长城电源维修关键测试点（Great Wall power supply maintenance key test point）

长城电源维修关键测试点（Great Wall power supply maintenance key test point） 长城电源维修关键测试点(Great Wall power supply maintenance key test point) Great Wall power supply maintenance key test point 1, KA7500B feet (power supply foot) : 12 normal data: + 12 v + 20 v or so, a few machines 26 v is normal. 2, LM339N 3 feet (foot) PG output: normal data: 5 v 3, KA7500B 8, and 11 feet (half bridge driver output) : normal data: + 2 v or so 4, KA7500B foot protection (foot) 4: normal data: 0 v, such as 3 v, the circuit in the protection state. 5, KA7500B 13, 14, 15 feet internal output pin (IC) : internal output 5 v. Great Wall series power supply IC exchange table: TL494 / KA7500B/BD494 / BDL494 S494PA/IR3M02 / MB3670 / MB3759 / MST894C/TL594 / ULN8186 DBL494 / ULS8194R/IR9494 / UPC494 UA494 / TL494CN Here's a little more detail: offline standby, two large after testing rectifier filter capacitor should have + 300 v dc voltage, ATX14 feet (the green line, PS) there should be a 5 v LM339 13 feet (PG) should be 0 v, ATX purple line should have a + 5 v, the other foot to 0 v. Short after green, black start power supply, ATX green line to 0 v, PG is 5 v, at the same time the output voltage of the ATX other foot should be normal. Continue to measure should be 12 feet of power supply of 7500 or 494 feet - 20 v dc 12 v power supply, 13, 14, 15 feet should have the 5 v output from the inside, 4 feet (dead zone, protect the foot) normal to 0 v, 8 feet, 11 feet should be 1.5-2 v driving voltage output. Which point voltage is wrong, check the relevant circuit, can find out the fault components. Some common faults add site: 1: insurance power cut Level before varistor, rectifier bridge, filter capacitor, two power tubes, backup power supply of power tube, first check whether the short circuit 2: electricity 300 v high voltage check 5 VSB for output, if yes, then check switch power supply voltage controller 4 foot, so 4 v, power protection, check the voltage sampling resistance and LM339, but experience, fast rectifier majority caused by short circuit protection, output filter capacitor explosion pulp caused as well. The other two control of power tube in the resistance and diode also want to check, the short circuit, switching power supply tube not their damage probability is not big, cannot be ignored, be sure to check it. Possible parts common protection problems: (1) 5 v12v fast rectifier circuit (2) one of the voltage sampling resistance or resistance change (3) the output filter capacitor explosion pulp (4) short circuit voltage output (5) the LM339 abnormal high and low level output 3:5 VSB no output (1) the starting resistance (a few hundred K size) burn out or resistance have change, the most damage (2) the power tube open circuit or short circuit (3) the back-up power supply pipe damage of peripheral resistance or diode (4) 5 VSB output rectifier diode short circuit (5) E junction after burn out small resistance values Step 3: this may be a little difficult for the little white people, for the master of the road is nothing, but for us this little white is difficult! I will say the process of my study, there are a lot of things that don't understand, I can only say a little bit! Must pay attention to is don't want disorderly open, because we are all idiots, dismantled can not restore back, can read circuit diagram is recommended to the pig's feet big stick son, look not to understand just follow me to do! First of all, the small white should know to understand the circuit board, the line is to see the front and back! Then have a piece of paper and a pen handy. Let's draw the line and draw it! Of course, the multimeter is adjusted to the resistance stall at any time! Because sometimes we look at the back line and we don't know if it's a line, We can test it with resistance. Now we are going to prepare the analysis. First of all, we have to analyze the 12V output, which is the yellow line, and we changed the good ATX from the yellow line to the output! We first to find it from the 12 v output area, see have solder will turn it over and see if it resistance, if it is followed, then each find a spot, you can draw it, on both symbols and words, anyway, you put the draw down towards the! Find resistance and then continue to follow, generally come with you and go out with a foot for 7500 b, if you find that from the 12 v output to 7500 b 1 foot line in the middle of more than one resistor you also need not afraid, no matter a few to last as long as you place them replaced with a precise resistance of 24 k 5 ring (blue) of the (in fact, the result is the 12 v output to 7500 b 1 foot just a 24 k precise resistance). Then we looked for the 5v output, with a resistance and a 1-foot connection of 7500B, and welded him into the empty space! When you find one foot, general will find that there are two or a resistor from 1 foot, and then connected to the 7500 b 7 feet (my is connected to the feet in the July, actually is also grounded)! If you find a resistance that has one foot attached to the 7 feet, then you don't have to do a few to weld it and replace it with 1 12k precision resistor (5 ring blue) If the above is smooth, we will look for the 2 feet of 7500B, and the 2 feet of 7500B should have a resistance and 7500B of 7 feet! Let's weld this resistor down! This resistor's position is left blank! Step 4: now we need a 10K precision potentiometer, which is actually a multi-circle winding resistance! If you can't find a normal 10k potentiometer, it's not too accurate. What I started with was the volume switch that was removed from the radio, and I could barely use it! The potentiometer usually has three joints. The first step is to find the median foot. The foot in the middle of the ordinary potentiometer is the middle one, and the precision potentiometer is the bottom of the foot. Then the middle leg leads to the 2 feet of 7500B, and then the other two joints receive 7500B 7 feet and 7500B 14 feet respectively! The reformed principle diagram So far, the transformation 0-15v has been completed. This is the process of my transformation. But I spent days trying to figure it out! Step 4: start testing, attach the load (I use the power fan), and then use the meter to test the voltage, test the voltage! If your circuit and mine should be okay, but if it's a lot worse, pray! I just say my process here, did not say must follow my do, give each small white a reference! Actually need not be afraid, because the overpressure protection of power supply is still in, if the voltage exceeds 15v, overpressure protection starts, at this time the output has no voltage! Re-plug it! Generally, I repair the power supply by fault phenomenon to analyze which part of the circuit has problems. Generally bad power supply has the following several kinds of fault imagination 1. After the electricity, the measurement is not 5vsb, the green line has no voltage 2. After power supply, there are 5vsb. The green line has 5v voltage, the fan is not working at short time, and there is no voltage output 3. After electricity, there are 5vsb. The green line has 5v voltage, and the fan will stop when it is short 4. After power supply, there are 5vsb. The green line has 5v voltage, and the fan will only stop when it is short 5. The voltage of each circuit is normal, but the load is light 6, the voltage is all normal, but the boot has a screeching 7, the voltage is either low or high 8. What's the problem with the power plug? If you don't take the computer, you have a single power Now, the problem of which part of the circuit is analyzed through the fault phenomenon 1 no 5vsb voltage Determining insurance is good (broken with 3.15 a 250v) / Make sure that the bridge/four rectifier diodes are good (the secondary tube is broken and must be changed in the same model or in four)] Make sure that two 200v/xxUF large capacitors are good (if the capacitor is out of the night, It must be bad in parallel with it. It is good to determine the auxiliary switch tube (the auxiliary switch tube is usually a triode or a field tube, if the breakdown of insurance must hang). Determine the diode, triode, and capacitance that are associated with the b pole/g of the switch tube If the above confirmation is finished, all are good words, the electricity test, the large capacitor has 150v voltage, the switch tube c pole/d is high pressure, may explain the auxiliary circuit hot ground Good. Now we're going to analyze the cold part It is good to determine the rectification diode of 5vsb It is good to determine the filter capacitance of 5vsb 2. After power supply, there are 5vsb. The green line has 5v voltage, the fan is not working at short time, and there is no voltage output Analysis: there are 5vsb voltage, indicating that the auxiliary circuit is normal, which should be a problem for the control output circuit, which is clear in 494 339ic Determine the 12v - 30v voltage of the 12 feet of 494 (this voltage is derived from the auxiliary transformer and can run down the circuit) Determine whether the 4-foot voltage of 494 is high level (high level protection) Determine whether the voltage of the 8 feet of 494 is 1.5-2v (the voltage of both feet should be equal) But it's important to point out that the three switch tubes C and D have a 494 8 and 11 feet that are likely to explode. 3. After electricity, there are 5vsb. The green line has 5v voltage, and the fan will stop when it is short Analysis: the fan turns around to indicate that there is a short circuit, power protection The key is to find the output voltage of the rectifier diode, capacitance 4. After power supply, there are 5vsb. The green line has 5v voltage, and the fan will only stop when it is short Analysis: the fan has been switched off a few times during the short time, indicating that the power supply has been overpressured, causing power protection The key is to find the rectifier diode of the output voltage of each channel, the capacitance, try to replace the optical coupling, 431, run down the circuit along the 1.2 feet of 494, and measure the resistance to them, the capacitance Whether it's damaged or not 5. The voltage of each circuit is normal, but the load is light Analysis: some of the components are aged and power is reduced Key suspects: 200v large capacitance, primary switch tube, output voltage of the rectifier tube, capacitance 6, the voltage is all normal, but the boot has a screeching The experience of the base predecessors is to change the output voltage capacitance 7, the voltage is either low or high The main suspect is the rectifier tube of the output voltage of each channel, capacitance. Try to replace the optical coupling, 431, along the 494 1.2 foot running line 8. What's the problem with the power plug? No computer, single power source (base ID: new). I think the green line is pulled down, and you can go down the green line to the line of 393, 494, and if the elements on the line are normal, you can try to change it to 393.494 Q: what is the problem with power plugging? If you don't take the computer, you have a single power According to my thinking, I should follow the green line to find the line between the two ics. If the elements on the line are good, try to exchange 494 and 393 There is an abnormal voltage when there is a scream. Check the 8 and 11 foot voltage of 494. What is the problem with power plugging? If you don't take the computer, you have a single power. Great Wall 300P4 auxiliary power to burn the necessary inspection components In practice, the Great Wall atx-300p4 - PFC power is burned by the auxiliary power supply tube, and I think it is necessary to check some of the components in the picture, as these are all possible problems 1) the resistance of 100K is like: may break 2) fixed capacitance 103/1kv parallel to 100K is now like: leakage 3) HER107 diode current image: reverse leakage 4) IN4148 diode current image: breakdown 5) 100 ohm current image: may break 6) electrolysis of 1UF / 50V is like: no capacity Have a comparison point, the question is, don't look at the main filter capacitance is no blister, the two 100 k equalizing resistance value can change or open circuit, so the main filter capacitor, and the two resistance, It is recommended to check, since I have found no small break in the voltage resistance of the capacitor Add the circuit diagram, the whole drawing actually has been downloaded from BBS, the symbol on the diagram is as follows: 1) R12 2) C31 microcomputer 3) D5 4) D7 5) R11 6) C32 3. Auxiliary power supply circuit As shown in figure 6, the output of the rectifier + 300 v dc bus voltage, the switch transformer primary (1) ~ (2) through T3 winding to the auxiliary power switch tube Q03 c, another path starting resistance R002 offer Q03 b a very positive bias voltage and starting current, make the Q03 began to conduction. Ic through the T3 primary (1) ~ (2) the winding, T3 (3) ~ (4) the feedback winding induced electromotive force (is), our fleet through positive feedback loops, D8, R06 to Q03 b very fast Q03 saturated conductivity, the Ic on the Q03 to maximum current, the current rate of change is zero, the D7 conduction, the resistance R05 sent out a comparison of voltage to the IC3 (photoelectric coupler Q817) (3) feet, at the same time T3 secondary windings to produce inductive electromotive force after D50, C04 rectifier filter, a path R01 current-limiting sent to (1) of the IC3's feet after another path R02 sent to IC4 (precision voltage regulator circuit TL431), because the Q03 saturation conduction from the secondary winding of induction electromotive force is smooth and stable, the output of IC4 K to the IC3 voltage change rate is almost zero, (2) feet make within the IC3 light-emitting diodes (leds) through the electric current is almost zero, as the light activated triode, resulting in Q1 closing. Feedback current through R06, R003, Q03 b and e greatly equivalent resistance of a capacitor charging our fleet, with our fleet charging voltage increases, the flow through Q03 b current gradually decreases, and the (3) ~ (4) on the feedback winding induction electromotive force began to decline, eventually make the T3 (3) ~ (4) the feedback winding induction electromotive force inverse (negative under positive), and with our fleet voltage superposition to Q03 b, make the b very potential negative and the switch tube Q03 for b very quickly without starting current. When switching tube Q03 cutoff, T3 - feedback windings, D7, R01, R02, R03, R04, R05, C09, IC3 and IC4 are composed of revibrator support. In the process of Q03 guide, the T3 primary winding converts the magnetic energy into the electric energy to provide voltage for the components in the circuit, while the control of the T3 feedback winding inducts the negative voltage, D7 guide and Q1 cut-off. When Q03 cut-off, T3 (4) side of the feedback winding induction positive voltage, D7 cut-off, T3 secondary windings on the two output induction electromotive force is positive, T3 magnetic energy into electrical energy stored after D50, C04 rectifier filter to provide a voltage change for IC4, conduction, makes the IC3 (1) and (2) feet inside the IC3 light-emitting diodes (leds) through the current increases, make the light activated triode, thus make Q1 conduction, to switch tube Q03 b provides the starting current, make the switch tube Q03 into conduction by the deadline. At the same time, the charging voltage of the positive feedback branch of C02 is based on the T3 feedback winding, R003, Q03's be pole equivalent resistance and R06 to form the discharge circuit. As the C41 charging current gradually decreases, the Ub potential of the switch tube Q03 increases, and when the Ub potential increases to the be pole opening voltage of Q03, Q03 again leads to the next period of oscillation. This cycle is repeated, forming a self-excited multivibrator. 8 D3 S (W5 e ^ - W % h (h) Q03 saturation during, T3 secondary winding induction electromotive force is negative, at the rectifier diode D9 and D50 cut-off, flows through the primary winding of the current flow in the form of magnetic energy stored in the auxiliary power supply transformer T3. When Q03 is switched to the cut-off by saturation, the induction electromotive force of the secondary winding two outputs is positive, and the magnetic energy stored in T3 is converted to power by D9, D50 rectifier output. The D50 rectifier output voltage is stabilized by the three-terminal voltage stabilizer 7805, and then the output + 5VSB after the inductive L7 filter. If the voltage is lost, the motherboard will not automatically wake up the ATX power supply. D9 rectifier output voltage supply IC2 (pulse width modulation integrated circuit KA7500B) of 12 feet (power input), the IC2 internal voltage, output voltage + 5 v from 14 feet, ATX power supply control circuit of relevant components of the operating voltage. T2 as the main power source excitation transformer, when vice power switch tube Q03 conduction, Ic through the T3 primary (1) ~ (2) winding, the T3 (3) ~ (4) the feedback winding induced electromotive force (is), and used for T2 primary, (2) ~ (3) winding induced electromotive force (negative under positive), the D5, D6, C8, R5 to Q02 b provides the starting current, make the main power switch tube Q02 conduction, an electrical current in the circuit, ensure the normal work of the whole circuit; At the same time, the feedback winding of T2 primary ~ is generated by induction electromotive force (upper and lower negative), D3 and D4 cut-off, and the main power supply switch tube Q01 is in the cutoff state. During the Q03 cutoff of the power switch tube, the working principle is the opposite of the above process, namely Q02 cutoff, Q01 work. Among them, D1 and D2 are continuation diode, which can provide continuous current during the closing and guiding of switch tube Q01 and Q02. Thus formed the main switch power it more resonant circuit, guarantee the normal work of the part to the T2 primary winding circuit, thus on T2 secondary windings produce inductive electromotive force to push the transistor Q3 and Q4 c, guarantee the stability of the whole driver circuit can continue to work, at the same time, again through the T2 primary winding reaction in T1 main transformer, switch power supply to the main power supply circuit to work, to provide load + 3.3 V, + 5 V, + 12 V working voltage