Two-Transistor Model of SCR (Two Transistor Analogy)

Published On: April 17, 2023By Tags: , ,

Two-Transistor Model of SCR (Two Transistor Analogy)

The operation of an SCR can also be explained very simply by considering it in terms of two transistors. This is known as the two-transistor analogy of the SCR. The SCR can be considered as an npn and a pnp transistor, where the collector of one transistor is attached to the base of the other and vice versa, as shown in Figure 19. This model is obtained by splitting the two middle layers of the SCR into two separate parts.

Figure 19

It is observed from the figure that the collector current of transistor T1 becomes the base current of transistor T2 and vice versa.

$latex \displaystyle {{I}_{{c1}}}={{I}_{{b2}}}\,\,\,\,and\,\,\,\,{{I}_{{b1}}}={{I}_{{c2}}}$

$latex \displaystyle {{I}_{k}}={{I}_{a}}+{{I}_{g}}$

Now, we have the relation from transistor analysis,

$latex \displaystyle {{I}_{{b1}}}={{I}_{{e1}}}-{{I}_{{c1}}}$

$latex \displaystyle {{I}_{{c1}}}={{\alpha }_{1}}{{I}_{{e1}}}+{{I}_{{co1}}}$

Ico1 is the reverse leakage current of the reverse biased junction J2 when the two outer layers are absent.

$latex \displaystyle \begin{array}{l}{{I}_{{b1}}}={{I}_{{e1}}}-{{\alpha }_{1}}{{I}_{{e1}}}-{{I}_{{co1}}}\\{{I}_{{b1}}}=\left( {1-{{\alpha }_{1}}} \right){{I}_{{e1}}}-{{I}_{{co1}}}\end{array}$

From Figure 19, it is evident that the anode current of the device becomes the emitter current of transistor T1 which is

$latex \displaystyle {{I}_{a}}={{I}_{{e1}}}$

$latex \displaystyle {{I}_{{b1}}}=\left( {1-{{\alpha }_{1}}} \right){{I}_{a}}-{{I}_{{co1}}}$

$latex \displaystyle {{I}_{{c2}}}={{\alpha }_{2}}{{I}_{{e2}}}+{{I}_{{co2}}}$

From Figure 19, it is also observed that the cathode current of the SCR becomes the emitter-current of transistor T2.

$latex \displaystyle {{I}_{k}}={{I}_{{e2}}}$

$latex \displaystyle {{I}_{{c2}}}={{\alpha }_{2}}{{I}_{k}}+{{I}_{{co2}}}$

$latex \displaystyle {{I}_{{b1}}}={{I}_{{c2}}}$

$latex \displaystyle \left( {1-{{\alpha }_{1}}} \right){{I}_{a}}-{{I}_{{co1}}}={{\alpha }_{2}}{{I}_{k}}+{{I}_{{co2}}}$

$latex \displaystyle \left( {1-{{\alpha }_{1}}} \right){{I}_{a}}-{{I}_{{co1}}}={{\alpha }_{2}}\left( {{{I}_{a}}+{{I}_{g}}} \right)+{{I}_{{co2}}}$

$latex \displaystyle \left( {1-{{\alpha }_{1}}-{{\alpha }_{2}}} \right){{I}_{a}}={{\alpha }_{2}}{{I}_{g}}+{{I}_{{co1}}}+{{I}_{{co2}}}$

$latex \displaystyle {{I}_{a}}=\frac{{{{\alpha }_{2}}{{I}_{g}}+{{I}_{{co1}}}+{{I}_{{co2}}}}}{{1-\left( {{{\alpha }_{1}}+{{\alpha }_{2}}} \right)}}$

Assuming the leakage current of transistors T1 and T2 to be negligibly small, we have

$latex \displaystyle {{I}_{a}}=\frac{{{{\alpha }_{2}}{{I}_{g}}}}{{1-\left( {{{\alpha }_{1}}+{{\alpha }_{2}}} \right)}}$

From the Equation, it can be analyzed that if (α1 + α2) = 1, the value of anode current Ia becomes infinite, that is, the anode current suddenly attains a very high value, approaching infinity. In other words, we can say that the device suddenly latches into the conduction (ON) state from the non-conduction (OFF) state. This characteristic of the device is known as its regenerative action. This can also be stated as the gate current Ig is of such a value that (α1 + α2) approaches the unity value, the device will trigger. This turn-on condition {(α1 + α2) ≥ 1} of the SCR can be satisfied in the following ways:

(a) If the temperature of the device is very high, the leakage current through it increases, which may then satisfy the required condition to turn it on.

(b) When the current through the device is extremely small, the alphas will be very small and the condition for breakover can be satisfied only by large values of hole multiplication factor Mp and electron multiplication factor Mn. Near the breakdown voltage of junction J2, the multiplication factors are very high and the required condition for breakover can be obtained by increasing the voltage across the device to VBO, which will close the breakdown voltage of junction J2.

(c) The required condition for breakover can also be realized by increasing α1 and α2. If a current Ig is injected into the base P in the same direction as the current Ia across J2, the current gain of the NPN transistor can now be increased independently of the anode to cathode voltage Va and current Ia, because α2 depends on (Ia + Ig) and α1 would still, depend on Ia. The total current gain will now depend on Ig and independent means of breakover is obtained. The presence of gate-current modifies the static V-I characteristics.

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