Simple DC voltage doubling circuit

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Detailed explanation of the principle of voltage doubler circuit

Note: To understand the voltage doubling circuit, we must first consider the charged capacitor as a power source. It can be connected in series with the power supply, just like the principle of connecting ordinary batteries in series.

DC half-wave rectifier voltage circuit 1) In the negative half cycle, that is, when A is negative and B is positive, D1 is turned on and D2 is turned off. The power supply charges capacitor C1 through D1. Under ideal conditions, D1 can be regarded as a short circuit in this half cycle. At the same time, capacitor C1 is charged to Vm. Its current path and the polarity of capacitor C1 are shown in the figure above (a). (2) In the positive half cycle, that is, when A is positive and B is negative, D1 is turned off and D2 is turned on. At this time, the voltage after the power supply and C1 are connected in series is 2Vm, so C2 is charged to the maximum value of 2Vm. Its current path and the polarity of capacitor C are shown in the figure above (b).

It should be noted that:

(1) In fact, the voltage of C2 cannot be charged to 2Vm within one and a half cycles. It must take several weeks for it to gradually approach 2Vm. For the sake of convenience, the following circuit description also makes this assumption.

(2) If the half-wave voltage doubler is used in a power supply without a transformer, we must connect a current limiting resistor in series with C1 to protect the diode from the inrush current when the power supply is first charged.

(3) If there is a load connected in parallel with the output of the voltage doubler, as would be expected, the voltage on capacitor C2 will drop during the negative half cycle (at the input) and then be charged back to 2Vm during the positive half cycle as shown in the figure below.

Therefore, the voltage waveform on capacitor C2 is a half-wave signal filtered by the capacitor filter, so this voltage doubler circuit is called a half-wave voltage circuit.

(4) During the positive half cycle, the maximum reverse voltage that diode D1 withstands is 2Vm. During the negative half cycle, the maximum reverse voltage that diode D2 withstands is also 2Vm. Therefore, a diode with PIV>2Vm should be selected in the circuit.

Figure 3 Output voltage waveform

Simple DC voltage doubling circuit

At the moment 1.5V is turned on, the 1.5V DC voltage charges C2 through the energy storage inductor L and R1. Since the voltage across the capacitor cannot change suddenly, the base voltage of VT1 is almost zero, so VT1 is turned on, which makes VT2 saturated and turned on. At this time, the current of L will gradually increase from small, and L converts electrical energy into magnetic energy and stores it. In this process, VD2 is cut off, Vo=0V, and the voltage stabilizing circuit composed of VT3, R2, VD3, and VD1 does not work. ②When the current in L no longer changes, the base potential of VT1 also increases to the maximum. At this time, VT1 turns from on to off, and VT2 is also cut off. Since the current flowing through L cannot change suddenly, a reverse induced electromotive force UL will be generated at both ends of L, with its polarity of negative on the left and positive on the right. This UL is connected in series with 1.5V DC to increase the voltage between the collector of VT2 and the ground. At this time, VD2 is turned on. When Vo is greater than 9V, VT3 is turned on. The higher Vo is, the higher the voltage applied to the base of VT1 through VT3, the smaller the conduction angle of VT1, the conduction angle of VT2 also decreases, and the current flowing through L also decreases, thereby controlling the amount of energy stored in L, and achieving dynamic stability of Vo at a certain voltage value (9V).

This DC double voltage circuit using 555 integrated circuit can generate a DC output voltage approximately twice the DC supply voltage.

mao45

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