Diode Applications Chapter 2 Overview Power Supply Half-wave rectifiers Full-wave rectifiers Line regulation Limiter Clamper DC Power Supply Basic components – – – – Transformer (not shown) Rectifier Filter Regulator 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)/p = 0.318 x Vp(out) [31.8 % of Vp] Vp(out) = Vp(in) – VBar For Silicon VBar = 0.7 V Vp(in) Vp(out) Half-wave Rectifier Vavg 2p 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: 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. 78-0.7 n = ½ = 0.5 Vsec = n.Vpri = 78 V Vout = Vsec – 0.7 = 77.3 V Full-wave Rectifier Note that the frequency is doubled Vavg = 2Vp(out)/p = 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? n=1:2=0.5 Vsec=n*Vpri=25 Vout = Vsec/2 – 0.7 PIV = Vsec-0.7=24.3 V Vsec 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 – When negative voltage D3 and D4 are forward biased Two diodes are always in series with the load – – D1 and D2 are forward biased 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 120Vr ms n=Vsec/Vpri = 0.110:1 Vp(sec) = (0.707)-1 x Vrms = 1.414(12)=17 V Vp(out) = V(sec) – (0.7 + 0.7) = 15.6 V through D1&D2 PIV = Vp(out) + 0.7 = 16.3 V Note: Vp-Vbr ; hence, always convert from rms to Vp Bridge Full-wave Rectifier - Comparison 12 0Vr 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 Filters -Ripple voltage depends on voltage variation across the capacitor - Large ripple means less effective filter peak-to-peak ripple voltage 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) 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 Negative limiter Remember: When positive voltage reverse biased No current no clipping! Positive limiter What if we mix these together? 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: Diode Characteristics (VRRM & IF(AV) )

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