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(A) Calculate the value of V_(0) and i if the silicon and germanium diode start conducting at 0.7 V and 0.3 Vrespectively. (B) If the Ge diode connection is now reversed, what will be the new values of V_(0) and i ? |
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Answer» Solution :Here cut-in voltage (thereshold voltge) for Si diode is 0.7V and cut-in voltage for Ge diode is 0.3V. BATTERY voltage = 12V Load resistance `R_(L)=5kOmega=5xx10^(3)Omega` (A) Current will flow earlier in Ge then Si diode. Forward bias voltage `=12.0-0.3` `=11.7V` `therefore ` Current `I_(F)` in `R_(L)=(11.7)/("Load resistance")` `=(11.7)/(5xx10^(3))` `=2.34xx10^(-3)A` `therefore I=2.34mA` (B) In REVERSE bias CONNECTION current will flowearlier in Si diode. `therefore ` Reverse bias voltage `=12.0-0.7` `=11.3V` `therefore` Current `I_(R )` in `R_(L )=(1.3)/("Load resistance")` `=(11.3)/(5xx10^(3))` `=2.26xx10^(-3)A` `therefore I_(R )=2.26 mA` |
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