Power Supplies. Linear Regulated Supplies Switched Regulated Supplies Batteries

Transcription

1 Power Supplies Linear Regulated Supplies Switched Regulated Supplies Batteries

2 Im Alternating Current The Power -Im π/2 π 2π π t Im Idc Direct Current Supply π/2 π 2 π πt -Im ٢

3 Exercise 1 The current in a 10 Ω resistor is 5*sin(314t) A Draw the waveform of the current Define and calculate the following values for the current: Peak Peak to peak Average Root Mean Square (RMS) Calculate the value of the power dissipated by the resistor How much would be the current if it would be DC to generate the same power on the resistor? ٣

4 Power supply terminology P AV 1 T 2 R = i Rdt T = 0 T Effective (rms) value of alternating component of the wave ripplefactor(r)= Average(dc) value of the wave V ML -V FL 100= R0 Load regulation (%)= x 100 V RL V O Input regulation (% /V I N )= x 100 V I N V O Efficiency ( η )= T 0 i 2 dt I 2 eff P P R OUT I N V = V I O N I T 2 eff = Irms = i dt = 0 x I x I ٤ = L I N 1 T average i 2

5 Im Alternating Current Im Idc Direct Current πt π /2 π 2 π π/2 π 2π πt -Im -Im AC Input AC line components Step-down transformer Rectifier Filter Regulator DC Output Block diagram of a linear regulated power supply ٥

6 Unregulated supply with ac line components (a transient suppressor and line filter) Fuse ac line filter ac line transformer Filter capacitor 127 Vrms 60 Hz + Unregulated dc output Input socket Transient suppressor Snubber Bridge rectifier ٦

7 Fuse ac line filter ac line transformer Filter capacitor 127 Vrms + Unregulated dc output 60 Hz Input socket Transient suppressor Snubber Bridge rectifier fast blow fuses cut the power as quick as they can slow blow fuses tolerate more short term overload wire link fuses are just an open piece of wire, and have poorer overload characteristics than glass and ceramic fuses ٧

8 Transient suppressor and line filter Fuse ac line filter ac line transformer Filter capacitor 127 Vrms 60 Hz + Unregulated dc output Input socket Transient suppressor Snubber Bridge rectifier ٨

9 Fuse ac line filter ac line transformer Filter capacitor 127 Vrms 60 Hz + Unregulated dc output Input socket Transient suppressor Snubber Bridge rectifier RC Snubbers ٩

10 Transformer V V = transforme r +V unregulate d ripple 2η +V diode ١٠

11 V transforme r V = unregulate d +V ripple 2η +V diodes Exercise 2 For transformer in a power supply Required average output voltage = 10 V Ripple voltage = 1 V Diode drops = 2 V Output current (average) = 1 A Efficiency (η) of the transformer = 0.8 Find the required output voltage of the transformer Find the input current of the transformer if the input voltage is 220 V Find the output power delivered by the power supply Find the power loss by the transformer ١١

12 Rectifier Diodes Diode Maximum Current Maximum Reverse Voltage 1N4001 1A 50V 1N4002 1A 100V 1N4007 1A 1000V 1N5401 3A 100V 1N5408 3A 1000V ١٢

13 Half-Wave Rectified (Single Diode Rectifier) Vo = Vm - Vd with Vd 1volt. Vdc = (Vm - Vd)/π, Vrms = (Vm - Vd)/2 yielding a ripple factor (r) = Vrms 60 Hz VmSinωt D1 R L - + Output: half-wave varying DC (Pulsating DC) (using only half the AC wave) ١٣

14 D1 D2 + R L - Vo = Vm - Vd with Vd 1volt. Vdc = 2*(Vm - Vd)/π, Vrms = (Vm - Vd)/ 2 yielding a much reduced ripple factor that is r = Full-wave rectified with a center tapped transformer Vm π/2 π 2π -Vm ١٤

15 Bridge Rectifiers ١٥

16 D1 D3 D4 D2 R L - + Full-wave rectified with a bridge Vm D1 D4 D2 D3 π/2 π 2 π rectifier -Vm ١٦ Pulsating DC

17 Smoothing Filters Types of smoothing filters Capacitive Inductive L Section π - Section ١٧

18 Capacitive filter I L Amplitude Light load Rectifier diodes Filter capacitor Heavy load Filtered DC Time C = smoothing capacitance in farads (F) Io = output current from the supply in amps (A) Vs = supply voltage in volts (V), this is the peak value of the unsmoothed DC f = frequency of the AC supply in hertz (Hz) ١٨

19 Effect of Capacitance on Ripple Amplitude Light load + + Vi C R L Vo - - T1 T Heavy load T2 Time Assuming that the load current stays constant, the ripple voltage (peak to variation at the top of the waveform) can be approximately from the charge lost by the capacitor as I L =C*Vr/T 1 yielding Vr=I L *T 1 /C. T 1 T; Vr=I L /2fC for full-wave rectified; I L = V dc /R L ; r=2400/r L C and Vdc=(Vi Idc/C) where C is in μf and f = 60 Hz ١٩

20 Some Facts on Smoothing Capacitors C r charging time T 2 larger currents to flow through the rectifier diodes. Eventually, rectifier diodes and the transformer will be afflicted by increased I 2 R heating. Care for: The ripple voltage we can tolerate value of C and its tolerance Polarity The maximum DC voltage that the capacitor can withstand (the working DC (WVDC)) >50% more than the maximum voltage is a good choice Appreciable series inductive components may not behave as an effective capacitive element for high frequency spikes Add a small parallel capacitor Add a small series resistance (to conduction resistance of the diode and wire resistance of the transformer) Improves the ripple factor considerably. Limit the forward current extend the life of diodes and transformer The charged capacitor retains some charge A (bleeder) resistor (around 1 kω, 0.25 or 0.5 W) connected across discharges the capacitor in a few seconds. If a led indicator is connected, then no need for such a resistor. ٢٠

21 Inductive Filters + + L Vi R Vo L - - Vi: Rectified input voltage Vo: Filtered output volt R L : Effective load resistance Inductive filters have better control of the ripple for large load currents. The inductor behaves as a short circuit for the DC component. Hence, when 2fL» R L the DC value of the output is approximately 2Vi/π and the ripple factor r 0.118R L /fl where R L is the effective load resistance, f is the frequency of the ripple and L is the inductance (in Henry). ٢١

22 L and π section filters + L + + L + C Vi R Vo Vi Vo L C 1 R - C2 L L - section r = 0.83/LC Vdc = 0.636Vm π - section r = 3300/C 1 R L C 2 L Vdc = Vm-4200*Idc/C ٢٢

23 Summary of filter responses L + + Vi L + + L + Vi R Vo L Vo C R L Vo Vi R - C L Vo Vi C C 2 R L - Inductive Capacitive L - section π - section Vi: Rectified input voltage; Vo: Filtered output volt. R L : Effective load resistance; Frequency: 60 Hz Effective (rms) value of AC part Ripple factor (r) = Average (dc) value of output R L r L R L C LC C 1 R L C 2 L Vdc ٢٣ 0.636Vm Vm-4200Idc/C Vm Vm-4200Idc/C

24 Exercise 3 A series R-L circuit has R = 0.1 kω and L = 10 mh. The circuit is excited by Vi = sin(1000t) V Draw the circuit diagram Calculate the voltages across R and L ٢٤

25 Duties for next lecture Study linear (dissipative) regulated power supplies from the lecture notes Solve the exercises in this lecture in detail at home and bring it to the next lecture for a discussion in class Be prepared for a quiz and active learning ٢٥