Detailed explanation of mobile phone charger circuit diagram

Mobile phone charger circuit diagram (I)

This circuit is a self-excited, flyback, transformer-coupled, PWM switching power supply; the power conversion process: alternating current (AC, input mains) → direct current (DC) → alternating current (AC, high frequency) → direct current (DC, output); the circuit consists of four major systems: rectification, oscillation, voltage regulation, and protection.

Input rectification and filtering circuit: It is composed of diode VD1 and electrolytic capacitor C1. It is a half-wave rectification circuit. It outputs pulsating DC voltage with a peak voltage of 311v. After capacitor filtering, it reaches a DC voltage of about 300v. VD1 is 1N4007, which is a common diode. The rectification current is 1A and the reverse voltage is 1000v. The withstand voltage of the electrolytic capacitor must be greater than 300v.

Oscillation circuit: It is composed of R2, VT1, L1, L2, C4 and R5. Without the feedback branch of L2, C4 and R5, transistor VT1 leads a dull pastoral life. It provides a suitable bias voltage through the bias resistor R2 to form a general amplification circuit. However, the intervention of the third party - the feedback circuit makes its life no longer peaceful, but turbulent - forming an oscillating current.

L2 is a feedback coil. From the relationship between the same-named terminals of L1 and L2 in the figure, it can be seen that the feedback is positive feedback, thus forming an oscillation circuit. Due to the existence of capacitor C4, the oscillation formed by the oscillation circuit is intermittent oscillation, not a sine wave;

Starting process: When the circuit is connected, the starting resistor R2 provides bias current for the circuit, so the collector of VT1 has current Ic passing through Ic. When the current of the collector coil L1 changes (0→increase), a self-induced electromotive force will be generated, with a direction of + up and - down. Because L2 and L1 are wound on the same magnetic core, L2 generates an induced electromotive force of + up and - down under the action of mutual inductance; diangon.com All rights reserved. It is equivalent to a power supply, which forms a loop for charging through C4, R5, and the emitter junction of the transistor VT1. Therefore, the emitter junction voltage Ube of the transistor VT1 adds an additional current on the basis of the original bias current. Ib increases, and Ic increases accordingly. The corresponding mutual inductance electromotive force of L2 increases further, and the feedback is strong, so a very steep output waveform is formed at the output end.

However, this increase will not increase indefinitely, because the charging properties of the capacitor are as follows: the moment of connection is equivalent to a short circuit, and then it slowly increases, and the charging current gradually decreases, so the voltage across capacitor C2 gradually increases, the polarity is right + and left -, and this gradually increasing voltage forms an obstacle to positive feedback, the so-called "bringing out the apprentice will starve the master", when it reaches a certain value, its negative voltage is opposite to the polarity of the emitter junction bias voltage of the transistor VT1, causing Ube to gradually decrease, and when it decreases to 0.5v, the transistor is cut off.

At the cut-off time, the voltage of capacitor C4 reaches 0, and the charging current is zero. It will not stop because of this, and will discharge as long as there is a chance. Its negative voltage creates conditions for the power supply voltage to charge it, so the power supply voltage reversely charges it through R2, which not only offsets its original charging voltage but also reversely charges it, making its voltage left + right -, and gradually increasing. When it increases by more than 0.5v, the transistor has the conditions for conduction again, and a new round of oscillation begins again, and this cycle repeats over and over again.

From the above analysis, we can know that transistor VT1 acts as a switch, sometimes conducting and sometimes cutting off, constantly oscillating.

When the transistor VT1 is cut off, a mutual inductance electromotive force of up + and down - will be generated at both ends of L3, and electric energy will be output. After rectification and filtering by the diode, output DC is formed; VD7 and R6 are the output indication circuit; there will be output only when it is cut off, and there will be no output when it is turned on. This is the origin of the flyback type.

Voltage stabilization circuit

It is composed of transistors VT2, VD3, C3, VD4, and VD5; VD5 is turned on when the switch transistor VT1 is turned off: L2 is +, C3 is +, and diode VD5 forms a loop; C3 voltage is + up and - down, voltage is 6v, the upper end ground potential is 0v, then the lower end potential is -6v, this is a sampling voltage, which is a standard value, to make VD4 conduct, the left end potential of VD4 is 0.2v. When the voltage increases, the voltage of capacitor C3 increases, that is, the lower end potential is lower than -6v, and the voltage across VD4 remains unchanged, so the left end potential is pulled down, lower than 0.2v, which pulls down the base potential of transistor VT1, shortens its saturation time, and achieves the purpose of voltage stabilization.

Protection Circuit

Short circuit protection: realized by the input protection resistor R1, but when a serious short circuit fault occurs in the power supply, R1 will sacrifice itself and cut off the circuit to avoid further damage;

R3, C2, and VD2 are peak absorption circuits used to protect transistor VT1. As we know, at the moment of cutoff, the transistor will generate a self-inductance electromotive force of down + up -, which exceeds 1000v after superimposing with the power supply voltage, far exceeding the reverse voltage of the transistor. Through this circuit, this part of the electric energy can form a loop and release it. At the same time, a changing current is formed during the release process, which can couple the energy to L3;

Overcurrent protection: R4 is a sampling resistor. When the current of transistor VT1 increases, the emitter voltage of transistor VT1 increases, making transistor VT2 conduct, lowering the base voltage of VT1 and shortening its saturation time, thus achieving the purpose of protecting the transistor.

Diodes VD1, VD2, VD6, VD5, and VD4 have different operating frequencies, so different diodes are selected. Fast recovery diodes are used for high frequencies. A high-frequency transformer is used as the transformer.

Mobile phone charger circuit diagram (II)

When analyzing a power supply, we often start from the input. The 220V AC input is rectified by a 4007 half-wave rectifier at one end and filtered by a 10uF capacitor at the other end after passing through a 10-ohm resistor. This 10-ohm resistor is used for protection. If a fault occurs later and causes overcurrent, this resistor will be burned out to avoid causing a larger fault. The 4007, 4700pF capacitor, and 82KΩ resistor on the right constitute a high-voltage absorption circuit. When the switch tube 13003 is turned off, it is responsible for absorbing the induced voltage on the coil, thereby preventing the high voltage from being applied to the switch tube 13003 and causing breakdown. 13003 is a switch tube (the full name should be MJE13003), with a withstand voltage of 400V, a maximum collector current of 1.5A, and a maximum collector power consumption of 14W. It is used to control the on and off between the primary winding and the power supply. When the primary winding is constantly on and off, a changing magnetic field will be formed in the switching transformer, thereby generating an induced voltage in the secondary winding. Since the same-name ends of the windings are not marked in the figure, it is impossible to tell whether it is a forward or flyback type.

However, from the structure of this circuit, it can be inferred that this power supply should be a flyback type. The 510KΩ on the left is a startup resistor, which provides the base current for the switch tube to start. The 10Ω resistor below 13003 is a current sampling resistor. The current is converted into a voltage (its value is 10*I) after sampling. This voltage is added to the base of transistor C945 after passing through diode 4148. When the sampling voltage is approximately greater than 1.4V, that is, the switch tube current is greater than 0.14A, transistor C945 is turned on, thereby lowering the base voltage of the switch tube 13003, thereby reducing the collector current, thus limiting the current of the switch to prevent the current from being too large and burning (in fact, this is a constant current structure, which limits the maximum current of the switch tube to about 140mA).

The induced voltage of the winding (sampling winding) at the lower left of the transformer is rectified by the rectifier diode 4148 and filtered by the 22uF capacitor to form a sampling voltage. For the convenience of analysis, we take one end of the emitter of the transistor C945 as the ground. Then the sampling voltage is negative (about -4V), and the higher the output voltage, the more negative the sampling voltage. After the sampling voltage passes through the 6.2V voltage regulator diode, it is added to the base of the switch tube 13003. As mentioned earlier, when the output voltage is higher, the sampling voltage is more negative. When it is negative to a certain extent, the 6.2V voltage regulator diode is broken down, thereby lowering the base potential of the switch 13003, which will cause the switch tube to disconnect or delay the conduction of the switch, thereby controlling the energy input into the transformer, and also controlling the increase of the output voltage, realizing the function of voltage stabilization output.

The 1KΩ resistor and the 2700pF capacitor in series at the bottom are the positive feedback branch, which takes the induced voltage from the sampling winding and adds it to the base of the switch tube to maintain oscillation. There is not much to say about the secondary winding on the right. It outputs a 6V voltage after being rectified by the diode RF93 and filtered by the 220uF capacitor. I didn't find any information about the diode RF93, but it is probably a fast recovery tube, such as a Schottky diode. Because the operating frequency of the switching power supply is relatively high, a diode with an operating frequency is required. Common Schottky diodes such as 1N5816 and 1N5817 can be used here instead.

Also because of the high frequency, the transformer must also use a high-frequency switching transformer. The core is generally a high-frequency ferrite core with a high resistivity to reduce eddy current.