A non-ideal diode is biased with a voltage of –0.03 V and a diode current of 1I is measured. The thermal voltage is 26 mV and the ideality factor for the diode is 15 /13. The voltage, in V, at which the measured current increases 1 to 1.5 I is closest to

The correct option/answer is –0.09.

To find the new voltage, we start with the standard diode current-voltage relationship given by the Shockley diode equation:
$$I=I_0\left(e^{\frac{V}{\eta V_T}}-1\right)$$ where:
• $I_0$ is the reverse saturation current,
• $V$ is the applied bias voltage,
• $V_T$ is the thermal voltage (26 mV = 0.026 V),
• $\eta$ is the ideality factor (15/13).

First, let's calculate the product of the ideality factor and the thermal voltage ($\eta V_T$):
$$\eta V_T=\frac{15}{13}\times 0.026\text{V}=15\times 0.002\text{V}=0.03\text{V}$$

For the initial state, the diode is biased at $V_1=-0.03\text{V}$ and the measured current is $I_1=I$:
$$I=I_0\left(e^{\frac{-0.03}{0.03}},-,1\right)=I_0\left(e^{-1}-1\right)$$ Using the approximation $e^{-1}\approx 0.3679$:
$$I\approx I_0\left(0.3679-1\right)=-0.6321I_0$$

We are looking for the new voltage $V_2$ at which the current increases in magnitude to $I_2=1.5I$:
$$I_2=1.5\times \left(-0.6321I_0\right)=-0.9482I_0$$

Now, we substitute $I_2$ and $V_2$ back into the diode equation:
$$-0.9482I_0=I_0\left(e^{\frac{V_2}{0.03}}-1\right)$$ Dividing both sides by $I_0$:
$$-0.9482=e^{\frac{V_2}{0.03}}-1$$

Rearranging the equation to solve for the exponential term:
$$e^{\frac{V_2}{0.03}}=1-0.9482=0.0518$$

Taking the natural logarithm of both sides:
$$\frac{V_2}{0.03}=ln\left(0.0518\right)\approx -2.96$$

Multiplying by 0.03 V gives:
$$V_2=-2.96\times 0.03\approx -0.0888\text{V}$$

This value is closest to –0.09 V.