Chapter 2. Diode Applications Outline:  Load-line Analysis  Logic Gates  Waveform Modification Circuits  Zener Diodes Diode Applications

Methods  Developing a working knowledge of a device through the use of appropriate approximation.  Avoiding an unnecessary level of mathematical complexity.

Diode Applications

Load-line Analysis Example: For the series diode configuration, employing the diode characteristics, determine: (1). VDQ and IDQ. (2). VR.

Diode Applications

Solution: For the knowledge of Circuit Analysis, we get: E  VD  VR  VD  I D R

So

E VD ID   R R

This means that ID is a function of VD , i.e. Diode Applications

I D  f (VD )

Furthermore, it’s a linear relationship. Recalling the diode characteristics , it’s also the relationship between ID and VD. So ID and VD should satisfy both the the network parameters and characteristics at the same time. Diode Applications

The diode characteristics is ready here. And our work is to draw the linear relationship between ID and VD on the same set of axes. For straight line, two points are sufficient to determine it. The first point is: I DQ

E 10V    20mA R VD 0V 0.5k Diode Applications

The other point is: VDQ  E I

D  0V

 10V

From the two points, we get the straight line. The straight line is called a load line because the intersection on the vertical axis is defined by the applied load R. Diode Applications

The intersection of the two curves will define the solution for the question and also define the current and voltage for the network. The intersection, the point of operation, is usually called the quiescent point, denoted by Q-point. The analysis method is therefore called loadline analysis. Diode Applications

From the intersection between the load line and the characteristic curve, we may read off that: I DQ  18.5mA VDQ  0.75V

Note here that, ID and VD are now referred to as IDQ and VDQ due to that operation point is Q-point. Diode Applications

The remaining problem is VR: VR  I D  R

 18.5mA  0.5k  9.25V

Or VR  E  VDQ  10V  0.75V  9.25V Diode Applications

Figure: Diode circuit Diode Applications

Figure: Characteristic Diode Applications

E/R Q-point

IDQ

Load-line

VDQ

E

Figure: Characteristic & load-line Diode Applications

Notes:  The characteristics are defined by the chosen device.  The accuracy is determined by the scales, but the one with much higher accuracy would be unwieldy.  Different approximation models, diode equivalent circuits, would be used depending on different degrees of accuracy required. Diode Applications

 The load line is determined solely by the applied network.  It’s feasible when nonlinear curve is involved and a “pictorial” solution is obtained without lengthy mathematical derivations.  The load-line analysis is also useful to BJT and FET in the following chapters. Diode Applications

OR Gates Assumptions:  10-V level is assigned a “1” and 0-V a “0” for Boolean algebra.  The focus is to determine the voltage level instead of Boolean algebra.  The voltage across the diode must be 0.7V positive to switch to the “ON” state. Diode Applications

Example: As shown in the figure, analyze the circuit to verify that input V1, V2 & output VO voltages conform to OR logic. Proof: There would be four cases: (1). When V 1 =10V, V 2 =0V, Diode Applications

Because V 2 =0V, D2 is in “OFF” state. Then D2 can be regarded as open circuit. Thus, V 1 =10V, D1 is in “ON” state. So we obtain: VO = V 1 - VD = 10-0.7=9.3V This means VO is in logic “1”. Diode Applications

(2). When V 1 =0V, V 2 =10V, The result is the same as in (1), in which VO is in logic “1”. (3). When V 1 =10V, V 2 =10V, The result is obvious that VO is almost 9.3V, and still in logic “1”.

Diode Applications

(4). When V 1 =0V, V 2 =0V, Apparently, that VO is 0V, and in logic “0”. Briefly, the I/O relation is: V2

V1

1 (10V)

0 (0V)

1 (10V)

1 (9.3V)

1 (9.3V)

0 (0V)

1 (9.3V)

0 (0V)

Diode Applications

Figure: OR gate Diode Applications

Figure: OR gate in case 1 Diode Applications

AND Gates Example: As shown in the figure, analyze the circuit to verify that input V1, V2 & output VO voltages conform to AND logic. Proof: There would be four cases: (1). When V 1 =10V, V 2 =0V, Diode Applications

Because V 1 =10V, D1 is in “OFF” state. Then D1 can be regarded as open circuit. Thus, V 2 =0V, D2 is in “ON” state. So we obtain: VO = V 2 + VD = 0+0.7=0.7V This means VO is in logic “0”. Diode Applications

(2). When V 1 =0V, V 2 =10V, The result is the same as in (1), in which VO is in logic “0”. (3). When V 1 =10V, V 2 =10V, Both D1 and D2 are in “OFF” state. There is no current in network. Thus, V O =10V, is in logic “1”. Diode Applications

(4). When V 1 =0V, V 2 =0V, Apparently, Both D1 and D2 are in “ON” state. VO = 0.7V, and in logic “0”. Briefly, the I/O relation is: V1

1 (10V)

0 (0V)

1 (10V)

1 (10V)

0 (0.7V)

0 (0V)

0 (0.7V)

0 (0.7V)

V2

Diode Applications

Figure: AND gate Diode Applications

Figure: AND gate in case 1 Diode Applications

Half-Wave Rectification  The input is the simplest time-varying signal, sinusoidal waveform.  The ideal model of diode is used, i.e., the “ON” state can be replace with short circuit and “OFF” state open circuit. Diode Applications

 The input sinusoidal waveform has peak value Vm, and period of T.  The average value of input sinusoidal waveform is zero.  The circuit is called half-wave rectifier and will generate a waveform vo which will have an average value.  The half-wave rectifier is useful in power supply circuits. Diode Applications

 During the first half cycle, the applied voltage vi is forward bias voltage and the diode is in “ON” state.  Replacing the diode with short circuit, the output vo is exact copy of the input signal.  During the 2nd half cycle, the applied voltage vi is reverse bias voltage and the diode is in “OFF” state. Diode Applications

 Replacing the diode with open circuit, the output vo is zero.  The figure shows the input & output signals for comparison purpose.  The output signal vo now has a net positive area above axis over a full period and an average value is 0.318Vm. Diode Applications

Figure: Half-wave rectification Diode Applications

Figure: Half-wave rectified signal Diode Applications

Full-Wave Rectification Bridge Network: With four diodes in a bridge configuration, the sinusoidal input can be used 100%. And this circuit is referred to as full-wave rectification.

Diode Applications

During the first half cycle, v i is positive. Thus, D2 and D3 are in “ON” state. And D1 and D4 are in “OFF” state Applying diode equivalent circuit in the bridge network, we obtain that: vo = v i Diode Applications

During the second half cycle, v i is negative. Thus, D2 and D3 are in “OFF” state. And D1 and D4 are in “ON” state Applying diode equivalent circuit in the bridge network, we obtain that: vo = -v i Diode Applications

So with bridge network, the two halves of sinusoidal waveform have been used. And the average value of full-wave rectification is 0.636Vm , twice that of half-wave rectification. Also, there are other configurations to fulfill full-wave rectification. Diode Applications

Figure: Full-wave rectification Diode Applications

Figure: Diode states in the first half of cycle Diode Applications

Figure: Diode equivalent circuit in the first half of cycle Diode Applications

Figure: Diode states in the 2nd half of cycle Diode Applications

Figure: Diode equivalent circuit in the 2nd half of cycle Diode Applications

Vm Vdc =0 V

Vm Vdc =0.636 Vm

Figure: Output of bridge network Diode Applications

Clippers Clippers are networks that employ diodes to “clip” away a portion of an input signal without distorting the remaining part of the applied waveform. The half-wave rectifier is the simplest form of diode clipper. There are two general categories of clippers: series & parallel. Diode Applications

Series Clippers The series configuration had been introduced in half-wave rectification. Moreover, depending on the orientation of the diode, the positive or negative region of the signal is “cut” off. It is also useful when applied with more types of input signals. Diode Applications

Figure: Series clipper Diode Applications

Figure: Signals which are clipped off Diode Applications

Example: As shown in the figure, an additional dc supply is added in the circuit. Plot the output waveform vo. Solution:  Without the dc supply, it is easy to analyze.  Without the diode, it is also simple. Diode Applications

 Introducing a new variable, vi’, the question can be divided into two steps. (1) The relationship between vi’ and vi . (2) The relationship between vo and vi’.

Diode Applications

 The first step: it’s easy to obtain: vi’= vi - V  The 2nd step: it’s a half-wave rectification. The output is the portion of vi’ which is above the abscissa.

Diode Applications

Figure: Series clipper with a dc supply Diode Applications

Figure: Analysis of clipper with a dc supply Diode Applications

Figure: Output of clipper with a dc supply Diode Applications

Parallel Clippers The simplest of parallel diode configuration is shown in the figure. As before, the equivalent circuit for “ON” state is short circuit and “OFF” state open circuit. Various types of input signals can be clipped off by parallel diode circuit. Diode Applications

Figure: Parallel clipper Diode Applications

Figure: Response to a parallel clipper Diode Applications

Example: As shown in the figure, an additional dc supply is added in the circuit. Plot the output waveform vo. Solution:  Pay attention to the positive terminal of diode, the red point, the voltage here is always 4V. Diode Applications

 Also the negative terminal of diode is just the output, the green point, the voltage here is vo.  The states of diode are determined by the voltage between the red and green points.  If vo > 4 V, the diode is in “OFF” state. Then the diode can be regarded as open circuit, so vo = vi . Diode Applications

That is , vo = vi ,when vi > 4V .  If vo < 4V ,The states of diode is in “ON”. So the diode can be regarded as short circuit, that is vo = 4 V. This means that vo can not be less than 4 V and should only equal to 4 V .  This diode condition is no-bias actually. Diode Applications

Figure: Parallel clipper with a dc supply Diode Applications

Figure: Output of parallel clipper with a dc supply Diode Applications

Clamper A clamper is a network constructed of a diode, a resistor, and a capacitor that shifts a waveform to a different dc level without changing the appearance of the applied signal.

Diode Applications

Note: The selected resistor and capacitor must satisfy that the time constant, τ = RC, is sufficiently large so that the capacitor does not discharge significantly during the interval which the diodes is non-conducting.

Diode Applications

Example: As shown in the figure, plot the output waveform vo of the clamper circuit. Solution:  While in the first half, vi = V, If assume the diode is in “ON” state.

Diode Applications

The capacitor will charge up instantaneously to voltage V. The output vo = 0. If assume the diode is in “OFF” state. The capacitor will also charge up to V. The output vo = 0. Actually, diode is non-biased. Diode Applications

 While in the 2nd half, vi = -V At the instant when vi switches to -V, the capacitor holds its original voltage V and keeps it due to a large τ. This means that vi - vo =V and vi = -V So we get vo = -2V. Also actually, the diode is in “OFF” state.

Diode Applications

Figure: Clamper and its input signal Diode Applications

Figure: Output of a clamper circuit Diode Applications

Zener Diodes As shown in the figure, each of the three portions of characteristics of Zener diodes has approximation model. (1) Forward-bias: There exists a negative power supply of about 0.7 V. This is only occasional case. Diode Applications

(2) Reverse-bias: The equivalent circuit is open circuit. (3) Zener region: The equivalent circuit is a power supply with potential of VZ. VZ can be used as a part of protection circuit. Diode Applications

Figure: Approximation model of Zener diodes Diode Applications

Example 2.17: (1) For the Zener diode regulator, determine VL , VR , IZ and PZ . (2) With RL = 3k, repeat the above problem. Solution: 1. Firstly, determine the state of the Zener diode. Diode Applications

Remove it from the network and calculate the voltage across the resulting open circuit. So we get:

RL V Vi R  RL

1.2k  16V 1k  1.2k

 8.73V Diode Applications

Since V=8.73V, less than VZ , so the Zener diode is in “OFF” state. So the equivalent circuit is open circuit. Then we get: VL  V  8.73V VR  Vi  VL  16V  8.73V  7.27V I Z  0 and PZ  0 Diode Applications

2. While RL = 3k, the same as before, determine the state of the Zener diode. RL V Vi R  RL

3k  16V 1k  3k

 12V

Diode Applications

Since V=12V, greater than VZ , so the Zener diode is in Zener region. So:

VL  VZ  10V VR  Vi  VL

 16V  10V  6V 6V VR  6mA IR   R 1k Diode Applications

VL 10V IL   3.33mA  RL 3k

IZ  IR  IL

 6mA  3.33mA  2.67mA

And the power dissipated in Zener diode is PZ  I R  VR

 2.67mA  10V  26.7mW Diode Applications

Figure: Zener diode regulator Diode Applications

Figure: Determining V for the regulator Diode Applications

Figure: “ON” state of Zener diode regulator Diode Applications

Summary of Chapter 2 • Load-line analysis, quiescent point • Simple logic gates: OR gate & AND gate • Half-wave rectification & full-wave rectification • Series clippers & parallel clipper • Clamper • Zener diodes Diode Applications