r = = = Maximum It is obvious that 1 has to be as high as possible while r should be as small as possible. We should choose 1= 1 - Tolerance, always keepg d that all resistors have a range of tolerance (maybe 1%). To get the best stabilization we should use -diodes with a low differential resistance r. Unfortunately these -diodes come with a very low value for I. Improvements If the resistor 1 is replaced by a nonlear current source with r 1 1 we almost get a horizontal le and changes of the put voltage don t affect the output voltage anymore. The regulation action would be ideal if the U/I-characteristic of the -diode was perfectly vertical. Because of the fite value for r this curve presents a certa clation and the voltage at the diode ends can vary a little when the load current I changes. If we use a current amplifier followg the -diode the load of the -diode will nearly be constant and the operation pot will be fixed. The clation of the -characteristic has less fluence on the output voltage. There remas only the shift of the -characteristic due to temperature. Usg a -diode workg between the avalanche and the field effect (5 V < U < 6 V) the -diode shows a imal temperature coefficient. 4/7

5 3. Equipment & Instruments: - Module No. : DL 3155M12 - Function Generator - Oscilloscope 4. Components List: 1 = 22 Ω - 1/2W - 5% 2 = 22 Ω - 1/2W - 5% 3 = 1 kω manual regulation trimmer V1 = -diode - 6.2V - 1W Electrical Diagram I Fig Procedure: Insert the Module 12 the console and set the ma switch to ON; Uo = f(i ) CHAACTEISTIC - connect the jack 3 of the power supply to the jack 1 of the regulator and set the switches S1 to OFF and S2 to ON; - sert the positive teral of the digital voltmeter, set to direct current, the jack 3 of the regulator and the other one the ground jack; - ead the value of the no-load voltage with the load 2-3 disconnected and write the value Tab. 5.1.; - sert the positive teral of the digital milliammeter the jack 3 of the regulator and the other one the jack 4; - adjust the 3 value so to read on the milliammeter a current of 6 ma; - read the voltage value on the digital voltmeter and write it Tab. 5.1; - epeat the procedure for all the current values written Tab. 5.1.; - draw Fig. 5.5a the output voltage diagram as a function of the load current Uo = f(i ); - move the positive teral of the digital milliammeter the jack 3 of the power supply and the other one the jack 1 of the regulator; - read the value of the put current and verify that this one remas constant with the load variation (defe the measured quantity the circuit diagram Fig. 5.4); 5/7

6 Uo = f(ui) CHAACTEISTIC - turn, completely counterclockwise the potentiometer 3 so to brg to the imum value the current on the load; - adjust the voltage +V for all the values written Tab.5.2. and for every value of the put voltage survey the correspondg value of the output voltage (defe the quantity +V the circuit diagram Fig. 5.4); - draw Fig.5.5-b the diagram of the output voltage as a function of the put voltage Uo = f(ui); 6. esults: U /V U /V I / ma U i / V Fig. 5.5 I / ma Uo / V Tab. 5.1 Uo = f (I load), Ui = const U i (+V) / V Uo / V Tab. 5.2 Uo = f (Ui), full load 7. Conclusion: In this conclusion you should compare the measured range of stabilization with the theoretical values you get out of the graphical solution /7

7 8. Questions: If the put voltage applied to the - regulator is lower than 6.2 V and there is no load what is the value for the output voltage? 1. The same as the put one 2. V 3..6 V V If the load resistance is disjoted and the voltage +V is equal to 15 V, the current the -diode is approximately: 1..6 m A 2. 1 m A 3. 2 m A 4. 4 m A The power dissipated from the -diode the circuit is higher when: 1. The -diode is short-circuited 2. The load resistance is disjoted 3. The load resistance is short-circuited 9. Design: In the followg design exercise, refer to the design that was completed Lab 2. A. Design a power supply circuit with a bridge full-wave rectifier, a filter and a regulator. The put is 24Vrms AC signal (5 Hz) and the output is a 4.7V DC signal. B. Build and test the circuit. (Do not use a transformer. Take a stepped-down signal directly from the function generator). C. Experimentally detere the range of load resistance such that your power supply circuit matas voltage regulation with +/- 2% of desired output voltage. 7/7