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ENCYCLOPEDIA OF RADIO ELECTRONICS AND ELECTRICAL ENGINEERING Calculation of power supplies. Encyclopedia of radio electronics and electrical engineering
Encyclopedia of radio electronics and electrical engineering / Beginner radio amateur The vast majority of amateur radio designs are powered by the mains through the power supply. It usually contains a network transformer T1 (Fig. 45), a diode rectifier VD1 -VD4 and a high-capacity oxide smoothing capacitor C1. Auxiliary, but necessary devices include the SA1 switch, the FU1 fuse and the on indicator - a miniature incandescent lamp HL1, with a rated voltage, a slightly higher voltage of the secondary winding of the transformer (lamps that burn with a short time last much longer). A voltage regulator, if present, is connected between the rectifier output and the load. The voltage at its output, as a rule, is less than Uout, and a noticeable power is spent on the stabilizer. Let's start with the calculation of the network transformer. Its dimensions and weight are completely determined by the power that the power supply should give: Рout = Uout ·Iout. If there are several secondary windings, then it is necessary to sum up all the powers consumed by each of the windings. To the calculated power, add the power of the indicator light Rind and the power losses on the rectifier diodes Rvyp = 2Upr Iout where Unp is the forward voltage drop across one diode, for silicon diodes it is 0,6 ... 1 V, depending on the current. Unp can be determined from the characteristics of the diodes given in the reference books. From the network, the transformer will consume power, somewhat larger than calculated, which is associated with losses in the transformer itself. There are "losses in copper" - for heating the windings when current passes through them - these are ordinary losses caused by the active resistance of the windings, and "losses in iron" caused by the work of remagnetization of the core and eddy currents in its plates. The ratio of the power consumed from the network to the output power is equal to the transformer efficiency η. The efficiency of low-power transformers is low and amounts to 60 ... 65%, increasing to 90% or more only for transformers with a power of several hundred watts. So, Рtr \uXNUMXd (Pout + Rind + Rvyp) / η Now it is possible to determine the cross-sectional area of the central rod of the core (passing through the coil) using the empirical formula: S2=Ptp. The designations of the magnetic circuits already contain data for determining the cross section. For example, W25x40 means the width of the central part of the W-shaped plate is 25 mm, and the thickness of the set of plates is 40 mm. Given the loose fit of the plates to each other and the insulation layer on the plates, the cross section of such a core can be estimated at 8 ... 9 cm2, and the power of the transformer wound on it - at 65 ... 80 W. The cross-sectional area of the central core of the transformer magnetic circuit S determines the next important parameter - the number of turns per volt. It should not be too small, otherwise the magnetic induction in the magnetic circuit increases, the core material enters saturation, while the idling current of the primary winding increases sharply, and its shape becomes non-sinusoidal - large current peaks appear at the tops of the positive and negative half-waves. The stray field and vibration of the plates sharply increase. The other extreme - an excessive number of turns per volt - leads to excessive consumption of copper and an increase in the active resistance of the windings. It is also necessary to reduce the diameter of the wire so that the windings fit in the window of the magnetic circuit. These issues are considered in more detail in [1]. The number of turns per volt n for factory transformers wound on a standard core of W-shaped plates is usually calculated from the ratio n \u45d (50 ... 2) / S, where S is taken in cm10. By determining n and multiplying it by the rated voltage of the winding, its number of turns is obtained. For secondary windings, the voltage should be taken XNUMX% more than the nominal voltage in order to take into account the voltage drop across their active resistance. All voltages on the transformer windings (UI and UII in Fig. 45) are taken in effective values.
The amplitude value of stresses will be 1,41 times higher. If the secondary winding is loaded on the bridge rectifier, then the voltage at the output of the rectifier Uout at idle is almost equal to the amplitude on the secondary winding. Under load, the rectified voltage decreases and becomes equal to: Uout = 1,41UII-2Unp-Ioutp.tp. Here rtp is the resistance of the transformer from the side of the secondary winding. With sufficient accuracy for practice, we can set rtp = (0,03 ... 0,07) Uout / Iout, and smaller coefficients are taken for more powerful transformers. Having determined the number of turns, one should find the currents in the windings. Secondary winding current Iii = Iind + Pout/UII. Active current of the primary winding (due to the load current) Iia = Ptr / UI. In addition, a reactive, "magnetizing" current also flows in the primary winding, creating a magnetic flux in the core, almost equal to the no-load current of the transformer. Its value is determined by the inductance L of the primary winding: Iip = Ui/2πfL In practice, the no-load current is determined experimentally - for a properly designed transformer of medium and high power, it is (0,1 ... 0,3) IiA. The reactive current depends on the number of turns per volt, decreasing as n increases. For low-power transformers, Iip = (0,5 ... 0,7) IiA is allowed. The active and reactive currents of the primary winding are added in quadrature, so the total current of the primary winding Ii2 = Iiai2 + Iipi2. Having determined the winding currents, it is necessary to find the wire diameter based on the permissible current density for transformers of 2 ... 3 A / mm2. The calculation is facilitated by the graph shown in Fig. 46 [2].
The possibility of placing the windings in the window is evaluated as follows: by measuring the height of the window (width of the coil), the number of turns of one layer of each winding is determined and then the required number of layers. By multiplying the number of layers by the diameter of the wire and adding the thickness of the insulating spacers, the thickness of the winding is obtained. The thickness of all windings should be no more than the width of the window. Moreover, since tight winding by hand is impossible, the resulting thickness of the windings should be increased by 1,2 ... 1,4 times. In conclusion, we present a simplified calculation of the rectifier (Fig. 45). The allowable direct average current of diodes in the bridge circuit must be at least 0,5Iout, in practice, diodes with a large forward current are chosen (for reliability). The permissible reverse voltage should not be less than 0,71 Uii + 0,5Uout, but since at idle Uout reaches 1,41Uii, it is advisable to choose the reverse voltage of the diodes not less than this value, i.e. the amplitude value of the voltage on the secondary winding. It is also useful to take into account possible fluctuations in the mains voltage. The amplitude of the rectified voltage ripple in volts can be estimated using a simplified formula: Upulse = 5Iout/S. The output current is substituted in amperes, the capacitance of the capacitor C1 is in microfarads. At load currents of several tens of milliamps or less, it is permissible to limit the simplest device with a zener diode. For high load currents, we recommend using a slightly more complex stabilizer, the circuit of which is shown in Fig. 47.
As you can see, here, to the simplest stabilizer on the elements R1, VD1, an emitter follower, assembled on a transistor VT1, is added. If in the simplest stabilizer the load current cannot be greater than the current of the zener diode, then here it can exceed the current of the zener diode by h21e times, where h21e is the static current transfer coefficient of the base of the transistor in a circuit with a common emitter. To increase it, a composite transistor is often used in place of VT1. The output voltage of the stabilizer is 0,6 V less than the stabilization voltage VD1 (1,2 V for a composite transistor). It is recommended to start the calculation of a stabilized power supply with a stabilizer. Based on the required voltage and load current, the transistor VT1 and the zener diode VD1 are selected. The base current of the transistor will be: Ib \u21d Iout / hXNUMXe. It will be the output current of the simplest stabilizer on the elements R1 and VD1. Then evaluate the minimum voltage at the output of the rectifier Uout-Upulse - it should be 2 ... 3 V more than the required voltage at the load even at the minimum allowable mains voltage. Further, the calculation is carried out in the described way. More advanced schemes and calculation of stabilizers are given in [3]. Self-test questions 1. Using the information from the previous sections (the impulse response of the RC circuit), derive the above formula for the amplitude of the ripple at the output of an unregulated rectifier. In this case, let the duration of the capacitor discharge to the rectifier load be 0,01 s (pulse frequency 100 Hz) and use the approximation et/RC - 1 - t/RC. 2. Having found an old mains transformer (it can be burned out), disassemble and unwind it, remembering or even writing down how it works (this will come in handy when making transformers yourself). Estimate the number of turns of the windings and the diameter of the wire. Calculate this transformer according to the described method and compare the results. 3. Calculate a fully regulated power supply for a voltage of 13,5 V and a current of 1 A. Replies The voltage waveform at the output of a full-wave rectifier without a smoothing capacitor is shown in Fig. 64 thin line. We see that the voltage pulsates from zero to Um with a frequency of 100 Hz. In the presence of a capacitor, it is charged at the peaks of the rectified voltage to a value slightly less than Um, and discharges between peaks. The average value of the rectified voltage is denoted as UO. pulsation amplitude - Upulse.
During the discharge of the capacitor, the voltage on it changes according to the law specified in the condition from the value UO + Upulse up to U valueO - ORpulse Therefore, one can write UO - ORpulse =(UO + Upulse)e-t/RC-(UO + Upulse).(1 - t/RC), where t = 0,01 s; R is the load resistance of the rectifier; C is the capacitance of the smoothing capacitor. Opening brackets, shortening UO and neglecting the term Upulset/RC due to its smallness (pulsation amplitude is less than UO) we get 2Upulse = UOt/RC. Note now that UO/R is equal to the load current I, and substitute t: Upulse = 5 10-3l/C, where all quantities must be substituted in basic units - volts, amperes and farads. If we take the current in milliamps, and the capacitance in microfarads, we get the above formula for the ripple voltage in volts: Upulse= 5 l/C. Literature
Author: V.Polyakov
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