In the following circuit the equivalent resistance between A and B is
Both the diodes are reversed biased so can be replaced by open circuit.
Req= 8+2+6= 16 ohm
If break down voltage of zener diode is 6 V then value of I2 will be :
Voltage across 500 ohm resistor = 12 Volt
I = 12/500 = 24 mA
I 1 = 6/1000= 6mA
I2= I - I1= (24-6 )mA= 18 mA
Determine V0 for the network
Both the diodes are forward biased so current will flow through the circuit
i= 10/4.5= 20/9 mA
V0= i R= (20/9)×3.3 Volt= 6.6 volt
When a PN junction diode is reverse biased :
Factual Question: When a PN junction diode is reverse biased potential barrier of depletion increases. Thus Electrons and holes move away from the junction depletion region.
Two PN-junctions can be connected in series by three different methods as shown in the figure. If the potential difference in the junctions is the same, then the correct connections will be :
In circuit 1. First Diode is forward biased and second diode is reverse biased. So, potential across 1st diode is less than 2nd diode.
In circuit 2: Both the diodes are forward biased so potential difference across them is equal
In circuit 3: Both the diodes are reverse biased so potential difference across them is equal to V/2.
The dominant mechanism for motion of charge carriers in forward and reverse biased silicon P-N junction are :
Theory: In forward biased mode diffusion current is more than drift current. Where as in reverse biased mode diffusion current decreased and become less than drift current.
In the given circuit
The current through the battery is :
Diode D2 and D3 are forward biased. Where as Diode D1 is reversed biased. No current will flow through D1.
Equivalent Resistance= 20/3 ohm
Current = Voltage / Resistance = 10/ (20/3) = 1.5 Ampere
Phodiode, zenerdiode and solar cell is used in:
A Photodiode and Zener Diode operate in Reverse Bias, while a Solar Cell operates in an Unbiased condition (zero external voltage).
Which of the following statement is not correct when a junction diode is in forward bias :
In forward bias, the applied field opposes the barrier field, pushing major carriers (electrons on n-side and holes on p-side) towards the junction, thereby decreasing the depletion width. Thus, statement (4) is incorrect.
The approximate ratio of resistances in the forward and reverse bias of the PN-junction diode is :
Forward resistance of a diode is very small (\(approx 10\Omega\)), while its reverse resistance is extremely high (\(approx 10^5 \Omega\)). The ratio \(R_f / R_r\) is of the order of \(10^{-4}\), which corresponds to \(1 : 10^4\).