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An ideal diode has no losses and negligible reverse recovery time.
A diode circuit is so arranged that when the switch is open it’s KVL gives
Ri+ 1/C ∫i dt = 0
When the switch is closed,
Ri+ 1/C ∫i dt = Vs
Vs is the dc supply voltage.
The diode is so connected that it is forward biased when switch is closed
The circuit is mostly likely be a
Examine the equation, the same current flows through R & C. Also when switch is open, the equation R.H.S is 0. Hence, all the elements are in series.
A circuit is so formed such that the sourceRCdiodeswitch are in series. Consider the initial voltage across the C to be zero. The diode is so connected that it is forward biased when the switch is closed. When the switch is closed,
Instant switch is closed, the current is maximum Vs/R than starts to reduce, whereas voltage starts to increase from 0 to Vs (Capacitor Charging).
For a series RC circuit, τ = RC.
A circuit is so formed such that sourceRLdiodeswitch are all in series. Consider the initial current in L to be zero. The diode is so connected that it is forward biased when switch is closed.
When the switch is closed,
Instant switch is closed, the current is minimum zero than starts to increase till it reaches a constant value Vs/R, whereas voltage starts to reduce from Vs to 0 (Inductor Charging).
In the figure shown below,
As the switch (shown in the green) is pressed, the voltage across the diode(ideally)
As the switch is pressed, current starts to flow & the whole supply voltage (ideally) appears across the load R & voltage across the diode is zero.
For a diode circuit the voltage across the capacitor is given by
Vc(t)= Vs(1e^{(t/RC)})
Then the initial rate of change of capacitor voltage is given by
Find d(Vc)/dt and put t = 0.
In the circuit show below,
The initial current through the inductor is zero. When the switch (shown in green) is closed, then the current through the inductor
Current increases from zero to maximum value gradually due to the L nature. The KVL when switch is closed gives,
Ri + Ldi/dt = Vs
Solve for i(t). Maximum value comes out to be Vs/R.
In the figure shown below,
When the switch is open, the voltage across the diode (ideally)
When the switch is open, the diode experiences all the supply voltage.
The time constant τ for a series RL circuit is R/L.
For the initially relaxed circuit shown below, KVL with switch in the closed position gives a certain equation. The Laplace of this equation will have the right hand side (RHS) as
When switch is closed,
Ldi/dt + 1/C ∫ idt = Vs
Laplace of the above gives,
L[sI(s)] + 1/C [I(s)/s] = Vs/s.
For the initially relaxed circuit shown below, the Laplace transform of the KVL when the switch is closed is
I(s) [ X ] = Vs/s
The value of X is
When switch is closed,
Ldi/dt + 1/C ∫ idt = Vs
Laplace of the above gives,
L[sI(s)] + 1/C [I(s)/s] = Vs/s.
For the initially relaxed circuit shown below, if Ω=1/√LC. Then the current is a function of ___
When switch is closed,
Ldi/dt + 1/C ∫ idt = Vs
Laplace of the above gives,
L[sI(s)] + 1/C [I(s)/s] = Vs/s
I(s) = Vs/(LΩ) * (Ω/Ω^{2} + s^{2})
Taking the inverse lapace gives,
I(t) = Vs * √ (L/C) * sin Ωt.
For the circuit shown below, the capacitor is initially charged to a voltage of Vo with the upper plate positive. After the switch (shown in green) is closed, the current through the load
The capacitor acts as a source. At instant switch is closed the current is maximum and than discharges till zero value through the load R.
For the circuit shown below, the capacitor is initially charged to a voltage of Vo with the upper plate positive. Switch is closed at t=0. The peak value of the current through the diode is
When switch is closed, the equation is
Ri + 1/C ∫idt = 0 (Voltage across capacitor cannot change instantaneously)
Solution of the above equation gives, Vo/R at t= 0.
When the switch (shown in green) is closed, the average current through the diode in the positive cycle is
The switch S.C’s the source.
When the switch is closed, the steady state current through the diode is
I = Vo/R
When the switch is open, the current through the diode in the positive cycle is
When the switch is open, the diode is forward biased and I = Vs/(Rd + R). Where, Rd is the diode resistance.
For the circuit shown in the figure below, consider the diode as an ideal diode & R.M.S value of source voltage as Vs.
The output voltage waveform at R is most likely to have
The diode S.C’s the load in the positive half cycle.
For the circuit shown in the figure below, V8 is AC voltage source with peak value Vm. The waveform of the load voltage at the resistor is
Diode is reversed biased in the positive half cycle. In the negative half cycle, apply KVL to get the value of peak voltage at the load.
For the circuit shown in the figure below, consider the diode as an ideal diode & rms value of source voltage as Vs.
The output voltage waveform at R will have
The diode short circuits the load in the negative half cycle. The peak value in the positive half is 1.414 x Vs.
For the circuit shown in the figure below, Vs is the ac voltage source with peak value Vm. The waveform of the load voltage at the resistor will have
Diode is reversed biased in the positive half cycle. In the negative half cycle, apply KVL to get the value of peak voltage at the load.
For the circuit shown below,
Vs=230V, Voltage drop across the diode (Vd) = 2V
The peak value of voltage at R in the positive and the negative half cycles are ___ & ___ respectively.
Load is S.C in the positive half cycle hence, voltage is zero. In negative half the peak value is (230 x 1.414) – 2(drop across the diode) = 323 Volts.
For the circuit shown in the figure below, V9 is the AC voltage source with peak value Vm. The waveform of the load voltage at the resistor has a
Diode is reversed biased in the negative half cycle. In the positive half cycle, apply KVL to get the value of peak voltage at the load. Peak value = Vm + V1, as V1 is aiding the positive anode of the diode.
The circuit shown below has the following parameters:
V1 = 8 Volts
V2 = 6 Volts
Vs = 10V/√2 (rms)
Voltage drop across D1 & D2 = 0.7 Volts
At the load (R), the peak value in the positive half cycle will be
In the positive half, only D1 is active. Hence use KVL, Vo = 8 + 0.7 Volts.
The circuit shown below has the following parameters:
Voltage drop across D1 & D2 = 0.7 Volts
V1 = 8 Volts
V2 = 6 Volts
Vs = 10V/√2 (RMS)
At the load (R), the peak value in the negative half cycle is
In the negative half, only D2 is active. Hence use KVL, Vo = 6 + 0.7 Volts.
Consider the diode to be an ideal one and Vo = Vr + Vdc during positive half cycle. Thus, during the negative half cycle
During the negative half diode is not conducting hence Vr (voltage across the resistor) = 0. Vo = Vdc.
When a diode is connected in series with an AC source & R load, the conduction time per cycle is
The diode is forward biased for half cycle (180 degrees) and reversed biased in the another 180 degrees. Hence, per cycle it only conducts for 180 degrees or π radians.
When diode is connected in series to an AC source & RL load, the conduction time for the diode
For an R – L load, the inductively load can make the diode to force conducted hence, the conduction time can be greater than 180 degrees.
For the circuit shown below,
Vdc = 50 V
Cutin voltage for D1 = 0.2 V
Cutin voltage for D2 = 0.6 V
R = 5KΩ
Current through D1 & D2 would be,
As the cutin voltage for D1 is lesser than D2, D1 would start conducting first & S.C D2. Hence, only D1 conducts with I = 50/5000 A.
5 videos39 docs63 tests

5 videos39 docs63 tests
