EEVblog Electronics Community Forum
Electronics => Beginners => Topic started by: alireza7 on July 29, 2019, 07:47:33 am
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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.
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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.
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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.
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.
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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.
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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.
is there any way to protect a boost converter from being in such condition?
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It is starving the boost converter of the input voltage....
is there any way to protect a boost converter from being in such condition?
While it is true that inserting 100R in series with the boost converter will result in a huge voltage drop, this should only cause it to stop working, not blow up the switch. In addition, any converter IC intended for boost (or flyback/buck-boost) applications should limit the switch duty cycle to no more than 90-95% (though that really needs to be 80% or less, as these topologies are very unstable at high duty cycle). Consequently, it sounds like this controller IC is either terminally stupid or else the switch current limit function isn't being used or working properly.
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After doing a lot of experiments I figured out in addition to 3x10uF output capacitors, there was a 220uF tantalum capacitor somewhere else in the secondary circuit. I removed that and the problem solved.
But I don't know why it cased TPS6102 burn. its datasheet said about minimum output capacitor but I didn't see anything about maximum output capacitor.
Why it is so?
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The greater the output capacitance, the high the inrush current and the greater the voltage drop across your ammeter. Did you build this circuit using a PCB using the recommended layout, or were you trying to use a breadboard? Switching converters rarely work correctly on breadboards.
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The greater the output capacitance, the high the inrush current and the greater the voltage drop across your ammeter. Did you build this circuit using a PCB using the recommended layout, or were you trying to use a breadboard? Switching converters rarely work correctly on breadboards.
I used a PCB using the recommended layout.
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Also, I found out even without ammeter and by just connecting the battery directly to the circuit this happens and IC burns if the output capacitor be large.