Question

(a) What will be the forward current at a forward voltage of $0.6V$?(b) What will be the increase in the current if the voltage across the diode is increased to $0.7V$?(c) What is the dynamic resistance?(d) What will be the current if reverse bias voltage changes from $1V$ to $2V$?

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Solution

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In a p-n junction diode, the expression for current is $I=I_{0}exp[2k_{B}TeV −1]$

where $I_{0}$ is the reverse saturation current = $5×10_{−12}A$

$T=$Absolute temperature = $300K$

$V$ is the voltage across diode.

For forward voltage $V=0.6V$

$I=5×10_{−12}×exp[1.376×10_{−23}×3001.6×10_{−19}×0.6 −1]=0.0256A$

(b)

In a p-n junction diode, the expression for current is $I=I_{0}exp[2k_{B}TeV −1]$

where $I_{0}$ is the reverse saturation current = $5×10_{−12}A$

$T=$Absolute temperature=$300K$

$V$ is the voltage across diode.

For forward voltage, $V=0.7V$

$I=5×10_{−12}×exp[1.376×10_{−23}×3001.6×10_{−19}×0.7 −1]=1.257A$

Hence, increase in current = $1.257A−0.0256A=1.23A$

(c)

In a p-n junction diode, the expression for current is $I=I_{0}exp[2k_{B}TeV −1]$

where $I_{0}$ is the reverse saturation current = $5×10_{−12}A$

$T=$Absolute temperature=$300K$

$V$ is the voltage across diode.

For forward voltage $V=0.6V$

$I=5×10_{−12}×exp[1.376×10_{−23}×3001.6×10_{−19}×0.6 −1]=0.0256A$

For forward voltage $V=0.7V$

$I=5×10_{−12}×exp[1.376×10_{−23}×3001.6×10_{−19}×0.7 −1]=1.257A$

Hence increase in current=$1.257A−0.0256A=1.23A$

Dynamic resistance = Change in voltage /Change in current

$=1.230.7−0.6 =0.081Ω$

(d)

If reverse bias voltage changes from $1V$ to $2V$, the current will be almost constant, because dynamic resistance in the reverse bias is infinite.

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