problem with boost converter - Page 1
Forum section: electronics / beginners

author: alireza7
author_rank: Regular Contributor
timestamp: July 29, 2019, 07:47:33 am
content: hiI 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. 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.

author: krish2487
author_rank: Frequent Contributor
timestamp: July 29, 2019, 09:13:11 am
content: two words - burden voltageRead 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.

author: alireza7
author_rank: Regular Contributor
timestamp: July 29, 2019, 09:25:25 am
content: Quote from: krish2487 on July 29, 2019, 09:13:11 amI 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.

author: krish2487
author_rank: Frequent Contributor
timestamp: July 29, 2019, 09:38:27 am
content: 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.

author: alireza7
author_rank: Regular Contributor
timestamp: July 29, 2019, 10:50:56 am
content: Quote from: krish2487 on July 29, 2019, 09:38:27 amis there any way to protect a boost converter from being in such condition?

author: MagicSmoker
author_rank: Super Contributor
timestamp: July 29, 2019, 11:21:26 am
content: Quote from: alireza7 on July 29, 2019, 10:50:56 amWhile 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.

author: alireza7
author_rank: Regular Contributor
timestamp: July 29, 2019, 02:11:43 pm
content: 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?

author: mikerj
author_rank: Super Contributor
timestamp: July 29, 2019, 05:14:44 pm
content: 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.

author: alireza7
author_rank: Regular Contributor
timestamp: July 29, 2019, 05:18:57 pm
content: Quote from: mikerj on July 29, 2019, 05:14:44 pmI used a PCB using the recommended layout.

author: alireza7
author_rank: Regular Contributor
timestamp: July 29, 2019, 05:21:55 pm
content: 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.

author: Jfragos
author_rank: Newbie
timestamp: June 19, 2025, 12:30:28 pm
content: Well well well. I recommend staying away from this chip.It burns if you apply a heavy load AND you supply it with a source that has resistance (OR BATTERY IR !!!) greater than approximately 0.5OhmSorry TI, I did better with a Chinese MT3608B.I hope I am not wrong but I have burned systematically 4 chips on the testing phase of a failed iteration of a project due to that reason. I attach the layout (4 layer) for any case of questioning my design. The brown layer is a complete GND plane.Second time a bite the dust with a modern TI boost DCDC regulator:'(

author: mtwieg
author_rank: Frequent Contributor
timestamp: June 19, 2025, 01:46:46 pm
content: Out of curiosity I looked up this chip on TI's user forums andyikes. While it's likely that following TI's application design exactly works fine, there's clearly something wrong if going a tiny bit off-script causes immediate failure.Always pays to do some googling when selecting a part like this...

author: Vovk_Z
author_rank: Super Contributor
timestamp: June 19, 2025, 02:26:25 pm
content: Quote from: alireza7 on July 29, 2019, 07:47:33 am(Resistor is possibly not 100 Ohm but 100 mOhm == 0.1 Ohm).Input voltage goes down under load current not quite low but still low, to the voltage instability region, when IC cant control itself and its switching transistor.It is designed in such a way that input voltage normally must move from zero to rated input working voltage fast enough. If not fast enough - you see this.For example, there are popular Chinese 400W step-up modules with TL494/SG3525, which often burn its power MOSFET if voltage became less then 7-6 VDC.There are two (actually three) main rules to measure current safely:1) Try to never use ammeter, until there is a very strong reason. Using ammeter leads to other bad possibilities to short something with its wires when you forget to resit DMM wires to voltage from current. (After measuring current immediately set DMM wires to voltage).2) Instead of ammeter - use voltmeter and Ohms Law to measure current on a well-known resistance (which is very often present in the circuit). The wires itself can be used as such shunt resistor too.3) To measure current and power consumption of something like this I just power such things from a bench power supply (which has a voltmeter and an ammeter obviously) so I almost never really have to use ammeter.

author: Jfragos
author_rank: Newbie
timestamp: June 21, 2025, 12:51:31 pm
content: Quote from: Vovk_Z on June 19, 2025, 02:26:25 pmIsn't the reason of a well specifiedUVLO(Under Voltage Lock Out) in such regulators the prevention of this?A regulator that has X Volts UVLO should ever get in that trap and burn?Bottom line: boost regulator projects are hard and often fail. Besides damaged chips you easily get voltage overshoots and other nasty staff (like a runaway effect of asking more current on input drops input voltage and that asks even more current).Boost is hard, buck is easy.I have seen many boost regulators that isolate the output from the input especially to go to heaven (or hell). Prefer the more traditional where the output is always connected to the input through the coil and the diode.

author: LinuxHata
author_rank: Frequent Contributor
timestamp: June 22, 2025, 06:19:12 am
content: This is common issue with boost ICs with MOSFET output stage.When mosfet opens, input voltage drops, so driver circuit can't close the mosfet, it remains open, and short circuits the battery, which might deliver current enough to burn the mosfet.The only solution is to use switches with BJT transistors, but these are expensive and quite rare - LT1370, as example.