Diode Applications

Chapter 2

Overview

Power Supply Half-wave rectifiers Full-wave rectifiers Line regulation Limiter Clamper

Basic components– Transformer (not

shown)– Rectifier– Filter– Regulator

DC Power Supply

Or Full-wave rectifier

Sine Wave

The sine wave is a common type of alternating current (ac) and alternating voltage.

The time required for a sine wave to complete one full cycle is called the period (T).

Frequency ( f ) is the number of cycles that a sine wave completes in one second.

– The more cycles completed in one second. The higher the frequency.

– Frequency is measured in hertz (Hz)

Relationship between frequency ( f ) and period (T) is:

f = 1/T

Peak-to-Peak / Average / RMS

The peak-to-peak value of a sine wave is the voltage or current from the positive peak to the negative peak.

The peak-to-peak values are represented as:

Vpp and Ipp

Where: Vpp = 2Vp and Ipp = 2Ip

The rms (root mean square) value of a sinusoidal voltage is equal to the dc voltage that produces the same amount of heat in a resistance as does the sinusoidal voltage.

Vrms = 0.707Vp

Irms = 0.707Ip

Half-wave Rectifiers

Forward biased

Half-wave Rectifiers

Reverse biased

Output result

Half-wave Rectifier

Note that the frequency stays the same Strength of the signal is reduced Vavg = Vp(out)/= 0.318 x Vp(out) [31.8 % of Vp] Vp(out) = Vp(in) – VBar For Silicon VBar = 0.7 V

Half-waveRectifier

Vp(out)Vp(in)

Vavg

2

Half-wave Rectifier - Example

Output:

Peak Inverse Voltage (PIV) – The peak voltage at which the

diode is reverse biased– In this example PIV = Vp(in)-

– Hence, the diode must be rated for PIV = 100 V

Draw the output signal– Vp(out) = Vp(in) – 0.7– Vavg = 99.3/– What happens to the

frequency?

Transformers (Review)

Transformer: Two inductors coupled together – separated by a dielectric When the input magnetic field is changing voltage is induced on the

second inductor The dot represents the + (voltage direction)

Applications: Step-up/down Isolate sources

Turns ratio (n) n = Sec. turns / Pri. turns = Nsec/Npri

Vsec = n. Vpri

depending on value of n : step-up or step-down

Center-tapped transformer Voltage on each side is Vsec/2

Half-wave Rectifier - Example

Example: – Assume that the input is a sinusoidal signal with Vp=156 V & T = 2

msec; assume Nsec:Npri = 1:2– Draw the signal– Find turns ratio; – Find Vsec;– Find Vout.

n = ½ = 0.5Vsec = n.Vpri = 78 V Vout = Vsec – 0.7 = 77.3 V

78-0.7

Full-wave Rectifier

Note that the frequency is doubled Vavg = 2Vp(out)/= 0.637 x Vp(out)

Full-wave Rectifier Circuit

Center-tapped full-wave rectifier– Each half has a voltage = Vsec/2

Only one diode is forward biased at a time

The voltages at different halves are opposite of each other

Full-wave Rectifier Circuit

Center-tapped full-wave rectifier– Each half has a voltage = Vsec/2

Only one diode is forward biased at a time

The voltages at different halves are opposite of each other

Full-wave Rectifier Circuit

Vout = Vsec /2 – 0.7 Peak Inverse Voltage (PIV)

– PIV = (Vsec/2 – 0.7)- (-Vsec/2) = Vsec – 0.7

Vout = Vsec/2 – 0.7

Assuming D2 is

reverse-biased

No current through D2

Full-wave Rectifier - Example

Assuming a center-tapped transformer Find the turns ratio Find Vsec Find Vout Find PIV Draw the Vsec and Vout What is the output freq? Vsec

n=1:2=0.5 Vsec=n*Vpri=25 Vout = Vsec/2 – 0.7 PIV = Vsec-0.7=24.3 V

Full-wave Rectifier - Multisim

XFrmr can be virtual or real Use View Grapher to see the details of your

results The wire-color can determine the waveform

color Make sure the ground is connected to the

scope.

Bridge Full-wave Rectifier

Uses an untapped transformer larger Vsec Four diodes connected creating a bridge

– When positive voltage D1 and D2 are forward biased

– When negative voltage D3 and D4 are forward biased

Two diodes are always in series with the load– Vp(out) = Vp(sec) – 1.4V– The negative voltage is inverted

The Peak Inverse Voltage (PIV)– PIV=Vp(out)+0.7

Bridge Full-wave Rectifier - Example

Assume 12 Vrms secondary voltage for the standard 120 Vrms across the primary

– Find the turns ratio– Find Vp(sec)– Show the signal direction when Vin is positive– Find PIV rating

n=Vsec/Vpri = 0.110:1Vp(sec) = (0.707)-1 x Vrms = 1.414(12)=17 VVp(out) = V(sec) – (0.7 + 0.7) = 15.6 V through D1&D2 PIV = Vp(out) + 0.7 = 16.3 V

120Vrms

Note: Vp-Vbr ; hence, always convert from rms to Vp

Bridge Full-wave Rectifier - Comparison

120Vr

ms

Vp(2)=Peak secondary voltage ; Vp(out) Peak output voltage ; Idc = dc load current

Make sure you understand this!

Filters and Regulators

Filters

Filters

Filters

-Ripple voltage depends on voltage variation across the capacitor- Large ripple means less effective filter

Filters

peak-to-peak ripple voltage

Too much ripple is bad! Ripple factor = Vr (pp) / VDCVr (pp) = (1/ fRLC) x Vp(unfiltered)VDC = (1 – 1/ fRLC) x Vp(unfiltered)

Filters

Diode Limiting

What is Vout?– Vout+ = Vin (RL)/(RL+R1) = 9.09– Vout- = -0.7

Forward biased when positive

Reverse biased when negative, hence voltage drop is only -0.7

So how can we change the offset?

Diode Limiting – Changing the offset

What if we mix these together?

Positive limiter

Negative limiter

Remember: When positive voltage reverse biased No current no clipping!

Diode Limiter

When the input signal is positive D1 is reversed biased; acting as positive limiter

Pos. Limiter

-VBIAS-0.7+VBIAS+0.7

Diode Clamper

It adds a dc level When the input voltage is negative, the

capacitor is charged – Initially, this will establish a positive dc

offset

Note that the frequency of the signal stays the same

RC time constant is typically much larger than 10*(Period)

Note that if the diode and capacitor are flipped, the dc level will be negative

Output:

DiodeCharacteristics (VRRM & IF(AV) )