In a p-n junction diode, the current I can be expressed as $$I = I_0\,exp\, \left( \dfrac{eV}{2K_BT} - 1 \right)$$, where $$I_0$$ is called the reverse saturation current, $$V$$ is the voltage across the diode and is positive for forward bias and negative for reverse bias, and $$I$$ is the current through the diode, $$k_B$$ is the Boltzmann constant $$(8.6 \times 10^{-5} eV/K)$$ and $$T$$ is the absolute temperature. If for a given diode $$I_o = 5 \times 10^{-12}$$$$ A$$ and $$T = 300\ K$$, then (a) What will be the forward current at a forward voltage of $$0.6\ V$$?(b) What will be the increase in the current if the voltage across the diode is increased to $$0.7\ V$$?(c) What is the dynamic resistance?(d) What will be the current if reverse bias voltage changes from $$1\ V$$ to $$2\ V$$?
(a)
In a p-n junction diode, the expression for current is $$I=I_0exp[\dfrac{eV}{2k_BT}-1]$$
where $$I_0$$ is the reverse saturation current = $$5\times 10^{-12}A$$
$$T=$$Absolute temperature = $$300\ K$$
$$V$$ is the voltage across diode.
For forward voltage $$V = 0.6\ V$$
$$I=5\times 10^{-12}\times exp[\dfrac{1.6\times 10^{-19}\times 0.6}{1.376\times 10^{-23}\times 300}-1] = 0.0256\ A$$
(b)
In a p-n junction diode, the expression for current is $$I=I_0exp[\dfrac{eV}{2k_BT}-1]$$
where $$I_0$$ is the reverse saturation current = $$5\times 10^{-12}A$$
$$T=$$Absolute temperature=$$300\ K$$
$$V$$ is the voltage across diode.
For forward voltage, $$V = 0.7\ V$$
$$I=5\times 10^{-12}\times exp[\dfrac{1.6\times 10^{-19}\times 0.7}{1.376\times 10^{-23}\times 300}-1]=1.257\ A$$
Hence, increase in current = $$1.257\ A-0.0256\ A=1.23\ A$$
(c)
In a p-n junction diode, the expression for current is $$I=I_0exp[\dfrac{eV}{2k_BT}-1]$$
where $$I_0$$ is the reverse saturation current = $$5\times 10^{-12}A$$
$$T=$$Absolute temperature=$$300\ K$$
$$V$$ is the voltage across diode.
For forward voltage $$V = 0.6\ V$$
$$I=5\times 10^{-12}\times exp[\dfrac{1.6\times 10^{-19}\times 0.6}{1.376\times 10^{-23}\times 300}-1]=0.0256\ A$$
For forward voltage $$V = 0.7\ V$$
$$I=5\times 10^{-12}\times exp[\dfrac{1.6\times 10^{-19}\times 0.7}{1.376\times 10^{-23}\times 300}-1]=1.257\ A$$
Hence increase in current=$$1.257\ A-0.0256\ A = 1.23\ A$$
Dynamic resistance = Change in voltage /Change in current
$$=\dfrac{0.7-0.6}{1.23}=0.081\Omega$$
(d)
If reverse bias voltage changes from $$1\ V$$ to $$2\ V$$, the current will be almost constant, because dynamic resistance in the reverse bias is infinite.
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