JUNCTION DIODE APPLICATIONS

SLIDES FOR PN JUNCTION DIODE APPLICATIONS

1. Chapter 2: Diode Applications Module-1 Basic Electronics

2. Load Line The fact that this current and the defined direction of conduction of the diode are a “match” reveals that the diode is in the “on” state and conduction has been established. The resulting polarity across the diode will be the forward-bias region. Applying Kirchhoff’s voltage law to the series circuit - E=V D +I D *R I D =E/R when V D =0V V D =E when I D =0A

3. Load Line The Load Line plots all possible current (I D ) conditions for all voltages applied to the diode (V D ) in a given circuit. E/R is the maximum I D and E is the maximum V D . Where the Load Line and the Characteristic Curve intersect is the Q point , which specifies a particular I D and V D for a given circuit.

4. Diode Approximations In Forward Bias: Silicon Diode: V D = .7V Germanium Diode: V D = .3V In Reverse Bias: Both diodes act like opens V D = source voltage and I D =0A

5. Diode in DC Series Circuit: Forward Bias The diode is forward biased. V D = .7V (or V D = E if E < .7V) [Formula 2.4] • V R = E – V D I D = I R = IT = V R /R [Formula 2.5] [Formula 2.6] If a diode is in the “on” state, one can either place a 0.7-V drop across the element, or the network can be redrawn with the V T equivalent circuit

6. Diode in DC Series Circuit: Forward Bias – Equivalent circuit

7. Diode in DC Series Circuit: Reverse Bias The diode is reverse biased. V D = E V R = 0V I D = I R = I T = 0A Mentally replacing the diode with a resistive element will reveal that the resulting current direction does not match the arrow in the diode symbol. The diode is in the “off” state. Due to the open circuit, the diode current is 0 A

8. Diode in DC Series Circuit: Reverse Bias – Equivalent Circuit

9. Diode in any DC Circuit Solve this circuit like any Series/Parallel circuit , knowing V D = .7V (or up to .7V) in forward bias and as an open in reverse bias. V D1 = V D2 = Vo = .7V V R = 9.3V Diodes in parallel are used to limit current: I R = E – V D R = 10V - .7V = 28mA .33k  I D1 = I D2 = 28mA/2 = 14mA

10. Diodes in AC Circuits The diode only conducts when it is in forward bias, therefore only half of the AC cycle passes through the diode.

11. Diodes convert AC to DC in a process called rectification . The diode only conducts for one-half of the AC cycle. The remaining half is either all positive or all negative. This is a crude AC to DC conversion. The DC Voltage out of the diode : V DC = 0.318Vm where V m = the peak voltage [Formula 2.7] Half Wave Rectifier

12. PIV (PRV) Because the diode is only forward biased for one-half of the AC cycle, it is then also off for one-half of the AC cycle . It is important that the reverse breakdown voltage rating of the diode be high enough to withstand the peak AC voltage. PIV (PRV) > V m [Formula 2.9] PIV = Peak Inverse Voltage; PRV = Peak Reverse Voltage V m = Peak AC Voltage

13. Full Wave Rectification The rectification process can be improved by using more diodes in a Full Wave Rectifier circuit. Full Wave rectification produces a greater DC output.

14. Full Wave Rectifier Circuits There are two Full Wave Rectifier circuits: Bridge Rectifier Center –Taped Transformer Rectifier

15. Bridge Rectifier Circuit Four diodes are required. V DC = 0.636 Vm [Formula 2.10]

16. Operation of the Bridge Rectifier Circuit For the positive half of the AC cycle: For the negative half of the AC cycle:

17. Center–Tapped Transformer Rectifier Circuit Two diodes and a center-tapped transformer are required. V DC = 0.636(V m ) Note that V m here is the transformer secondary voltage to the tap.

18. Operation of the Center–Tapped Transformer Rectifier Circuit For the positive half of the AC cycle: For the negative half of the AC cycle:

19. Note: V m = peak of the AC voltage. Be careful, in the center tapped transformer rectifier circuit the peak AC voltage is the transformer secondary voltage to the tap. Rectifier Circuit Summary Rectifier Ideal V DC Realistic V DC Half Wave Rectifier V DC = 0.318V m V DC = 0.318V m - .7V Bridge Rectifier V DC = 0.636(V m ) V DC = 0.636(V m ) – 2(.7V) Center- Tapped Transformer Rectifier V DC = 0.636(V m ) V DC = 0.636(V m ) - .7V

20. There are a variety of diode networks called clippers that have the ability to “clip” off a portion of the input signal without distorting the remaining part of the alternating waveform. Depending on the orientation of the diode, the positive or negative region of the input signal is “clipped” off. There are two general categories of clippers: series and parallel. The series configuration is defined as one where the diode is in series with the load, while the parallel variety has the diode in a branch parallel to the load. Clippers

21. Diodes “clip” a portion of the AC w ave. Although first introduced as a half-wave rectifier (for sinusoidal waveforms), there are no boundaries on the type of signals that can be applied to a clipper. The diode “clips” any voltage that does not put it in forward bias. That would be a reverse biasing polarity and a voltage less than .7V for a silicon diode. Clipper Diode Circuit

22. Steps to analyze the network in presence of DC supply:- Make a mental sketch of the response of the network based on the direction of the diode and the applied voltage levels. The direction of the diode suggests that the signal vi must be positive to turn it on. The dc supply further requires that the voltage vi be greater than V volts to turn the diode on. The negative region of the input signal is “pressuring” the diode into the “off” state, supported further by the dc supply. In general, therefore, we can be quite sure that the diode is an open circuit (“off” state) for the negative region of the input signal. Clippers

23. Steps to analyze the network in presence of DC supply:- Determine the applied voltage (transition voltage) that will cause a change in state for the diode. For the ideal diode the transition between states will occur at the point on the characteristics where v d = 0 V and i d = 0 A. Applying the condition i d = 0 at v d = 0 to the will result in the configuration of Fig., where it is recognized that the level of vi that will cause a transition in state is v i = V Clippers

24. Steps to analyze the network in presence of DC supply:- Be continually aware of the defined terminals and polarity of v o . When the diode is in the short-circuit state, such as shown in Fig., the output voltage v o can be determined by applying Kirchhoff’s voltage law in the clockwise direction: v i -V- v o = 0 v o =v i -V Clippers

25. Steps to analyze the network in presence of DC supply:- It can be helpful to sketch the input signal above the output and determine the output at instantaneous values of the input. Keep in mind that at an instantaneous value of vi the input can be treated as a dc supply of that value and the corresponding dc value (the instantaneous value) of the output determined. Clippers

26. By adding a DC source to the circuit, the voltage required to forward bias the diode can be changed. Variations of the Clipper Circuit

27. By taking the output across the diode, the output is now the voltage when the diode is not conducting. A DC source can also be added to change the diode’s required forward bias voltage. Changing Output Perspective

28. Clipper Circuits Summary

29. Clipper Circuits Summary

30. A diode in conjunction with a capacitor can be used to “clamp” an AC signal to a specific DC level. Clamper Diode Circuits

31. A diode in conjunction with a capacitor can be used to “clamp” an AC signal to a specific DC level. Clamper Diode Circuits

32. The input signal can be any type of waveform: sine, square, triangle wave, etc. You can adjust the DC camping level with a DC source. Variations of Clamper Circuits

33. Summary of Clamper Circuits

34. The Zener is a diode operated in reverse bias at the Zener Voltage (V z ). Zener Diode

35. Zener Calculations Determine the state of the Zener: if Vi  Vz , then the Zener is biased “on” ; the Zener is at Vz if Vi < Vz , then the diode is biases “off” ; Vz = Vi For Vi  Vz: The Zener voltage [Formula 2.16] The Zener current 2.18] The Zener Power I Z = I R - I L [Formula P Z = V Z I Z [Formula 2.19] For Vi < Vz: The Zener acts like an open.

36. Zener Calculations

37. The size of the load resistor affects the current in the Zener. RL is too large Not enough current through the Zener and it is biased “off”. The minimum current for a Zener is given as IZK in the data sheets. [Formula 2.25] [Formula 2.26] RL is too small Too much current in the Zener and it avalanches and is quickly destroyed. The maximum current for a Zener is given as I ZM in the data sheets. [Formula 2.21] [Formula 2.20] Load Resistance in a Zener Circuit I Lmin  I R - I ZM R Lmax  V Z I Lmin R L R LMIN V Z I Lmax  V L  Vi  V Z RV Z R LMIN 

38. Voltage Multiplier Circuits Voltage multiplier circuits use a combination of diodes and capacitors to step up the output voltage of rectifier circuits. Voltage Doubler • Voltage Tripler Voltage Quadrupler

39. Slide 27 Voltage Doubler This half-wave voltage doubler’s output can be calculated as V out = VC2 = 2V m V m = peak secondary voltage of the transformer.

40. Operation of a Voltage Doubler Circuit The 1 st capacitor charges up to V m during the positive half of the cycle, then the 2 nd capacitor charges up to V m in the same polarity as the 1 st capacitor, finally the output is the sum of the voltages across both capacitors: Vout = 2V m

41. Voltage Tripler and Quadrupler Circuits By adding more diode-capacitor networks the voltage can be increased.

42. Practical Applications of Diode Circuits Rectifier Circuits Conversions of AC to DC for DC operated circuits Battery Charging Circuits Simple Diode Circuits Protective Circuits against Overcurrent Polarity Reversal Currents caused by an inductive kick in a relay circuit Zener Circuits Overvoltage Protection Setting Reference Voltages