Lead Acid Battery Charger Solution

One set of Zhejiang olonc wle-36v lead-acid battery charger with input AC220V / 80va and output DC36V / 1.6A. The charger failed to charge. Open the cover box to see that the workmanship is good. Two large-area aluminum door shaped radiators almost cover the component surface of the circuit board to dissipate heat for the power switch tube and fast recovery diode. It is found that there are many liquid leaks around the bottom of electrolytic capacitors and switching transformers after the power bridge is completed, and there are also liquid leaks around other electronic components to varying degrees. The mains fuse on the circuit board has been removed and seems to have been checked. First, clear the liquid leakage on the printed board, draw the printed circuit board according to the component layout, and sort out the electrical schematic diagram (see the attached figure) to analyze the circuit for your reference for maintenance and commissioning.

The 220V mains power supply passes through the fuse F101 and the power input filter composed of resistor R101, capacitor CX1 and inductance l101 to enhance the anti-interference performance. The thermistor ntc1 is connected in series to form a current surge suppression circuit, and then outputs a voltage of about 290v after filtering by the bridge rectifier kbp210 and electrolytic capacitor C101. One is sent to the primary L1 winding of switching transformer T1, and the other is sent to the grid of switching tube Q101 after being divided by R102 and R110 resistors to start the initial conduction. At the same time, the other winding L2 of T1 induces positive and negative voltages, This pulse voltage is sent to the gate of Q101 and accelerates the conduction of Q101. Since the drain current of Q101 does not change after saturation conduction, the induced voltage of L2 disappears, and the capacitor C104 changes from charging to discharging, so that the gate potential of Q101 gradually decreases, the current decreases, and Q101 quickly enters the cut-off state; Then the switch tube is started again, and the above process is repeated to form self-excited oscillation again and again. During the cut-off period of Q101, L3 winding of T1 outputs DC 45V voltage through fast diode d210. One circuit charges the battery and the other circuit supplies power to pin of voltage comparison integrated circuit lm358. Meanwhile, L4 winding of T1 outputs DC 20V voltage through rectifier diode D202 to supply power to precision adjustable reference three terminal integrated voltage regulator u201, integrated circuit IC lm358 and charging indication LED circuit. The voltage comparison circuit is composed of u201 and IC-2; The voltage stabilizing circuit of battery charger is composed of optocoupler HS2, diode d102, triode Q102 and q103. The resistors R105 and r108 of Q101 emitter, triode Q102 and q103 form overcurrent protection. The voltage stabilizing tubes Z101, Z201, z202 and z203 in the circuit are the overvoltage protection of each branch. The peak pulse absorption circuit is composed of D101, R111 and C102 to protect the safety during the cut-off period of Q101.

First, install the 250V fuse of power supply 2a, replace the damaged electronic components bridge bd101, electrolytic capacitor C101, capacitor C102 and triode Q102. Because the resistance color code of resistor r104 is not clear, select 330 . Use jl294 digital display transistor DC parameter test table to test the regulated value of each regulated tube Z101, Z201, z202 and z203, which are normal. Other components such as optocoupler HS2 and u201 were inspected and no abnormality was found. It indicates that the output circuit of the charger is basically normal.

In order to prevent the hidden trouble still exists, do not weld the switch tube Q101 and voltage stabilizing tube Z101, connect the isolation transformer and autotransformer at the mains power input of the power supply, and connect a voltmeter to the resistor R110 to monitor the partial voltage of resistors R102 and R110. First reduce the municipal voltage test. If there is no problem, adjust it to the rated voltage of 220V. The partial voltage of measuring resistors R102 and R110 is 17V, which is normal. After power failure, weld the voltage stabilizing tube Z101, conduct power on test, measure the voltage stabilizing value of Z101 to 15V, and check the potentials of Q102 and q103. No problems are found. At this time, a new switch tube Q101 (k2485n channel fet) can be disconnected and welded, and the output end is connected with a 36w60w bulb as the load. In the power on test, the load bulb does not light up and makes a "squeak" sound. The grid of Q101 is measured to be 2.2V, and the voltage after d102 rectification is measured to be 8V. The startup test is still invalid after C104 and short circuit thermistor are replaced. So weld down Q101, test with dm-1552c electronic laboratory special power supply, adjust the voltage of the instrument to 6V, and adjust the current to the minimum display of 0.04a, then connect the red lead of the instrument to the C pole of Q101 and the black lead to the S pole for test, and then use the gear rxl0 of 47 multimeter to measure that the forward and reverse resistance between D-S of this tube is 2 , which is normal. After the test, the switch can be started at the voltage of 6V and 0.04a. Therefore, it is thought that the performance of the newly purchased switch is different from that of the original one, and whether the starting current of the switch provided by the resistance R102 (1m ) is too small. It is decided to calculate the partial voltage resistance of R102 and rll0 based on the circuit voltage of 290v. Finally, R102 is 700k and R110 is 42K . Turn on the power and start the load bulb, Measure that the total mains input current of the charger is 0.36A, and it is normal. Check that the heat dissipation tube on Q101 and fast diode d210 is normal, and the fault is eliminated.