Properties of electrical signals


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1 DC Voltage Component (Average voltage) Properties of electrical signals v(t) = V DC + v ac (t) V DC is the voltage value displayed on a DC voltmeter Triangular waveform DC component Halfwave rectifier output Fullwave rectifier output Triangular waveform AC component v ac (t) = v(t) V DC = v(t) V m / 2 Effective Current Average power over a load resistor Effective current is given by I rms is the current value displayed on a AC ammeter 1
2 Effective Voltage Average power over a load resistor Example: v(t) = A + B cos(wt) V rms =? Effective voltage is given by V rms is the voltage value displayed on a AC voltmeter Triangular waveform RMS value AC Triangular Waveform RMS value Halfwave rectifier output RMS value Fullwave rectifier output RMS value Halfwave rectifier output AC component RMS value Fullwave rectifier output AC component RMS value 2
3 Half Wave Rectifier Half Wave Rectifier Output Diodes convert AC to DC in a process called rectification. The diode only conducts for onehalf of the AC cycle. The remaining half is either all positive or all negative. This is a crude AC to DC conversion. The DC Voltage out of the diode : V DC = 0.318Vm where V m = the peak voltage PIV (PRV) Because the diode is only forward biased for onehalf of the AC cycle, it is then also off for onehalf of the AC cycle. It is important that the reverse breakdown voltage rating of the diode be high enough to withstand the peak AC voltage. Full Wave Rectification The rectification process can be improved by using more diodes in a Full Wave Rectifier circuit. Full Wave rectification produces a greater DC output. PIV (PRV) > V m PIV = Peak Inverse Voltage PRV = Peak Reverse Voltage V m = Peak AC Voltage Half Wave Rectifier PIV (halfwave rectifier) = V m Center Tapped Transformer Rectifier Circuit Full Wave Rectifier Circuits There are two Full Wave Rectifier circuits: Center Taped Transformer Rectifier Bridge Rectifier Two diodes and a centertapped transformer are required. V DC = 0.636(V m ) Note that V m here is the transformer secondary voltage to the tap. 3
4 Operation of the Center Tapped Transformer Rectifier Circuit For the positive half of the AC cycle: Bridge Rectifier Circuit PIV (centertapped) = 2V m For the negative half of the AC cycle: Four diodes are required. V DC = Vm Operation of the Bridge Rectifier Circuit For the positive half of the AC cycle: Rectifier Circuit Summary PIV (bridge) = V m For the negative half of the AC cycle: Note: V m = peak of the AC voltage. Practical Applications of Diode Circuits Rectifier Circuits Conversions of AC to DC for DC operated circuits Battery Charging Circuits Power Supply Diagram Simple Diode Circuits Protective Circuits against Overcurrent Polarity Reversal Currents caused by an inductive kick in a relay circuit Zener Circuits Overvoltage Protection Setting Reference Voltages 4
5 Voltage Regulation Filter Circuits The amount of variation in DC output voltage due to varying loads from noload to fullload is called voltage regulation. Voltage Regulation: V NL = noload voltage V FL = fullload voltage The output from the rectifier section is a pulsating DC. The filter circuit reduces the peaktopeak pulses to a small ripple voltage. Ripple Factor Halfwave Rectifier Ripple Factor DC output: AC Ripple output: Ripple Factor: (Note V m is the peak rectifier output voltage.) Fullwave Rectifier Ripple Factor After the filter circuit a small amount of AC is still remaining. The amount of ripple voltage can be rated in terms of Ripple Factor (r). DC output: AC Ripple output: Ripple Factor: V r(rms) = RMS value of the AC ripple voltage The fullwave rectifier has a significantly lower ripple factor. Types of Filter Circuits Capacitor Filter Capacitor Filter RC Filter Capacitor Filter 5
6 Capacitor Filter Output Ripple Voltage with a Capacitor Filter A capacitor significantly reduces the AC content of the rectified signal. For light load (r < 6.5%), ripple voltage: The larger the capacitor the smaller the ripple voltage. DC Output with a Capacitor Filter Diode Ratings with Capacitor Filter A capacitor increases the DC output. DC Output: Note: V n = peak rectified voltage I DC is the load current in ma. The size of the capacitor increases the current drawn through the diodes. The larger the capacitance, the greater the amount of current. Peak Current vs. Capacitance: The capacitor reduces the ripple factor. Ripple Factor with a Capacitor Filter C = capacitance V = change in capacitor voltage during charge/discharge t = the charge/discharge time A smaller capacitor will reduce the peak current through the diodes. Ripple factor for light load (r < 6.5%): Exercise (Midterm 1, ) Output voltage: (a) small C (b) large C Since the average current drawn from the supply must equal to the average diode current during the charging period, the following relation can be used 6
7 RC Filter Circuit Additional RC Filter Adding an RC section will further reduce the ripple voltage and decrease the surge current through the diodes. DC Operation Since both capacitors are opencircuit for DC operation, the resultant output voltage is AC Operation Due to voltage divider action of the capacitor AC impedance and the load resistor, the AC component voltage over the load is given by where V DC is the voltage on C 1 SIMPLIFICATION Example: f ripple = 50Hz, C 2 =10 F, R L =2K If R L >> X C e.g. R L 5X C then Z X C Consequently, V' r (rms) could be written as If R L >> X C and R >> X C (generally this won t hold) then V' r (rms) may be written as e.g. R L > 5X C and R > 6X C , thus the assumption Z = X C holds as R L > 5X C 7
8 Ripple Voltage in an RC Filter Circuit Relation between r' and r The ripple voltage is significantly reduced by the addition of an RC circuit. Ripple Voltage: V r(rms) = ripple voltage after the RC filter V r(rms) = ripple voltage before the RC filter R = resistor in the added RC filter X C = reactance of the capacitor in the added RC filter This RC filter gives better results for light loads as the capacitor filter. For heavy loads, it shows high ripples and low voltage regulation Example: C 1 =15 F, R=500, C 2 = 10 F, R L =5K, V DC =150V, V AC (rms) =15V. (f mains =50Hz) Find the DC and AC voltages over the load Find the ripple factors, %r and %r' values Find the voltage regulation factor %VR. Ripples are 3 times smaller Ripple factor dropped by 3 times 13.6V voltage drop  filters Resistor R in the RC filter is replaced by inductor L, as the DC resistance R l of the coil is small and but the AC reactance X L is high. Example: For the filter shown in the figure, the output DC voltage and current are given as 200V and 50mA. V DC(C1) = 220V, V r(c1) = 12V (rms) and the frequency of the ripple voltage f ripple = 100Hz. In order to satisfy r 0.02 (%2) find R L, R l, L and C 2. NOTE: R l denotes the DC resistance of the coil. 8
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