Switching Power Supply Debugging: 10 Most Common Problems

1. Transformer saturation

Transformer saturation phenomenon

When the transformer (and switch tube) is turned on with a high voltage or low voltage input (including light, heavy, or capacitive loads), an output short circuit, a dynamic load, or a high temperature, the current through the transformer (and switch tube) grows nonlinearly. When this happens, the peak current value can't be predicted or regulated, which can lead to current overstress and damage to the switch tube as a result of the overvoltage.

Current waveform at transformer saturation

Current Waveform at Transformer Saturation

It is prone to saturation:

1) The inductance of the transformer is too large;

2) The number of laps is too small;

3) The saturation current point of the transformer is smaller than the maximum current limiting point of the IC  ;

4) No soft start.

Solution:

1) Reduce the current limiting point of the IC;

2) Strengthen the soft start, so that the current envelope through the transformer rises more slowly.

Current Envelope

2. Vds is too high

Stress requirements for Vds:

The maximum value of Vds should not exceed 90% of the specified specification under the worst conditions (highest input voltage, heaviest load, worst ambient temperature, power up, or short circuit test).

Ways to reduce Vds:

1) Reduce the platform voltage: reduce the number of the turns ratio of the primary and secondary sides of the transformer;

2) Reduce the spike voltage:

1. Reduce leakage inductance:

When the switch is turned on, the transformer's leakage inductance stores energy, which is the primary cause of the peak voltage. The peak voltage can be reduced by lowering the leakage inductance.

2. Adjust the absorption circuit:

①Use 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

Reason and solution:

1) The internal MOSFET  losses are too large:

The switching loss is too high, and the transformer's parasitic capacitance is too high, resulting in a huge cross area between the MOSFET  's turn-on and turn-off currents and Vds. Solution: increase the transformer winding distance to reduce interlayer capacitance, or add a layer of insulating tape (interlayer insulation) between the layers if the winding is wound in many layers.

2) Poor heat dissipation:

The pins that carry heat to the PCB  and the copper foil on it are responsible for a considerable portion of the IC's heat. More solder should be applied and the copper foil's area should be extended as much as possible.

3) The air temperature around the IC  is too high:

The IC  should be in a place with good airflow and should be kept away from parts where the temperature of the parts is too high.

4. No load, light load can not start

Phenomenon:

No-load, light-load can not start,  Vcc repeatedly jumps back and forth from the start-up voltage and the shutdown voltage.

reason:

During no-load and light-load conditions, the induced voltage of the  Vcc winding is too low, and it enters a state of repeated restarts.

Solution:

Increase the number of Vcc winding turns, lower the Vcc current limiting resistance, and correctly add dummy loads. Please refer to the procedure for stabilizing Vcc if Vcc becomes too high under heavy load after increasing the number of Vcc winding turns and reducing the Vcc current limiting resistor.

5. Can not be reloaded after startup

Reason and solution:

1) Vcc is too high on reload

The induced voltage of the Vcc winding is high when the load is heavy, and when the Vcc is too high and hits the IC  's OVP point, the  IC 's overvoltage protection is triggered, resulting in no output. The IC  will be damaged if the voltage increases higher than it can withstand.

2) Internal current limit is triggered

1. The current limiting point is too low

If the current limiting point is too low under heavy load and capacitive load, the current passing through the MOSFET  is limited and insufficient, resulting in insufficient output. Increase the current-limiting pin resistance and the current-limiting point as a remedy.

2. The current rising slope is too large

If the rising slope is excessively steep, the current's peak value will be higher, and the internal current limiting protection will be triggered more easily. Increasing the inductance without saturating the transformer is the solution.

6. Large standby input power

Phenomenon:

Vcc is insufficient at no load and light load. This situation will cause the input power to be too high at no load and light load, and the output ripple will be too large.

reason:

The reason for the excessive input power is that when the Vcc is insufficient, the IC  enters a condition of repeated starting, which regularly necessitates high voltage to charge the Vcc capacitor, resulting in the startup circuit being lost. If a resistor is connected in series between the starting pin and the high voltage, the resistor's power consumption will be higher at this time, thus the starting resistor's power level should be sufficient.

The power  IC hasn't entered  Burst Mode or has, but the Burst frequency is too high, the switching durations are too long, and the switching loss is too high.

Solution:

Adjust the feedback parameters to reduce the feedback speed.

7. Excessive short-circuit power

Phenomenon:

When the output is shorted, the input power is too high and Vds is too high.

reason:

There are many repeated pulses when the output is short-circuited, and the peak value of the switch tube current is very large at the same time, causing the input power to be too high and the switch tube current to store too much energy in the leakage inductance, causing the Vds to be high when the switch tube is turned off.

When the output is short-circuited, the switch tube can stop working for one of two reasons:

1) Triggering OCP in this way can make the switch action stop immediately

1. Trigger the OCP of the feedback pin;

2. The switch action stops;

3. Vcc drops to IC off voltage;

4. Vcc rises back up to the IC startup voltage and restarts.

2) Trigger internal current limit

When this method is used, the available duty cycle is limited, and the switching action is stopped when Vcc reaches the lower limit of  UVLO ; however, the time it takes for Vcc to reach the lower limit of  UVLO is longer, so the switching action is maintained for a longer period of time, and the input power is higher.

1. Trigger the internal current limit, the duty cycle is limited;

2. Vcc drops to IC off voltage;

3. The switch action stops;

4. Vcc rises back up to the IC startup voltage and restarts.

Solution:

1) Reduce the number of current pulses so that the feedback pin's OCP is activated when the output is short-circuited, allowing the switching process to be immediately stopped and the number of current pulses to be reduced. This means that the voltage on the feedback pin should rise faster when a short circuit occurs. As a result, the feedback pin's capacitance should be kept to a minimum.

2) Reduce the peak current.

8. No-load, the light-load output ripple is too large

Phenomenon:

Vcc is insufficient at no load or light load.

reason:

When Vcc is insufficient, the oscillator IC works sporadically with a long duration between the startup voltage (such as 12V) and the shutdown voltage (such as 8V), providing energy to the output for a short time before stopping for a long time, causing the capacitor stored to discharge. The output voltage will decline because there is insufficient energy to maintain the output stable.

Solution:

It is guaranteed that Vcc can be supplied stably under any load conditions.

Phenomenon:

The frequency of intermittent operation is too low in Burst Mode, and the energy of the output capacitor cannot be kept stable at that frequency.

Solution:

Under the condition that the standby power consumption requirements are met, raise the frequency of intermittent operation and the output capacitance somewhat.

9. Overload and capacitive load cannot be started

Phenomenon:

It can start with a light load, and it can start with a heavy load once it has been begun, but it cannot start with a heavy load or a big capacitive load.

General design requirements:

Regardless of whether the input voltage is the lowest or the lowest (such as 10000uF), the output voltage must rise to a steady value within 20mS.

Reasons and solutions (if Vcc is within its typical operating range):

The capacitive load C=10000uF is used as an example for analysis in the following.

The output must rise to a stable output voltage (e.g. 5V) within 20 milliseconds, according to the specification.

E=0.5*C*V^2

The higher the capacitance C, the more energy must be transported from the input to the output in less than 20 milliseconds.

Chip FSQ0170RNA

The whole area S of the darkened region of the chip FSQ0170RNA, as shown in the picture, is the required energy. 1)Raise the peak current limit point I limit, which allows for a greater inductor current Id: increase the resistance connected to Pin4, the shunt from the internal current source Ifb is smaller, and utilize the  PWM  comparator as the current limit reference voltage. The positive input voltage will rise, allowing more current to pass through the MOSFET  /transformer, increasing the amount of energy available.

2) Increase the time it takes to transmit energy when you first start, i.e., extend the  Vfb rise time (before reaching the OCP protection point).

Extend the Rise Time of Vfb

The inductor current control on this FSQ0170RNA chip is based on  Vfb as the reference voltage, and the waveform of the Vfb voltage is proportional to the inductor current envelope. Controlling Vfb's rise time can regulate the inductive envelope's rise time, which increases the time for energy transfer.

The IC's OCP function is achieved by sensing when Vfb reaches  Vsd  (such as 6V). As a result, the rising time of Vfb can be lengthened to minimize the Vfb slope.

The energy transfer time is insufficient when the output voltage does not reach the usual value and the feedback pin voltage Vfb has reached the protection point. When a heavy load or capacitive load is applied, the output voltage rises slowly, the voltage applied to the optocoupler is low, the current through the optocoupler diode is low, and the optocoupler photosensitive tube is in a high resistance state (tends to be turned off) for an extended period of time. The capacitor linked to the feedback pin is charged faster by the IC's internal current source. The MOSFET  will be turned off if Vfb climbs to the protection point (such as 6V) within this time period. The startup fails because the output cannot reach the normal value.

Solution:

The feedback pin voltage Vfb is still less than the protection point when the output voltage reaches a normal value. Keep Vfb away from the protection point and gently climb, or lessen the increasing slope of Vfb to give the output ample time to rise to its usual value.

1. Increasing the feedback capacitor (C9) reduces the rising slope of Vfb from the D line to the A-line, as illustrated in the figure. However, if the feedback capacitor is too big, it will interfere with the normal functioning condition, slow down the feedback, and increase the output ripple. As a result, the capacitance cannot fluctuate too much.

2. Connect a capacitor (C7) in series with the voltage regulator tube (D6, 3.3V) in parallel with the feedback pin due to the deficiency of method A. When Vfb is careful: This approach will not disrupt the normal job, as stated in line B.

1) Increase the feedback pin capacitance (including the voltage regulator tube string capacitance), which has minimal influence on the problem of a very large capacitive load;

2) Increase the peak current limit point I limit, as well as the OCP point in steady-state. Under the conditions of capacitive load and minimum input, the transformer must be checked to see if it would saturate.

3) If you want to keep the current limit point, you should increase R10C11; however, in the case of a super capacitive load (10000uF), the rise time of 5Vsb may increase to more than 20mS, so this method should be checked to see if the dynamic response is too affected. ;

4) The bias resistor R10 of 431 is too small, and C11 of 431 in parallel should be larger;

5) In order to ensure the rise time, increasing the OCP point and increasing the R10×C11 method may be used at the same time.

10. No-load, light-load output debounce

Phenomenon:

When the output is no-load or light-load, the input voltage is turned off, and the output (such as 5V) may have a voltage bounce waveform as shown in the figure below.

Voltage Bounce Waveform

reason:

The 5V output will diminish when the input is shut off, as will Vcc, and the IC will stop operating. When there is no load or a light load, however, the voltage of the massive PC power supply and the enormous capacitor cannot fast drop, and the high-voltage start pin can still receive a significant voltage. The current restarts the IC.  and the 5V is output debounced once more.

Solution:

When the voltage of the huge capacitor drops to a still quite high level, it is insufficient to supply enough start-up current to the IC.  thus a large current-limiting resistor is connected in series with the start-up pin.

The enormous capacitor voltage has no effect on the starting before it is connected to the rectifier bridge. The start pin voltage might rapidly drop when the input voltage is shut off.