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1.
Transformer saturation
Transformer saturation phenomenon
When starting up under high or low voltage input (including light load, heavy load, capacitive load), output short circuit, dynamic load, high temperature, etc., the current through the transformer (and switch tube) will increase nonlinearly. When this phenomenon occurs , the peak value of the current cannot be predicted and controlled, which may lead to current over-stress and the resulting over-voltage damage of the switching tube.
Current waveform when transformer is saturated
Saturation is likely to occur:
1) The inductance of the transformer is too large;
2) The number of turns is too few;
3) The saturation current point of the transformer is smaller than the maximum current limit point of the IC;
4) There is no soft start.
Solution:
1) Reduce the current limit point of the IC;
2) Strengthen the soft start to make the current envelope through the transformer rise more slowly.
Stress requirements for Vds:
Under the worst conditions (highest input voltage, maximum load, highest ambient temperature, power supply start-up or short-circuit test), the maximum value of Vds should not exceed 90% of the rated specification.
Ways to reduce Vds:
1) Reduce the platform voltage: reduce the ratio of primary and secondary turns of the transformer;
2) Reduce peak voltage:
a. Reduce leakage inductance:
The leakage inductance of the transformer stores energy when the switch tube is turned on, which is the main reason for generating this peak voltage. Reducing the leakage inductance can reduce the peak voltage.
b. Adjust the absorption circuit:
① Use TVS tubes;
② Use slower diodes, which can absorb a certain amount of energy (peaks);
③ Inserting damping resistors can make the waveform smoother and help reduce EMI.
3. IC temperature is too high
Causes and solutions:
1) The internal MOSFET loss is too large:
the switching loss is too large, and the parasitic capacitance of the transformer is too large, resulting in a large cross area between the MOSFET's turn-on and turn-off current and Vds. Solution: Increase the distance between the transformer windings to reduce the interlayer capacitance. Just like when the windings are wound in multiple layers, add a layer of insulating tape (interlayer insulation) between the layers.
2) Poor heat dissipation:
A large part of the heat of the IC is guided to the PCB and the copper foil on it through the pins. The area of the copper foil should be increased as much as possible and more solder should be applied.
3) The air temperature around the IC is too high:
The IC should be in a place with smooth air flow and should be away from parts with too high temperatures.
4. Unable to start under no load or light load
Phenomenon:
It cannot start at no load or light load, and Vcc repeatedly jumps back and forth from the startup voltage and the shutdown voltage.
Reason:
At no-load or light load, the induced voltage of the Vcc winding is too low, and it enters a repeated restart state.
Solution:
Increase the number of Vcc winding turns, reduce the Vcc current limiting resistor, and add a dummy load appropriately. If you increase the number of Vcc winding turns and reduce the Vcc current limiting resistor, Vcc becomes too high under heavy load. Please refer to the method of stabilizing Vcc.
5. Cannot reload after startup
Causes and solutions:
1) Vcc is too high under heavy load. Under heavy
load , the induced voltage of the Vcc winding is high. When Vcc is too high and reaches the OVP point of the IC, the overvoltage protection of the IC will be triggered, causing no output. If the voltage rises further and exceeds the IC's ability to withstand it, the IC will be damaged.
2) Internal current limit is triggered
a. The current limit point is too low.
When under heavy load or capacitive load, if the current limit point is too low, the current flowing through the MOSFET will be limited and insufficient, resulting in insufficient output. The solution is to increase the current limiting pin resistance and raise the current limiting point.
b. The rising slope of the current is too large.
If the rising slope is too large, the peak value of the current will be larger, which will easily trigger the internal current limiting protection. The solution is to increase the inductance without saturating the transformer.
6. High standby input power
Phenomenon:
Vcc is insufficient at no load or light load. This situation will cause the input power to be too high and the output ripple to be too large at no load or light load.
reason:
The reason why the input power is too high is that when Vcc is insufficient, the IC enters a repeated starting state and frequently requires high voltage to charge the Vcc capacitor, causing losses in the starting circuit. If there is a resistor in series between the startup pin and the high voltage, the power consumption on the resistor will be larger at this time, so the power level of the startup resistor must be sufficient.
The power IC has not entered Burst Mode or has entered Burst Mode, but the Burst frequency is too high, the switching times are too many, and the switching loss is too large.
Solution:
Adjust the feedback parameters to reduce the feedback speed.
7. Short circuit power is too large
Phenomenon:
When the output is short-circuited, the input power is too large and Vds is too high.
Reason:
When the output is short-circuited, there are many repetitive pulses and the peak value of the switch tube current is very large. As a result, the input power is too large and the switch tube current stores too much energy on the leakage inductance, causing Vds to be high when the switch tube is turned off.
There are two possibilities for the switch to stop working when the output is short-circuited:
1) Triggering OCP can stop the switching action immediately
a. Trigger the OCP of the feedback pin;
b. The switching action stops;
c. Vcc drops to the IC shutdown voltage;
d. Vcc rises to the IC startup voltage again and restarts.
2) Trigger internal current limit
When this method occurs, the available duty cycle is limited, and the switching action is stopped by relying on Vcc to drop to the UVLO lower limit. However, the Vcc drop time is longer, that is, the switching action is maintained for a longer time, and the input power will be larger.
a. The internal current limit is triggered and the duty cycle is limited;
b. Vcc drops to the IC shutdown voltage;
c. The switching action stops;
d. Vcc rises to the IC startup voltage again and restarts.
Solution:
1) Reduce the number of current pulses so that the OCP of the feedback pin is triggered when the output is short-circuited, which can quickly stop the switching action and reduce the number of current pulses. This means that when a short circuit occurs, the voltage at the feedback pin should rise faster. Therefore, the capacitance of the feedback pin should not be too large;
2) Reduce the peak current.
8. No-load, light-load output ripple is too large
Phenomenon:
Vcc is insufficient at no load or light load.
Reason:
When Vcc is insufficient, the IC oscillates between the startup voltage (such as 12V) and the shutdown voltage (such as 8V) intermittently with a long period, providing energy to the output for a short period of time, and then stops working for a longer period of time, causing the capacitor to The stored energy is not enough to maintain output stability, and the output voltage will drop.
Solution:
Ensure that Vcc can be supplied stably under any load conditions.
Phenomenon:
In Burst Mode, the frequency of intermittent operation is too low. This frequency is too low, and the energy of the output capacitor cannot remain stable.
Solution:
Slightly increase the frequency of intermittent operation and increase the output capacitance while meeting the standby power consumption requirements.
9. Cannot start under heavy load or capacitive load
Phenomenon:
It can start under light load, and it can also start under heavy load after starting, but it cannot start under heavy load or large capacitive load.
General design requirements:
Regardless of heavy load or capacitive load (such as 10000uF), the input voltage is still the lowest, and the output voltage must rise to a stable value within 20mS.
Reasons and solutions
(provided that Vcc is within the normal working range):
The following takes the capacitive load C=10000uF as an example for analysis.
According to the specifications, there must be enough energy to make the output rise to a stable output voltage (such as 5V) within 20mS.
E=0.5*C*V^2
The larger the capacitance C, the greater the energy that needs to be transferred from input to output within 20mS.
Taking the chip FSQ0170RNA as an example, as shown in the figure, the total area S of the shaded part is the required energy. To increase the area S, the method is:
1) Increasing the peak current limit point I_limit allows a larger inductor current Id to flow: increase the resistance connected to Pin4, and the shunt from the internal current source Ifb is smaller, making the PWM comparator used as the current limit reference voltage The voltage at the positive input will rise, allowing more current to pass through the MOSFET/transformer, which can provide more energy.
2) When starting, increase the time for energy transfer, that is, extend the rise time of Vfb (before reaching the OCP protection point).
For this FSQ0170RNA chip, the inductor current control is based on Vfb as the reference voltage, and the waveform of the Vfb voltage is proportional to the envelope of the inductor current. Controlling the rise time of Vfb can control the rise time of the inductor envelope, that is, increasing the time for energy transfer.
The OCP function of the IC is implemented by detecting Vfb reaching Vsd (such as 6V). Therefore, to reduce the Vfb slope, the rise time of Vfb can be extended.
When the output voltage does not reach the normal value, if the feedback pin voltage Vfb has risen to the protection point, the energy transfer time is not enough. When heavy load or capacitive load is started, the output voltage is established slowly, the voltage applied to the optocoupler is low, the current passing through the optocoupler diode is small, and the optocoupler photosensitive tube remains in a high-resistance state (tends to turn off) for a long time. The internal current source of the IC charges the capacitor connected to the feedback pin faster. If Vfb rises to the protection point (such as 6V) during this period, the MOSFET will turn off. The output cannot reach the normal value and the startup fails.
Solution:
When the output voltage reaches the normal value, the feedback pin voltage Vfb is still less than the protection point. Make Vfb rise slowly away from the protection point, or extend the time for the feedback pin Vfb to rise to the protection point, that is, reduce the rising slope of Vfb, so that the output has enough time to rise to the normal value.
A. Increasing the feedback capacitor (C9) can reduce the rising slope of Vfb, as shown in the figure, changing from line D to line A. However, if the feedback capacitance is too large, it will affect the normal working state, reduce the feedback speed, and increase the output ripple. So this capacitance cannot change too much.
B. Due to the shortcomings of method A, connect a capacitor (C7) in series with the voltage regulator tube (D6, 3.3V) in parallel to the feedback pin. This method will not affect normal operation, as shown on line B, when Vfb
be careful:
1) Increasing the feedback pin capacitance (including the voltage regulator string capacitance) will have little effect on solving the problem of large capacitive loads;
2) Increasing the peak current limit point I_limit also increases the OCP point in steady state. It is necessary to check whether the transformer will saturate under capacitive load and minimum input conditions;
3) If you want to maintain the current limit point, you must make R10×C11 larger, but in the case of ultra-large capacitive load (10000uF), the rise time of 5Vsb may increase to more than 20mS. This method requires checking whether the dynamic response is too affected. ;
4) The bias resistor R10 of 431 is too small, and the C11 of 431 in parallel should be larger;
5) In order to ensure the rise time, the methods of increasing the OCP point and increasing R10×C11 may be used at the same time.
10. Output bounce at no load or light load
Phenomenon:
When the output is no-load or light-loaded, and the input voltage is turned off, the output (such as 5V) may have a voltage bounce waveform as shown in the figure below.
Reason:
When the input is turned off, the 5V output will drop, Vcc will also drop, and the IC will stop working. However, when no load or light load is applied, the voltage of the huge capacitor of the PC power supply cannot drop quickly, and it can still provide a relatively high voltage to the high-voltage start-up pin. The large current causes the IC to restart, 5V is output again, and bounces.
Solution:
Insert a larger current-limiting resistor in series with the startup pin, so that when the voltage of the large capacitor drops to a relatively high level, it is not enough to provide sufficient startup current to the IC.
Connect the startup before the rectifier bridge, and the startup will not be affected by the large capacitor voltage. When the input voltage is turned off, the startup pin voltage can drop rapidly.
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