Analog Electronics - Study Mode

[#201] An n-channel JFET has $${I_{{ ext{DSS}}}} = 2,{ ext{mA}}$$ xa0 and V p = -4 v. Its transconductance g m = in (mA/V) for an applied gate to source voltage V GS = -2 v is
Correct Answer

(B) 0.5

[#202] In a common emitter, unbypassed resister provides
Correct Answer

(C) negative voltage feedback

Explanation

Solution: Option A: Voltage shunt feedback is incorrect because voltage shunt feedback involves taking a portion of the output voltage and feeding it back in parallel with the input. This is not what an unbypassed resistor in a common emitter configuration does. Option B: Current series feedback is incorrect because current series feedback involves taking a portion of the output current and feeding it back in series with the input. This is not characteristic of an unbypassed resistor in a common emitter amplifier. Option C: Negative voltage feedback is correct because the unbypassed resistor in a common emitter amplifier provides negative voltage feedback. This resistor, typically placed in the emitter leg of the transistor, causes a decrease in the gain by reducing the overall amplification and stabilizing the output, leading to better linearity and reduced distortion. Option D: Positive current feedback is incorrect because positive current feedback would increase the overall gain and instability, which is not the role of an unbypassed resistor in a common emitter configuration. Conclusion: The correct answer is Option C: Negative voltage feedback because the unbypassed resistor in a common emitter amplifier provides negative voltage feedback, which stabilizes the gain and reduces distortion in the amplifier.

[#203] The current gain of a BJT is
Correct Answer

(C) $${{ ext{g}}_{ ext{m}}}{ ext{ }}{{ ext{r}}_pi }$$
(G) $${{ ext{g}}_{ ext{m}}}{{ ext{r}}_pi }$$

Explanation

Solution: Option A: $$g_m r_o$$ is incorrect because this expression represents the voltage gain in a transistor, not the current gain. The term $$r_o$$ is the output resistance, and multiplying $$g_m$$ (transconductance) with $$r_o$$ gives the voltage gain, not the current gain. Option B: $$frac{g_m}{r_o}$$ is incorrect because this expression represents the inverse relationship between transconductance and output resistance, which is related to the voltage gain of the amplifier, not the current gain. Option C: $$g_m r_pi$$ is correct because this is the standard expression for the current gain of a BJT in terms of the transconductance ($$g_m$$) and the base-emitter resistance ($$r_pi$$). The current gain is the product of these two parameters. Option D: $$frac{g_m}{r_pi}$$ is incorrect because this expression does not correctly represent the current gain of a BJT. Instead, it gives a ratio that relates to the voltage gain and other characteristics of the transistor, but not the current gain. Conclusion: The correct answer is Option C: $$g_m r_pi$$ because the current gain of a BJT is given by the product of transconductance ($$g_m$$) and base-emitter resistance ($$r_pi$$).

[#204] The current gain of a bipolar transistor drops at high frequencies because of
Correct Answer

(A) Transistor capacitances

Explanation

Solution: Option A: Transistor capacitances is correct because at high frequencies, the capacitances present in a bipolar transistor (such as base-emitter and base-collector junction capacitances) start to have a significant effect. These capacitances limit the current gain, causing a drop in the transistor's performance at high frequencies. The capacitive effects become more dominant as frequency increases, leading to a reduction in the current gain. Option B: High current effects in the base is incorrect because high current effects in the base generally affect the transistor's thermal stability and the possibility of transistor saturation. While they can affect the performance, they are not the primary reason for the drop in current gain at high frequencies. Option C: Parasitic inductive elements is incorrect because while parasitic inductive elements may have some impact on the high-frequency behavior of a transistor, they do not directly cause the current gain to drop. The primary limiting factor at high frequencies is the capacitance, not inductance. Option D: The early effect is incorrect because the Early effect refers to the variation of the transistor's collector current due to changes in the collector-base voltage. While the Early effect can influence the transistor's operation, it is not the main cause for the reduction in current gain at high frequencies. Conclusion: The correct answer is Option A: Transistor capacitances because at high frequencies, the transistor's junction capacitances play a significant role in limiting the current gain, leading to a reduction in performance.

[#205] Generally, the gain of a transistor amplifier falls at high frequencies due to the
Correct Answer

(A) Internal Capacitance of the device