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Regulated Linear Power Supply
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Dual-Output Adjustable Linear Regulated Power Supply
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Block Diagram
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Simple Description Transformer: “Downconvert” the AC line voltage to a smaller peak voltage Vm, usually about 2-3 Volts larger than the ultimately desired DC output. Diode Rectifier Circuit: produce a waveform with large DC component. Filter: smooth out the rectified sinusoid. Regulator: eliminate residual ripple
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c03f34 Half-Wave Rectifier Tin
Cause of ripple: the capacitor is discharged for almost an entire period.
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c03f36 inversion Ripple Reduction: Do not allow the capacitor to discharge so frequently
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An Inverting Half-Wave Rectifier
If Vin >0, D1 and D2 are off. If Vin <0, D1 and D2 are on and Vout>0.
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An Non-Inverting Half-Wave Rectifier
If Vin >0, D1 and D2 are on, Vout>0. If Vin <0, D1 and D2 are off.
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Full-Wave Rectifier Inverting Non-Inverting
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Full-Wave Rectifier Alternative Drawing
Full-Wave A.K.A. Bridge Rectifier Non-Inverting Inverting
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Using Constant Voltage Diode Model
Vout=-Vin-2VD, on Vout=Vin-2VD, on
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Input versus Output |Vin|<2VD,on
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Addition of the Smoothing Capacitor
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Modification of Ripple Estimation Formula
Turn-on voltage 1/2 to account for inversion of negative peaks.
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Maximum Reverse Voltage
VB=VD,on Vp is the amplitude of Vin VA=VP VAB=VP-VD,on Maximum reverse voltage is approximately Vp
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Compare Maximum Reverse Bias Voltage to Half-Wave Rectifier
A reverse diode voltage must sustain larger reverse bias voltage
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Currents as a function of time
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Application of Bridge Rectifier
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Summary
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A Voltage Regulator Circuit
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Schematic
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Full-Wave Rectifier Alternative Drawing
Full-Wave A.K.A. Bridge Rectifier Non-Inverting Inverting
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Rectified Waveform
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Smoothing Capacitor
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LM317/LM337
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Without Load Resistance (CL=100 uF)
9.36 V and V
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Load Assumption Assume that the load draws 0.2 A of current
Load resistance: 9.36V/0.2A=36.5 Ohms
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Each capacitor is 100 uF. Resistor is 37.5 Ohms
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Use Reload update curves
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Increase capacitor from 100
uF to 470 uF. Reduced ripples
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Extra Slides
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I-V Characteristic of LED
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