Shockley Diode

Electron Devices

Chapter 6: Thyristors, UJT and Optoelectronic Devices

Shockley Diode

• The Shockley diode is a thyristor. It has two terminals, the anode and the cathode and four semiconductor layers. Fig. 6.1.1 (a) shows the basic construction showing p‒n‒p‒n structure and three junctions of Shockley diode. The Fig. 6.1.1 (b) shows the two transistor analogy while Fig. 6.1.1 (c) shows the symbol of the Shockley diode.

• Fig. 6.1.2 shows the equivalent circuit of the Shockly diode using two transistors. The collector of Q₁ and base of Q₂ are connected. J₁ is emitter base junction of Q₁. J₂ is the common connected base collector junction between Q₁ and Q₂. J₃ is base emitter junction of Q₂. For the linear operation, J₁ and J₃ must be forward biased as the base emitter junctions while the junction J₂ must be reverse biased as collector base junction.

Basic Operation

• Consider the biasing provided to the Shockley diode as shown in the Fig. 6.1.3 (a). The various currents, due to the biasing arrangement are shown in the Fig. 6.1.3 (b).

• As per the requirement, junctions J₁, J₃ are forward biased while the junction J₃ is reverse biased. So both the transistors are operating in linear region.

• Expression for I_A: For low forward biased voltages let us obtain the expression for I_A using standard transistor relations. Considering the reverse leakage currents I_CBO1 and I_CBO2 through reverse biased collector base junctions, we can write,

 I_B1 = I_E1‒I_C1 ‒I_CB01

But

I_C1= α₁I_E1

I_B1 = I_E1 ‒ α₁I_E1 ‒ I_CB01 = (1‒α₁)I_E1‒I_CB01

Now

I_E1 = I_A = Anode current

I_B1 = (1‒α₁)I_A ‒ I_CB01

For transistor Q₂,

 I_C2 = α₂I_E2 +I_CBO2

But

I_E2 = I_K = Cathode current

I_C2 = α₂I_K + I_CBO2

But

I_C2 = I_B1

         ………….. Connected together

(1‒α₁)I_A ‒ I_CB01 = α₂I_K +I_CBO2

But

I_A = I_K

         ……………Current entering and leaving

 (1‒α₁)I_A ‒ I_CB01 = α₂I_A +I_CBO2

I_A [1‒ α₁‒ α₂] = I_CB01 + I_CBO2

 I_A =  [ I_CB01 + I_CBO2 ] / [1‒(α₁+α₂)]

where α₁ and α₂ are the d.c. current gains for the transistors, the values of which are very small. The reverase currents I_CBO1 and I_CBO2 are also small. Hence for small bias levels, the diode is said to be OFF which is its forward blocking region.

Characteristics of Shockley Diode

• We have seen that when the Shockley diode is forward biased, it is essentially OFF and acts as an open switch. This region for small forward bias level for which the diode remains OFF is called forward blocking region. The device has very high forward resistance, ideally ∞. The region continues from V_AK = 0 to a specific voltage called forward breakover voltage V_{BR (F)}. This is shown in Fig. 6.1.4.

• As V_AK is gradually increased from zero, the anode current increases gradually. Thus values of α₁ and α₂ also increase. At a particular point, α₁+a₂ =1 and hence denominator of equation of I_A becomes zero. This current is denoted as I_S and called switching current. The corresponding voltage V_AK is called forward breakover voltage. The transistors Q₁ and Q₂ are driven into saturation. Due to this, the V_AK suddenly decreases to the low value which is equal to V_BE +V_CE(sat). The I_A further increases. This region is called forward conduction region of the Shockley diode. The resistance of diode is very small in this region and the device acts as a closed switch. To turn OFF the device, the anode current must be reduced below a specific value, called holding current, I_H. When I_A is reduced below I_H, the device becomes OFF and enters into forward blocking region.

• The current at which the device switches from the forward blocking region to the forward conduction region, is called switching current, I_S. The value of I_S is always less than I_H.

Ex. 6.1.1: A certain Shockley diode is in its forward blocking region. The d.c. a values for the two transistors are α₁ = 0.3 and α₂ = 0.4. The leakage currents for both are 125 nA. Calculate the anode current. If the anode to cathode voltage is 15 V, calculate forward resistance of the diode.

Solution:

α₁= 0.3, α₂ = 0.4, I_CBO1 = I_CBO2 = 125 nA, V_AK = 15 V

I_A =  [ I_CB01 + I_CBO2 ] /  [ 1‒(α₁+α₂) ]

= [125×2×10^‒9]  / [1‒(0.3+0.4)]

= 0.833 μ Α

and

R_f = V_AK / I_A

= 15 / 0.833×10^‒6

= 18 MΩ

Thus the forward resistance of shockley diode in the forward blocking region is very very high and it acts as an open switch.

Ex. 6.1.2: Find the value of anode current for the device shown in the Fig. 6.1.5 which is ON. V_BR(F) = 40 V. Assume V_BE = 0.71V and V_CE(sat) = 0.15 V for the internal transistor. Also find the forward resistance of the diode.

Solution:

V_AK = V_BE + V_CE(sat)

= 0.71+ 0.15 = 0.86 V

         ... Device ON

V_RS = V_BIAS ‒V_AK

= 55‒0.86 = 54.14 V

I_A = V_RS / R_S

= 54.14 / 12×10³

= 4.511 mA

R_f = V_AK / Ι_Α

= 0.86 / 4.511×10^‒3

= 190.64 Ω

So R_f in ON state is very low.

Application

• Fig. 6.1.6 shows the use of the Shockley diode in a relaxation oscillator.

• When the switch S is closed, the capacitor starts charging through R. When the voltage across capacitor reaches to breakover voltage of diode, the diode becomes ON and acts as closed switch. The capacitor discharges rapidly through the diode. When the device current becomes less than the holding current, it turns OFF and the capacitor starts its charging again.

• The waveform of the voltage across the capacitor C is shown in Fig. 6.1.7.

• The capacitor will not discharge completely but a voltage level corresponding to switching current denoted as V_S which is slightly more than 0 V.

Review Questions

1. Explain the construction and working of Shockley diode.

2. Draw and explain the characteristics of Shockley diode stating its applications.