The switching power supply is a power supply that uses modern power technology to control the turn-on and turn-off time ratio of the switching transistor to maintain a stable output voltage. The switching power supply is composed of pulse width modulation (PWM) control (metal oxide semi-field effect transistor). The development direction of modern power electronics technology is changing from traditional power electronics, which mainly solves problems with low-frequency technology, to modern power electronics, which mainly solves problems with high-frequency technology. In the application of power electronics technology and various power supply systems, switching power supply technology is at the core. The following is a brief analysis of the difficulties in debugging switching power supplies for your reference.
What are the difficulties in debugging switching power supplies? Answers to the eight most common questions
Table of Contents
1. 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.
It is easy to produce saturation situations:
1) The inductance of the transformer is too large;
2) Too few turns;
3) The saturation current point of the transformer is smaller than the maximum current limit point of the IC;
4) No soft start.
Solution:
1) Reduce the current limit point of the IC;
2) Strengthen soft start to make the current envelope through the transformer rise more slowly.
2. Vds is too high
Stress requirements for Vds:
Under the worst conditions (highest input voltage, maximum load, highest ambient temperature, power startup or short circuit test), the maximum value of Vds should not exceed 90% of the rated specifications.
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 the 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 tube;
② Use a slower diode, which itself can absorb a certain amount of energy (spike);
③ Inserting a damping resistor 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 conducted 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 kept away from parts that are too hot.
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 shutdown voltage.
reason:
At no-load or light load, the induced voltage of the Vcc winding is too low and 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 during heavy load
Under heavy load, the Vcc winding induced voltage is high, causing Vcc to be too high and reaching the IC’s OVP point, which will trigger the IC’s overvoltage protection, 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
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 current rising slope 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 current is very large. This causes the input power to be too large and the switch current to store too much energy in the leakage inductance, causing Vds to be high when the switch 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 switch 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 happens, 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. Trigger internal current limit, duty cycle is limited;
b.Vcc drops to the IC shutdown voltage;
c. The switch action stops;
d.Vcc rises to the IC startup voltage again and restarts.
8. No-load and light-load output rebound
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 there is no load or light load, the voltage of the huge PC power supply capacitor cannot drop quickly, and it can still provide a large voltage to the high-voltage startup pin. The 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.
