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problem with boost converter

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alireza7:
hi
I use tps61022 and my design is exactly like the recommended design in the datasheet.

I connect a battery to its input and it works fine and gives me correct voltage (5.25V) and sufficient current. but when I want to measure the input current by putting an ampere meter on the input of the circuit, the tps61022 burns immediately and its SW pin connects to the ground  |O. the series resistor of amper meter is 100 ohm.
why it is so?
did you have such an experience?
do you have any idea?
I will be grateful if you help me with this problem.

krish2487:
two words - burden voltage


Read up on the term and what it means in context to an ammeter.. and it is pretty plain why the boost converter burns up.
:-)


You are pretty close to figuring it out yourself.. you have all the information you need... you just need to put the peices together.

alireza7:

--- Quote from: krish2487 on July 29, 2019, 09:13:11 am ---two words - burden voltage


Read up on the term and what it means in context to an ammeter.. and it is pretty plain why the boost converter burns up.
:-)


You are pretty close to figuring it out yourself.. you have all the information you need... you just need to put the peices together.

--- End quote ---

I expect that the chip just fails to start up with a 100 ohms series resistance in series with its input as this DCDC converter will need a high current to start up and only lower it's current consumption after it has started up properly. The 100 ohm at the input should just prevent a proper startup. but it causes permanent damage.

krish2487:
It is starving the boost converter of the input voltage...
depending on your load current, the (load current + quiescent current ) * 100 ohms will be dropped across the ammeters internal resistance..


and 3.6 - (the voltage across the 100 ohm resistor) is what is actually available to the boost converter to boost upto your output voltage...
If this value is too low, then the boost converter tends to go almost 100% duty cycle, which is basically the boost switch shorting the output.

alireza7:

--- Quote from: krish2487 on July 29, 2019, 09:38:27 am ---It is starving the boost converter of the input voltage...
depending on your load current, the (load current + quiescent current ) * 100 ohms will be dropped across the ammeters internal resistance..


and 3.6 - (the voltage across the 100 ohm resistor) is what is actually available to the boost converter to boost upto your output voltage...
If this value is too low, then the boost converter tends to go almost 100% duty cycle, which is basically the boost switch shorting the output.

--- End quote ---

is there any way to protect a boost converter from being in such condition?

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