When tracking down a power supply failure, be sure to localize and fix the problem and not just replace the faulty component. For example: a fuse that always blows indicates that there is damage to another component in the circuit or a burnt resistor indicates that a transistor or capacitor has experienced a short circuit and so on. The steps that can be taken are as follows:
a. Visual Inspection
The troubleshooting should begin with a good visual inspection of the power supply. Check the fuses or reset the circuit breakers and look for burned, broken, charred or cracked components. These components should be replaced first. If the power supply is still on, touch the series pass-through transistor, voltage regulator or other active components to see if they are hotter than they should be. Some components are usually warm. Be careful when doing this step. Use a temperature gauge if possible.
b. Voltage Measurement
For practical purposes, remove the load from the power supply and measure the output voltage. If the measured voltage is correct, the problem is probably with the load and not the power supply. This troubleshooting technique is called divide and conquer. Start at the output of the suspected circuit, and if you get the correct voltage, continue this initial step by dividing the circuit into logical parts. The problem may be in a previous part or stage. For example, if the primary fuse in the power supply is blown, you need to remove the regulator part from the rectifier section and then see if the circuit blows the fuse again. This will tell you whether the problem is in the regulator section or not.
Figure 6.20: Several Steps of Visual Inspection and Voltage Measurement
Oscilloscope measurements can also be used, especially when the power supply is oscillating. This type of damage is usually caused by bypass capacitors located near the regulator IC or the deviation amplifier (depending on the type of regulator circuit used).
c. Current Measurement
Current measurement can show whether the current limiting circuit is working or not, and whether each pass transistor is supplying the load properly or only one transistor is working. If an ammeter is not available, you can place a high-power resistor of approximately 0.1? across the part through which the current passes. Measure the voltage across the resistor and then calculate the current through it using Ohm's law (I=E/R), where I is the current in amperes. E is the voltage in volts and R is the resistance in ohms.
d. Damage that usually occurs
- Components: Rectifier diodes, regulator ICs, series pass transistors or filter capacitors are shorted or open. Replace them as needed, but be sure to find the source of the problem before repairing the power supply.
- Voltage regulation is not correct: Check the regulator, voltage reference component (zener diode) or deviation amplifier (IC Op-Amp) in figure 6.14. If after the load is removed the output voltage is zero, check the part of the circuit that is not working properly.
- Oscillating power supply: Check the IC bypass capacitor if an IC voltage regulator is used (C=500pF in figure 6.18). If a transistor or op-amp is used, check the other bypass or stabilizing capacitors of the drift detector or drift amplifier.
- Overheating of the series bypass transistor: Check the series bypass transistor. If more than one bypass transistor is used and connected in parallel (See Figure 6.21), make sure that the transistors are compatible (One of the transistors may supply more current than the other transistors and cause excessive heat). Also, the heat generated during the equipment operation can be caused by changes in the resistor value.
- Series bypass, the current limiting circuit will not work, so the bypass transistor will become excessively hot. This can damage the transistor. If the bypass transistor is driven by a regulator IC, then excessive heat in the bypass transistor can occur if the thermal sensing IC is damaged.
e. Component Replacement
If you replace components, make sure that:
- Replacement components have appropriate value. For example, when replacing a capacitor, make sure not only that the value in microfarads is correct but also that it has the correct voltage.
- Replacement component specifications regarding current, power and tolerance. For example, each transistor will have different current and voltage specifications. They may also have power specifications which are usually smaller than the maximum voltage and current specifications.
- Never replace protective components such as fuses with other components that do not match the amperage. Using a fuse with a current rating that is too high will endanger the equipment and is a very big chance for damage to occur.
- If you are replacing a circuit on a PCB, make sure to use a soldering iron that is hot enough to melt the solder, but remember not to overheat it as this will damage the PCB. The copper layers on a multilayer PCB may require more heat, because the conductor and ground tracks are in the middle layer of the PCB. In this case, make sure that all layers are removed from the solder, otherwise it is possible to damage the copper layer in the middle of the PCB, if you forcefully remove the component. To protect the internal parts, cut off the damaged part and solder the new part to the protruding end of the PCB.
Basics of Electronic Circuit Damage Tracking
Basics of Electronic Circuit Damage Tracking
1. Circuit & Reading Test
- An electronic circuit is a collection of components connected together to perform a specific electronic function.
- Each component plays a role in circuit operation
- If a component is damaged, operations will drop drastically.
2. R1 Open Circuit
If R1 is open or open circuit, Q1 does not get forward bias current, as a result the collector voltage of Q1 will increase and Q2 will work, so that the Relay is active continuously. Damaged components show some specific symptoms, these symptoms are often used to determine the type of component and the type of damage.
This reading shows that Q1 is not conducting, since the base voltage of Q1 is zero volts, leaving R1 in an open circuit state.
The voltage measured with a multimeter at the test point of the circuit above, if it works properly without any input, is:
| Titik Uji | 1 | 2 | 3 |
|-----------|---------|---------|--------|
| Tegangan | + 0.7 V | + 0.1 V | + 24 V |However, with R1 open (open circuit) the meter reading changes to:
| Titik Uji | 1 | 2 | 3 |
|-----------|-----|---------|----------|
| Tegangan | 0 V | + 0.7 V | + 0.15 V |3. Components & Common Damage
| KOMPONEN | JENIS KERUSAKAN |
|---------------------------------------|-------------------------------------------------------------------------------------------------|
| Resistor | Nilainya membesar atau open |
| Resistor variabel | Terbuka atau mekanisme arus yang dihasilkan putus-putus (intermittent) |
| Induktor (termasuk trafo) | Lilitan open (putus) atau short (hubung singkat) |
| Katup terminoik piranti semikonduktor | Lilitan open (putus) atau short (hubung singkat), koil short dengan bingkai / frame (inti besi) |
| Dioda, Transistor, FET, SCR, dsb. | Filamen terbuka, elektroda short (grid dengan katoda), emisi (pancaran) rendah, short. |The first example of damage is given a linear series regulator circuit as shown in Figure 6.21. The way this circuit works is as follows: Tr2 and Tr3 as series control elements in a darlington connection. A full load current of 1 Ampere flows through Tr3 when the current at the base of Tr3 is around 40 mA. This current is obtained from Tr2 which itself requires a base current of between 1 and 2 mA. Tr1 functions as an error amplifier, where the inverting input is the base of Tr1 and the non-inverting input is the emitter which is kept constant by a 5.6 Volt zener. During normal conditions the base voltage of Tr1 is approximately 0.6 Volts higher than the emitter (6.2 Volts), therefore the voltage across R4 is also 6.2 Volts. If R3 is set to 1 K???? then the total voltage drop across R3 and R4 is 10 Volts.
Figure 6.21: Linear Series Regulator Circuit Using Darlington System Transistor.
If the output voltage drops due to an increase in load, then there will also be a decrease in voltage at the base of Tr1, while the emitter voltage is kept constant by the zener 5.6 Volts, then the voltage value of the base emitter Tr1 will decrease, so that Tr1 will not be increasingly on which makes the current from R2 will increasingly turn on Tr2 and also Tr3 which tends to correct the output voltage to return to 10 Volts again. Likewise, if the output voltage increases due to a decrease in load, the opposite process will occur automatically.
The normal condition voltages measured when the circuit is fully loaded with 1 Ampere are as follows:
If one of the components is damaged, there will be differences in the measurements, for example:
Here it can be seen that at TP 1 = 0 Volt, then the damage is a short-circuited zener diode, which will make the voltage at TP 2 small so that Tr2 and Tr3 are increasingly off and result in a very small output voltage.
Other damages are given the following measurement results:
Here it can be seen that TP 3 = 0 Volt, so the damage is R3 open (remember not R4 short circuit, because the damage resistance is never short circuited. See Chapter 4.3), which causes Tr1 to be off so that Tr2 and Tr3 are very on so that the output voltage is large and cannot be controlled.
Other measurement results are:
Because TP 2, 3, and 4 = 0 Volts, it means Tr2 and Tr3 are not working, this is due to two possibilities, namely R2 is open or C1 is short circuited.
And another measurement result is given:
From TP 2 is very large and the output result = 0 Volt, it can be ascertained that Tr2 is damaged, the base emitter connection is open. The second example is a simple inverter circuit as shown in Figure 6.22 below.
Figure 6.22: Inverter Circuit for Low Power
The way this circuit works is as follows:
Input 6 Vdc in the switch with frequency determined by Q1 and Q2 (astable multivibrator), connected to the CT of the transformer. The primary CT transformer is given 12-0-12 and the secondary 120 Volts. This signal is used to work Q3 and Q4 to conduct. When Q1 is off, its collector voltage rises and causes current to pass through the base of Q4 (conductor) so that current flows through half the wave in the primary winding.
In the next half wave of the astable, Q1 conducts so Q4 is off. At the same time, Q2 is off so Q3 conducts. Current now flows in the opposite direction through the half wave in the primary winding, so that ac is formed & this is induced into the secondary output of ±100 Vrms when the load current is 30 mA. The frequency is ±800 Hz. The function of R5 and C3 is as a filter to reduce the amplitude of the spike when the transistor changes from conduct to off or vice versa.
TP1 & TP4 maximum 0.8 V in the form of square wave. So in working condition from TP 1 to 6 is in the form of square wave signal. For the damages below shows that the output voltage of the secondary part is absent, and the voltage measured on the TP-TP is DC voltage.
- In measurement A because the voltage TP1 = TP4, TP2 = TP3 and TP5 = TP6, it means there is no short circuit damage. Because the astable does not work, the damage is C1 or C2 is open.
- In measurement B, it is seen that TP1 = 0, that means there is a short circuit with the ground related to TP1, namely Q1 collector and emitter are short circuited or Q4 base and emitter are short circuited. Damage is impossible R1 is open because if R1 is open there must be a small voltage on TP1, as in measurement C (on TP4).
- In the C measurement, it can be seen that TP4 is smaller than TP1, and this is caused by R4 being open.
- In the D measurement, it is seen that TP2 = 0, this means that there is a short circuit in the TP2 area, namely Q1 base and emitter are short circuited. But R2 can also be open. And in this damaged condition, Q4 becomes hot because Q4 becomes a continuous conductor.
So from the two examples of the circuit above, the most important thing is to first know how the circuit works. So that when there is damage and we take measurements, we can immediately find out which area is wrong (localize) and then determine the damaged components in that area. It takes a little analysis and logic and flying hours to become an expert in this matter.
Below is given table 6.1 which shows some of the damages and symptoms that occur in a regulated power supply.
Table 6.1: Common faults in regulated power supplies
Figure 6.29: One of the Computer Power Supply Models
How Inverter Board Parts Work
For newer LCD Monitor designs, the inverter board is combined with the power board as shown in the photo above. Older LCD Monitors have the inverter board separate from the power board as shown below.
There are four types of inverter designs (topologies) used in LCD Monitors.
- Buck Royer inverter
- Push pull inverter (Direct Drive)
- Half bridge inverter and (Direct Drive)
- Full bridge inverter (Direct Drive).
Numbers 2, 3 and 4 are called Direct Drive because they eliminate the need for inductors (buck chokes) and resonant capacitors found in conventional Royer Oscillators. In other words, the Direct Drive architecture reduces component count, lowers manufacturing costs and most importantly improves transformer design that optimizes performance.
1] Buck Royer Inverter
Block diagram of Buck Royer inverter
A Basic Schematic of a Buck Royer Circuit