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Diode Applications Chapter 2

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**Overview Power Supply Half-wave rectifiers Full-wave rectifiers**

Line regulation Limiter Clamper

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**DC Power Supply Basic components Transformer (not shown) Rectifier**

Filter Regulator Or Full-wave rectifier

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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

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**Peak-to-Peak / Average / RMS**

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 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

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Half-wave Rectifiers Forward biased

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Half-wave Rectifiers Reverse biased Output result

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**Half-wave Rectifier For Silicon VBar = 0.7 V**

Note that the frequency stays the same Strength of the signal is reduced Vavg = Vp(out)/p = x Vp(out) [31.8 % of Vp] Vp(out) = Vp(in) – VBar For Silicon VBar = 0.7 V Half-wave Rectifier Vp(out) Vp(in) Vavg 2p

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**Half-wave Rectifier - Example**

Draw the output signal Vp(out) = Vp(in) – 0.7 Vavg = 99.3/p What happens to the frequency? 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 Output:

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**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

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**Half-wave Rectifier - 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 = 0.5 Vsec = n.Vpri = 78 V Vout = Vsec – 0.7 = 77.3 V 78-0.7

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**Full-wave Rectifier Note that the frequency is doubled**

Vavg = 2Vp(out)/p = x Vp(out)

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**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

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**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

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**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

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**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

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**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.

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**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

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**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 120Vrms n=Vsec/Vpri = 0.110:1 Vp(sec) = (0.707)-1 x Vrms = 1.414(12)=17 V Vp(out) = V(sec) – ( ) = 15.6 V through D1&D2 PIV = Vp(out) = 16.3 V Note: Vp-Vbr ; hence, always convert from rms to Vp

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**Bridge Full-wave Rectifier - Comparison**

120Vrms Vp(2)=Peak secondary voltage ; Vp(out) Peak output voltage ; Idc = dc load current Make sure you understand this!

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**Filters and Regulators**

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Filters

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Filters

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Filters

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

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**Filters Too much ripple is bad! Ripple factor = Vr (pp) / VDC**

Vr (pp) = (1/ fRLC) x Vp(unfiltered) VDC = (1 – 1/ fRLC) x Vp(unfiltered)

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**Diode Limiting What is Vout? Vout+ = Vin (RL)/(RL+R1) = 9.09**

Forward biased when positive Reverse biased when negative, hence voltage drop is only -0.7 So how can we change the offset?

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**Diode Limiting – Changing the offset**

Negative limiter Remember: When positive voltage reverse biased No current no clipping! Positive limiter What if we mix these together?

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Diode Limiter When the input signal is positive D1 is reversed biased; acting as positive limiter Pos. Limiter +VBIAS+0.7 -VBIAS-0.7

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**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:

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**Diode Characteristics (VRRM & IF(AV) )**

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AC DC: Using a full-wave diode rectifier circuit (used in the music system final project) The 20:1 turns ratio transformer here reduces the rms voltage.

AC DC: Using a full-wave diode rectifier circuit (used in the music system final project) The 20:1 turns ratio transformer here reduces the rms voltage.

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