Pure `Si` at `500K` has equal number of electron `(n_(e))` and hole `(n_(h))` concentration of `1.5xx10^(16)m^(-3)`. Dopping by indium. Increases `n_(h)` to `4.5xx10^(22) m^(-3)`. The doped semiconductor is ofA. `n`-typpe with electron concentration `n_(e)=2.5xx10^(23)m^(-3)`B. `p`-type having electron concentration `n_(e)=5xx10^(9)m^(-3)`C. `n`-type with electron concentration `n_(e)=5xx10^(22m^(-3)`D. `p`-type with electron concentration `n_(e)=2.5xx10^(10)m^(-3)`

Pure `Si` at `500K` has equal number of electron `(n_(e))` and hole `(

1.	Pure `Si` at `500K` has equal number of electron `(n_(e))` and hole `(n_(h))` concentration of `1.5xx10^(16)m^(-3)`. Dopping by indium. Increases `n_(h)` to `4.5xx10^(22) m^(-3)`. The doped semiconductor is ofA. `n`-typpe with electron concentration `n_(e)=2.5xx10^(23)m^(-3)`B. `p`-type having electron concentration `n_(e)=5xx10^(9)m^(-3)`C. `n`-type with electron concentration `n_(e)=5xx10^(22m^(-3)`D. `p`-type with electron concentration `n_(e)=2.5xx10^(10)m^(-3)`
Answer» Correct Answer - B `n_(i)^(2)=n_(e)n_(h)` `(1.5xx10^(16))^(2)=n_(e)(4.5xx10^(22))` `n_(e)=0.5xx10^(10)=5xx10^(9)` `n_(h)=4.5xx10^(22)` `n_(h)gtn_(e)(p-type)`

Discussion

No Comment Found

Related InterviewSolutions

The depletion layer in diode is `1 mum` wide and the knee potential is `0.6 V`, then the electric field in the depletion layer will beA. `5xx10^(6)Vm^(-1)`B. `5xx10^(-7)Vm^(-1)`C. `5xx10^(5)Vm^(-1)`D. `5xx10^(-1)Vm^(-1)`
The depletion layer in diode is `1 mum` wide and the knee potential is `0.6 V`, then the electric field in the depletion layer will beA. zeroB. `0.6Vm^(-1)`C. `6xx10^(4) V//m`D. `6xx10^(5) V//m`
The knee voltage of a p-n junction diode is 0.8 V and the width of the depletion layer is `2muA` . The electric field in the depletion layer isA. 4 MV/mB. 0.4 MV/mC. 4 KV/mD. 0.4 KV/m
The junction diode in the following circuit requires a minimum current of `1 mA` to be above the knee point `(0.7 V)` of its `I-V` characterstic curve. The voltage across the diode is independent of current above the knee point. If `V_(B)=5 V`, then the maximum value of `R` so that the voltage is above the knee point, will be A. `4.3 k Omega`B. `860 k Omega`C. `4.3 k Omega`D. `860 Omega`
In the study of transistor as an amplifier , `alpha=l_C/l_E` and `beta=I_C/I_B` where `I_C,I_E` and `I_B` are the collector, emitter and base currents respectively. The correct relation between `alpha` and `beta` is given byA. `beta=(1-alpha)/alpha`B. `beta=alpha/(1-alpha)`C. `beta=(1+alpha)/alpha`D. `beta=alpha /(1+alpha )`
A silicon diode is connected to a load resistance `R_(L)` as shown in the figure. If `V_(in)=15V` and `R_(L)=10kOmega` (a) If barrier voltage, `V_(B)=0.7V` then calculate (i) output voltage across `R_(L)` (ii) current in diode and ltbRgt (iii) forward resistance. (b) If diode is assumed ideal, then what will be (i) output voltage and (ii) output current in diode?
For the transistor circuit shown in figure , evaluate `V_E , R_B and R_E`. Given `I_C = 1 mA`, ` V_(CE) = 3V , V_(BE) = 0.5 V , V_(CC )= 12 V and beta = 1000`
(i) calculate the value of output voltage `V_(0)` and current `I` if silicon diode and germanium diode conduct at `0.7V` and `0.3 V` respectively. Fig. (ii) If now germanium diode is cinnected to `12 V` in reverse polarity, find new values of `V_(0)` and `I`.
The emitter of transistor is doped the heaviest because itA. receives the inputB. is supplier of charge carriersC. dissipates minimum powerD. should have low resistance
A device whose one end is connected to `-ve` terminal and other end connected to `+ve` terminal. If both ends are interchanged with suppy then current is not flowing then device will be-A. P-N junctionB. TransistorC. Zener diodeD. Triode

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment