Rectifier circuits & DC power supplies


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1 Rectifier circuits & DC power supplies Goal: Generate the DC voltages needed for most electronics starting with the AC power that comes through the power line? 120 V RMS f = 60 Hz T = 1667 ms) = )sin How to take timevarying voltage with an average value of 0 and turn it into a DC voltage? EE 201 rectifiers 1
2 120 V RMS transformer peak rectifier regulator DC load transformer : reduces AC amplitude to something safe and manageable V peak from the transformer will be a few volts bigger than the desired DC voltage peak rectifier : breaks up the AC waveform and produces a V DC V peak regulator : Refines the output of the rectifier optional) Issues: Total power Efficiency Cost Load regulation Does VDC change as the load draws different amounts of current?) Line regulation Does VDC change if the input AC amplitude changes?) EE 201 rectifiers 2
3 Halfwave rectifier ) = sin V p = 3 V V S v R Resistor represent a load We are trying to deliver DC power to the load Diode is off until V S > 07 V Current flows when diode is in forward conduction The output tracks the input during positive half cycle EE 201 rectifiers 3
4 The diode turns off when V S < 07 V It stays off during the negative half cycle of the sinusoid V S > 0: V S < 0: v R t) = 0: ) sin ) 0! To get the negative half of the cycle, turn the diode around EE 201 rectifiers 4
5 Time delay Note that since the diode will not turn on until the sinusoid goes above 07 V, there is time delay before the rectifier turns on It is a simple matter to determine the delay time, using the onoff diode model: 3 = sin 2 VS Vo 1 diode off again = arcsin V S, V o V) 01 output follows input diode initially off If f = 60 Hz T = 1667 ms) and V p = 3 V, t = 062 ms 23 time There is a similar time offset at the other end of the positive half cycle The effect of the time offset become negligible if V P >> 07 V EE 201 rectifiers 5
6 Peak rectifier Add a largish capacitor after the diode, in parallel with the load V S i D C C i C v R V S i C ir i R v R Initially, diode is on & cap charges to V P  07 V While V S < v C, diode is off! Cap discharges through load EE 201 rectifiers 6
7 V S i D C i C ir v R Diode stay off until V S comes back around and becomes bigger than v C Then diode comes on again and recharges the capacitor When V S falls to less than v C, the diode turn off again, and the cycle continues EE 201 rectifiers 7
8 Not a a perfect DC voltage at output There is some variation ripple) around an average value 0 0 )= )=[ ] exp [ ] exp = t1 T ) =[ t1 = time when diode conducts again EE 201 t1 T ) ] ) exp ) rectifiers 8
9 Example 1 = ) sin VS vo C T = 1667 ms C = 100 µf R = 5000! Find the average value of vo and the ripple voltage Repeat for R = 1000! and 200! =[ ] =[ exp ] exp ) ) = 047 V EE 201 )= ) = R = 1 k! R = 200! Vripple = 219 V Vripple = 809 V Vo avg) = 132 V Vo avg) = 102 V = Drawing more current causes the ripple to increase and VDC to droop Can fight this with more capacitance rectifiers 9
10 Example 2 = ) sin VS vo C T = 1667 ms R = 1000! Find the capacitance so that the ripple will be no bigger than 1 V What is the DC voltage? =[ = ] ln )= exp = ) = ln = = 397 µf What capacitance is needed to limit the ripple to 01 V? C = 4000 µf!!! EE 201 rectifiers 10
11 Fullwave rectifier With a few more diodes, we can rectify the entire sinusoidal input V S v v R R or V S The diodes are in a bridge configuration During the positive half cycle of the input, diodes 1 and 2 will be forward biased Current will flow from the positive source through those diodes and the resistor to generate a positive voltage across the resistor During the negative half cycle of the input, diodes 3 and 4 will be forward biased Current will flow from the negative source through those diodes and the resistor to generate a positive voltage across the resistor, again EE 201 rectifiers 11
12 v R 1 4 V S i R V S v R 2 i R 3 Note that there are no two diode drops in the conduction paths) Also, the frequency is effectively doubled EE 201 rectifiers 12
13 Fullwave peak rectifier Placing a capacitor in parallel with the load, turns the circuit into a fullwave peak rectifier It behaves essentially the same as the halfwave peak rectifier except with twice the frequency half the period) ) = sin V p = 8 V 1 V S 3 R 4 C v R 2 The ripple voltage is calculated in exactly the same way, except that the period is cut in half frequency doubled) =[ ] exp Same as doubling capacitance! EE 201 rectifiers 13
14 Example 3 You want to use a wall transformer that produces 10V RMS at the secondary to generate a DC voltage The desired voltage DC should be greater than 12 V and it should be able to supply at least 50 ma while keeping the voltage ripple to less than 5% Design the rectifier to meet these goals Note: f = 60 Hz) 10 V RMS 141 V amplitude effective R L V o / I o = 120 V / 50 ma) = 240! Note: This would be the minimum value of effective resistance If we choose C to meet the ripple requirement, then we will still be safe if we use a slightly higher V o Two options: halfwave or fullwave rectifier Try both Halfwave: V o max) = V p 07 V = 134 V V ripple 067 V = ln = 1350 µf V o avg) = V o max) V ripple / 2 = 1306 V EE 201 rectifiers 14
15 Fullwave: V o max) = V p 207 V) = 1274 V V ripple 064 V = ln ) = µ V o avg) = V o max) V ripple / 2 = 1242 V Either approach will work and meet the requirements The fullwave version uses extra diodes, but only half the capacitance Since diodes are nearly free pennies per piece), but big capacitors are relatively expensive, the fullwave circuit will actually cost less than the halfwave This is why fullwave rectifiers are used more commonly than halfwave rectifiers Component manufactures supply fullwave bridge rectifiers packaged as single unit with the transformer sinusoid as input the rectified waveform as the output EE 201 rectifiers 15
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