Strange problem in high frequency control of PMOS field effect tube

jake93

Strange problem in high frequency control of PMOS field effect tube [Copy link]

 

There was a problem when controlling the PMOS at high frequency. A 5V, 62.5KHZ sine wave signal was input to the base control terminal of the transistor, and a sine wave signal could also be measured at the collector of the transistor; but when it reached the gate of the field effect tube , the measured signal became a stable voltage signal, and the field effect tube was in a semi-conducting state, causing serious heat generation. I suspect that it was caused by the junction capacitance of the field effect tube, but as a junior, I have no evidence. If any senior can help me analyze what caused it and how to solve it, I would be grateful.

maychang

"A 5V, 62.5KHZ sine wave signal is input to the base control terminal of the transistor, and a sine wave signal can also be tested at the collector of the transistor."

The text in the picture also says it is a sine wave, but it is drawn as a rectangular wave.

maychang

"But when it reaches the gate of the field effect tube , the measured signal becomes a stable voltage signal, the field effect tube is in a semi-conducting state, and the heat is serious."

Resistors R16, R14 and transistor collector output resistance form a low-pass filter with the MOS tube gate input capacitance. Therefore, the MOS tube gate voltage is approximately a stable DC.

maychang

"If any senior can help me analyze the cause and how to solve it, I would be very grateful."

I would like to ask you: why do you input 62.5kHz to the base of Q4? This is not the way to control MOS tubes.

wenyangzeng

The resistance of R16 is too large, and the voltage divided by R14 is not enough to pull down the G pole of the MOS tube to 0. The MOS tube is in the amplification area, and the tube power consumption is very large. It is recommended that the resistance of R16 be about one tenth of R14.

chunyang

If it is a switching circuit, R16 should be removed. The voltage divider value of R14 and R16 is too high. What function do you want to achieve?

chunyang

Could it be that the OP wants to use this circuit to output a sine wave?

不亦心

wenyangzeng published on 2020-5-8 20:41 The resistance of R16 is too large, and the voltage divided by R14 is not enough to pull down the G pole of the MOS tube to 0. The MOS tube is in the amplification area and the tube power consumption is very large. It is recommended that the resistance of R16 be 10% of R14...

This is pure nonsense, it will make MOS die faster

jake93

maychang published on 2020-5-8 19:22 "A 5V, 62.5KHZ sine wave signal is input to the base control terminal of the transistor, and a sine wave signal can also be tested at the collector of the transistor" ...

I am very sorry, I wrote it wrong here, it should be square wave, 62.5K square wave is used for charging control, only part of the drawing is intercepted here

jake93

maychang published on 2020-5-8 19:25 "But when it reaches the gate of the field effect tube, the measured signal becomes a stable voltage signal, the field effect tube is in a semi-conducting state, and the heat is serious" ...

This circuit is used for charging control, and only a part of the drawing is captured here. Is there any solution to this problem?

jake93

chunyang posted on 2020-5-8 21:05 Could it be that the OP wants to use this circuit to output a sine wave?

I'm very sorry, I wrote it wrong here, it should be square wave, the control signal output by the charging chip, which realizes constant current charging through PWM control field tube switch

jake93

chunyang posted on 2020-5-8 21:03 If it is a switch circuit, R16 should be removed. The voltage divider value of R14 and R16 is too high. What function does the OP want to achieve?

The input voltage is 30V, the maximum Vgs of the field tube can withstand is -20V, and R16 cannot be removed. This circuit is used for constant current charging control. Only a part of the drawing is captured here.

jake93

wenyangzeng published on 2020-5-8 20:41 The resistance of R16 is too large, and the voltage divided by R14 is not enough to pull down the G pole of the MOS tube to 0. The MOS tube is in the amplification area and the tube power consumption is very large. It is recommended that the resistance of R16 be 10% of R14...

The input voltage is 30V, Vgs can withstand a maximum of -20V, and the voltage divider of R16 cannot be too small. I tried reducing both resistors at the same time.

maychang

jake93 posted on 2020-5-9 08:15 I am very sorry, I wrote it wrong here, it should be square wave, 62.5K square wave is used for charging control, only part of the drawing is intercepted here

Charging control should be to control the charging current. To control the charging current, PWM control can be used, which is actually a switching power supply with a nearly controllable constant current output.

62.5kHz, this frequency is not very low. To control the MOS tube to output a rectangular wave, the driver must ensure that the MOS tube is fully turned on and off, which is not a very simple thing. It is not realistic to rely on such large resistors as R14 and R16. Even if the resistance is reduced to hundreds of ohms, the rising and falling edges of the current in the MOS tube are not steep enough.

maychang

jake93 posted on 2020-5-9 08:17 This circuit is used for charging control. Only a part of the drawing is captured here. Is there any solution to this problem?

62.5kHz, cycle is 16us. The time left for the rising edge and falling edge is roughly only 1~2us, that is, the MOS tube must go from off to fully on or from fully on to off within 1~2us.

jake93

maychang published on 2020-5-9 08:28 Charging control should be to control the charging current. To control the charging current, PWM control can be used, which is actually a switch with an approximately controllable constant current output...

Constant current charging is achieved by adjusting the PWM pulse width. I reduced both resistors and tried again

jake93

maychang published on 2020-5-9 08:32 62.5kHz, the cycle is 16us. The time left for the rising edge and the falling edge is roughly only 1~2us, that is, the MOS tube must go from shutdown to power on within 1~2us...

Looking at the field control data, td(on), tr, td(off), and tf are all less than 100ns. Does this mean that the field control can be turned on or off within 200ns?

maychang

jake93 posted on 2020-5-9 08:17 This circuit is used for charging control. Only a part of the drawing is captured here. Is there any solution to this problem?

Turning on or off within 1 to 2 us means charging or discharging the gate input capacitor of the MOS tube within 1 to 2 us.

I haven't checked the IRF4905 manual, but I estimate that the gate input charge of the tube needs to be 5 to 20nC. If it takes 1us to complete the charge or discharge, the average current will be 5nC/1us=5mA or 20nC/1us=20mA. And your discharge current is at most 5V/51kΩ=0.1mA.

In order to turn off and on quickly, the charging and discharging current must be increased.

To this end, a transistor should be used to directly drive the gate of the MOS tube, with a resistor of at most several tens of ohms connected in series. For specific methods, please refer to the internal circuit of the switching power supply control chip.

maychang

jake93 posted on 2020-5-9 08:17 This circuit is used for charging control. Only a part of the drawing is captured here. Is there any solution to this problem?

In addition, an inductor-capacitor filter circuit should be added to the output of the MOS tube, and a freewheeling diode should be added. For specific methods, refer to the Buck circuit.

maychang

jake93 posted on 2020-5-9 08:41 Looking at the field control data, td(on), tr, td(off), tf are all less than 100ns. Does this mean that the field control can be turned on or off within 200ns? ...

It can be turned on or off within 200ns. But you must pay attention to the test conditions, especially the gate drive signal shape and the drive signal internal resistance.