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Power chip burnt out [Copy link]

16.8V power input, battery powered, now the 4.4V and 3.3V outputs will occasionally burn out, then the 4.4V will power the MC20 module, the MC20 module will be powered on and off regularly, now I don't know why the power chip burns out, please help me analyze it

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I have also encountered this in my previous company. The load was a GSM module. When the circuit was powered on, the instantaneous current was so large that the power chip burned out.  Details Published on 2020-8-20 16:03
 
 

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It is necessary to understand the load size of the MC20 module and the instantaneous current of startup and shutdown

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When MC20 starts, there is a current of 1.8A, but the upper limit of the chip is 3A. I asked a master and he said that it is because the current will be amplified at the moment of startup, exceeding the range, and all will burn, but not every piece will burn, only about 10%  Details Published on 2020-8-20 09:11
 
 
 

2w

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Damage to the peripheral components of both chips may damage the chips.

In addition, as qwqwqw2008 said on the second floor, you should figure out the instantaneous current of your MC20 module when it starts and shuts down.

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The peripherals are not damaged, only the chip is broken  Details Published on 2020-8-20 09:12
 
 
 

2w

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I am interested in this part of the circuit in the picture.

4V4_VDD is obviously a DC signal. The purpose of this circuit may be to light up the LED when the 4V4_VDD power supply is "on". Then why is C25 connected in series with the base of Q3? After C25 is connected in series, 4V4_VDD "on" can only light up the LED for a few seconds or longer, and cannot keep LED1 lit. Doesn't this mean that LED1 cannot indicate the working status of the 4V4_VDD power supply?

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This is just a power-on indication, it does not need to be on all the time  Details Published on 2020-8-20 09:08
 
 
 

1w

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What is the phenomenon of burning?

The inductor used in the circuit is 15uH, and the parameters seem inappropriate.

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There is a direct short circuit, and the power chip does not work.  Details Published on 2020-8-20 09:11
 
 
 

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maychang posted on 2020-8-19 12:21 I am interested in this part of the circuit in the figure. 4V4_VDD is obviously a DC signal. The purpose of this circuit may be to "turn on" the 4V4_VDD power supply...

This is just a power-on indication, it does not need to be on all the time

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qwqwqw2088 posted on 2020-8-19 11:48 It is necessary to understand the load size of the MC20 module and the instantaneous current of startup and shutdown

When MC20 starts, there is a current of 1.8A, but the upper limit of the chip is 3A. I asked a master and he said that it is because the current will be amplified at the moment of startup, exceeding the range, and all will burn, but not every piece will burn, only about 10%

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dcexpert posted on 2020-8-19 13:38 What is the phenomenon of burning? The inductor used in the circuit is 15uH, and the parameters seem inappropriate.

There is a direct short circuit, and the power chip does not work.

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maychang posted on 2020-8-19 12:12 Damage to the peripheral components of the two chips may damage the chips. In addition, as qwqwqw2008 said on the second floor, you should understand your MC20 module...

The peripherals are not damaged, only the chip is broken

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I have also encountered this in my previous company. The load was a GSM module. When the circuit was powered on, the instantaneous current was so large that the power chip burned out.

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How did you solve this?  Details Published on 2020-8-21 15:17
 
 
 

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aallian posted on 2020-8-20 16:03 I also encountered this in my previous company. The load was a gsm module. When the circuit was powered on, the instantaneous current was so large that the power chip burned out.

How did you solve this?

This post is from Power technology
 
 
 

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