Zener diodes as voltage regulators A Circuit diagram symbol for a zener diode. Zener diodes are used in circuits to maintain a fixed voltage across a load. A zener diode is a diode that breaks down at a specific voltage when it is reverse biased. This means that a zener diode will stop a reverse current from flowing through it until the reverse voltage applied across it reaches a fixed value known as the breakdown voltage (VZ). They are designed to ‘breakdown’ in a reliable and non-destructive way so that they can be used in reverse to maintain a fixed voltage across their terminals. Zener diodes can be distinguished from ordinary diodes by their code and breakdown voltage which are printed on them. Zener diode codes begin BZX ... or BZY ... Their breakdown voltage is printed with V in place of a decimal point, so 4V7 means 4.7V. The minimum voltage available is 2.7V. The diagram below shows the voltage versus current characteristic for a zener diode and how they are connected, with a resistor in series to limit the current. (a) The transfer characteristic of a zener diode (b) A zener diode in series with a current limiting resistor. Jacaranda Physics 2, 2nd edition © John Wiley & Sons Australia, Ltd 2004 1 Key term A zener diode is a diode that breaks down at a specific voltage when it is reverse biased. A zener diode can maintain the voltage at its breakdown voltage as long as the supply voltage is above this value. When a zener diode is forward biased it behaves like an ordinary silicon diode. When the zener diode is reverse biased with a voltage less than VZ it blocks the current, like an ordinary silicon diode. When the reverse bias voltage reaches VZ the voltage is limited to around this value, but it can conduct very large currents. If a zener diode is placed in parallel with one of the resistors in a voltage divider circuit, no current flows through the zener diode when the voltage across the resistor is less than the breakdown voltage. When the voltage is equal to the breakdown voltage, some current passes through the resistor, the rest passes through the diode. The result is that the voltage across the diode and the resistor remains constant. This is shown in figure U9.4. A zener diode regulates the voltage drop across a load resistor at 4.7 V as long as Vin is greater than 4.7 V. If the input voltage increases, the current through the voltage divider increases and any increase in voltage happens across the other resistor. Sample problem A 4V7 zener diode is used as a voltage regulator as shown in the diagram below. The input voltage comes from a rectifier and has a ripple that ranges between 8.5 V and 9.3 V. The resistor R1 has a resistance of 100 Ω. Jacaranda Physics 2, 2nd edition © John Wiley & Sons Australia, Ltd 2004 2 (a) Calculate the peak to peak value of the ripple voltage. (b) What is the value of the voltage drop across the load for this input voltage signal? (c) What happens to the magnitude of the current through resistor R1? Solution (a) Vr(p-p) = Vmax – Vmin Vr(p-p) = 9.3 V – 8.5 V Vr(p-p) = 0.8 V (b) 4.7 V (c) The voltage drop across the zener diode and the load remains constant at 4.7 V. The voltage drop across resistor R1 varies between 3.8 V and 4.6 V. I =V/R, so the current varies between 0.038 A and 0.046 A. Additional questions Understanding (before question existing question 17 on page 229) 1. (Before existing question 16 on page 229) (a) What happens to the voltage of an unregulated DC power supply vary when the load applied across its output terminals is decreased? (b) Why does this occur? 2. (Before existing question 17 on page 229) Why can zener diodes be used as voltage regulators? 3. What is the breakdown voltage of a zener diode? 4. Describe a circuit containing a zener diode that can provide a constant voltage across a load from a rectifier. Application Questions 5 to 7 refer to the following information and circuit diagram. A student sets up the circuit shown in the figure below. 5. Vin is 40 V, R1 is 50 Ω, RL is 100 Ω and the breakdown voltage of the zener diode is 20 V. (a) What is the voltage drop across the load resistance RL? (20 V) (b) Calculate the current through the load resistor RL. (0.20 A) (c) What is the voltage drop across R1? (20 V) (d) Calculate the current through R1. (0.40 A) (e) What is the current through the zener diode? (0.20) Jacaranda Physics 2, 2nd edition © John Wiley & Sons Australia, Ltd 2004 3 6. R1 is 4.0 kΩ, RL is 10 kΩ, and the breakdown voltage of the zener diode is 30 V. The input voltage ranges between 70 V and 100 V. (a) What is the range of voltage drops across R1? (40 V to 70 V) (b) Calculate the maximum and minimum currents through R1? –2 (Imax = 1.75 × 10 A (17.5 mA), Imin = 0.010 A (10 mA)) (c) What is the current in the load resistor? (3.0 × 10–3 A (3.0 mA)) (d) Calculate the maximum and minimum currents through the zener diode. –2 (Imax = 1.45 × 10 A (14.5 mA), Imin = 0.007 A (7 mA) ) 7. The voltage drop across RL is 12 V, Vin has values between 20 V and 35 V. The minimum current through RL is 100 mA. and the minimum current through the zener diode is 8.0 mA. (a) What is the rating of the zener diode? (12 V) (b) What is the minimum voltage drop across R1? (8 V) (c) What is the minimum current through R1? (108 mA) (d) Calculate the value of RL. (120 Ω) (e) Calculate the value of R1. (74 Ω) 8. Consider the characteristic curve for a zener diode shown below. (a) What is the current when the diode has a forward bias of 0.8 V? (80 mA) (b) What is the breakdown voltage of this diode? (25 V) (c) What is the power dissipated in this diode when it carries a reverse current of 100 mA? (2.5 W) (d) Describe how this diode could be used to provide a steady voltage of 25 V across a load from an unregulated DC supply. Jacaranda Physics 2, 2nd edition © John Wiley & Sons Australia, Ltd 2004 4.