Lecture 7 Diode Circuit Models and Applications. 1. i. i. 3. Graphical analysis using load-line method. i. Linear Circuit. I and V must satisfy both the linear And non-linear circuit equations. +. -. Nonlinear element. For the linear circuit.
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Problem: Find Q-point
Given data:V=10 V, R=10kW.
To define the load line we use,
VD= 5 V, ID =0.5 mA
These points and the resulting load line are plotted.Q-point is given by intersection of load line and diode characteristic:
Q-point = (0.95 mA, 0.6 V)
Problem: Find Q-point for given diode characteristic.
Given data: IS =10-13 A, n=1, VT =0.0025 V
is load line, given by a transcendental equation. A numerical answer can be found by using iterative method.
Use the technique in the previous slide
Q-point = ( 0.9426 mA, 0.5742 V)
Since, usually we don’t have accurate saturation current and significant tolerances for sources and passive components, we need answers precise up to only 2or 3 significant digits.
CH3 Diode Circuits
For positive half-cycle of input, source forces positive current through diode, diode is on, vo = vs.
During negative half cycle, negative current can’t exist in diode, diode is off, current in resistor is zero and vo =0 .
Using constant voltage model CVD model, during on state of diode vo =(VP sinwt)- Vd,on. Output voltage is zero when diode is off.
Often a step-down transformer is used to convert 120 V-60 Hz voltage available from power line to desired ac voltage level as shown.
Time-varying components in circuit output are removed using filter capacitor.
Select a diode that has at least 50% higher PIV
As input voltage rises during first quarter cycle, diode is on and capacitor (initially discharged) charges up to peak value of input voltage.
At peak of input, diode current tries to reverse, diode cuts off, capacitor discharges exponentially through R. Discharge continues till input voltage exceeds output voltage which occurs near peak of next cycle. Process then repeats once every cycle.
This circuit can be used to generate negative output voltage if the top plate of capacitor is grounded instead of bottom plate. In this case, Vdc = -(VP - Von)
Finding the Ripple Voltage
Output voltage is not constant as in ideal peak detector, but has ripple voltageVr.
Diode conducts for a short time DT called conduction interval during each cycle and its angular equivalent is called conduction angleθc.
Problem: Find dc output voltage, output current, ripple voltage, conduction interval, conduction angle.
Given data: secondary voltage Vrms 12.6 (60 Hz), R= 15 W, C= 25,000 mF, Vd = 1 V
Using discharge interval T=1/60 s,
Peak inverse voltage (PIV) rating of the rectifier diode gives the breakdown voltage.
When diode is off, reverse-bias across diode is Vdc - vs. When vs reaches negative peak,
PIV value corresponds to minimum value of breakdown voltage for rectifier diode.
In rectifiers, nonzero current exists in diode for only a very small fraction of period T, yet an almost constant dc current flows out of filter capacitor to load.
Total charge lost from capacitor in each cycle is replenished by diode during short conduction interval causing high peak diode currents.
Full-wave rectifiers cut capacitor discharge time in half and require half the filter capacitance to achieve given ripple voltage. All specifications are same as for half-wave rectifiers.
Reversing polarity of diodes gives a full-wave rectifier with negative output voltage.
PIV = 2vs-vD
Requirement for a center-tapped transformer in the full-wave rectifier is eliminated through use of 2 extra diodes.All other specifications are same as for a half-wave rectifier except PIV.
PIV = vs-vD
As input voltage rises, diode is on and capacitor (initially discharged) charges up to input voltage minus the diode voltage drop.
At peak of input, diode current tries to reverse, diode cuts off, capacitor has no discharge path and retains constant voltage providing constant output voltage
Vdc = VP – Vd,on.