Power supplies with a capacitor voltage divider. Scheme, description

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Power supplies with a capacitor voltage divider. Encyclopedia of radio electronics and electrical engineering

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Encyclopedia of radio electronics and electrical engineering / Power Supplies

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A network power supply with a quenching capacitor (Fig. 1), in fact, is a voltage divider, in which the upper shoulder is a capacitor, and the lower one is a complex nonlinear diode-resistor-capacitor circuit. This determines the disadvantages (and advantages, of course) of such devices.

Power supplies with capacitor voltage divider
Ris.1

In order for the source to operate in a wide range of load current with high efficiency, it is enough to make the input voltage divider purely reactive, for example, capacitor (Fig. 2). It allows you to additionally stabilize the output voltage of the source with a series-connected compensation or switching regulator, which cannot be done in a conventional source with a quenching capacitor. As shown in the article by S. Biryukov "Calculation of a network power supply with a quenching capacitor" - "Radio", 1997, No. 5, p. 48-50, - a series stabilizer can only be used if the voltage at its input is limited, which again significantly reduces the efficiency.

Power supplies with capacitor voltage divider
Ris.2

It is advisable to use a source with a capacitor voltage divider for joint operation with switching regulators. It is ideal for a device that consumes low current for a long time, but at some point requires a sharp increase in it. An example is an apartment watchdog on microcircuits "MOS with an actuating unit on a relay and an audible signaling device.

The current drawn by the capacitor divider will have a phase shift of 90 degrees. relative to the mains voltage, so the voltage divider on the reactive elements does not require cooling. Based on the foregoing, it seems that the current through the divider can be chosen arbitrarily large. However, an unjustified increase in the divider current will lead to active losses in the wires and to an increase in the mass and volume of the device. Therefore, it is advisable to take the current through the voltage divider within 0,5 ... 3 of the maximum load current.

The calculation of a source with a capacitive divider is simple. As follows from formula (2) in the mentioned article, the output voltage Uout and the total output current (zener diode and load Iout) of the source according to scheme 1,a are related as follows:

Iout = 4fC1(2Uc-Uout).

This formula is also suitable for calculating a source with a capacitor divider, in it you just need to replace C1 with the total capacitance of capacitors C1 and C2 connected in parallel, shown in Fig. 2. a Uc - on Uc2x (voltage across capacitor C2 at RH = °°), i.e. Uc2x = = Uc-C1 / (C1 + C2). Then 1out = 4f(C1+C2)xx[Uc-C1-i/2/(C1+C2)-Unbix] or after obvious transformations 1out = 4f-C1 [Uc^2 --out(1+C2/C1) ].

Since the voltage drop across the diodes of the bridge Ud at small values of Kout becomes noticeable, we finally get 1out \u4d 1f-C2 [Uc ^ / 2- (Cout + 1Cd) (2 + + C1 / CXNUMX)].

It can be seen from the formula that at Рн=0 (i.e., at Uout=0), the current Iout, if we neglect the voltage drop across the diodes, remains the same as for the power source assembled according to scheme 1,a. The voltage at the output without load decreases: Uauxx = =Uc-C1^/2/(C1+C2)-2Un.

The capacitance and operating voltage of the capacitor C2 is selected based on the required output voltage - the ratio of the capacitance values C1 / C2 is inversely proportional to the values of the voltage falling on C1 and C2. For example, if C1" = 1 μF, and C2 = 4 μF, then the voltage Uc1 will be equal to 4/5 of the mains voltage, and Uc2 \u5d Uc / 220, which, with a mains voltage of Uc \u186d 44 V, corresponds to 1,5 and XNUMX V. It must be taken into account that that the amplitude value of the voltage is almost XNUMX times higher than the current one, and select capacitors for the appropriate rated voltage.

Despite the fact that theoretically capacitors in an AC circuit do not consume power, in reality, some heat can be generated in them due to the presence of losses. You can check in advance the suitability of the capacitor for use in the source by simply connecting it to the mains and estimating the case temperature after half an hour. If the capacitor C1 has time to noticeably warm up, it should be considered unsuitable for use in the source.

Special capacitors for industrial electrical installations practically do not heat up - they are designed for high reactive power. Such capacitors are used in fluorescent lamps, in ballasts of asynchronous electric motors, etc.

Power supplies with capacitor voltage divider
Ris.3

Below are two practical power supply circuits with a capacitor divider: a five-volt general purpose one (Fig. 3) for a load current of up to 0,3 A and an uninterruptible power supply for quartz electronic-mechanical watches (Fig. 4).

Power supplies with capacitor voltage divider
Ris.4

The voltage divider of a five-volt source consists of a paper capacitor C1 and two oxide capacitors C2 and C3, which form a non-polar shoulder with a capacity of 100 microfarads. The polarizing diodes for the oxide pair are left-handed bridge diodes according to the scheme. With the ratings of the elements indicated in the diagram, the circuit current (at Rn = 0) is 600 mA, the voltage across the capacitor C4 in the absence of load is 27 V.

Electronic-mechanical watches are usually powered by a single galvanic cell with a voltage of 1,5 V. The proposed source generates a voltage of 1,4 V at an average load current of 1 mA. The voltage removed from the C1C2 divider rectifies the node on the elements VD1, VD2, C3. Without load, the voltage across capacitor C3 does not exceed 12V.

Author: O. Khovaiko, Moscow; Publication: N. Bolshakov, rf.atnn.ru

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