Discussion
For a given unijunction transistor (UJT) circuit peak voltage (Vp) is ..........
Diode Voltage VD= 0.6V
Intrinsic stand of ratio (η) = 0.6
Symbol carrier usual meaning.
Answer: D
The peak voltage is 2.4 V.
The UJT fires when the emitter voltage exceeds the voltage at the internal junction of its two base resistances by one diode drop. That junction sits at the fraction \(\eta\) (the intrinsic standoff ratio) of \(V_{BB}\):
\[ \begin{aligned} V_{p} &= \eta V_{BB} + V_{D} \\ &= 0.6 \times 3 + 0.6 = 2.4\ \text{V} \end{aligned} \]
Leaving out the diode drop would give 1.8 V, which is not offered. The other values do not fit \(\eta V_{BB} + V_{D}\) with \(\eta = 0.6\): 3.4 V would need \(\eta \approx 0.93\), while 1.4 V and 1.0 V would need \(\eta\) of only about 0.27 and 0.13.
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