1
\$\begingroup\$

Basically the LTC3603 seems to have internally shorted itself. I'm using a auto generated circuit, realised with Eagle CAD.

I'm trying to design a high current switching power supply with the Linear Technology's LTC3603. I used the companies software tools to create the required circuit and used Eagle CAD to create the circuit and ordered up a test PCB.

When I received the PCB I first buzzed it out to check that the circuit was as I expected it. Then I soldered on all the components and again buzzed it out to make sure that everything as connected and that their were NO shorts. With that all done I connected the circuit to a desktop power supply giving about 12V DC to the circuit.

My circuit didn't work at all. The switcher just seemed to connect the input 11.7 Volts straight to the output without out ever switching. I was only turning on power briefly and taking measurements and then turning off power. I didn't get any burning but after a few of these measurement operations the LTC3603 seems to have shorted itself Internally.

I'd happily post my eagle files for the test design but not sure where to do that. I'm just wondering if anybody would recognise or suggest why the device seems to have simply shorted itself out? Obviously even though the circuit was not switching, it was wrongly supplying 11.7V output from 11.7V Input. Maybe that left the mosfets on too long in the chip so maybe my question should be why it never switched?

Data sheet example circuit - replace with actual circuit when available.

enter image description here

LTC3603 data sheet here

\$\endgroup\$
1
  • \$\begingroup\$ If you can post the photos anywhere with a link somebody will add them to your post. You need a few points of "rep" before you can do that. If unable to post anywhere email them to me and I will add them - NOT my stack exchange email- normalise this -> nzphotosnz at gmail \$\endgroup\$
    – Russell McMahon
    Commented Oct 1, 2014 at 10:48

2 Answers 2

1
\$\begingroup\$

Aragh! Sorry My Bad. Just been over the data sheet again. In my design I left out the Cap CBST, between Boost and SW Pins. :-(

Sorry I asked the question as I'm sure I'll have to correct that issue before I can proceed. Thanks for all your comments and info on this.

\$\endgroup\$
1
  • 1
    \$\begingroup\$ That should be a comment on the question or in the question body. | Cboost is to provide gate drive voltage for the high side internal MOSFET(s). Without it the high side FETS would turn on either weekly or not at all. \$\endgroup\$
    – Russell McMahon
    Commented Oct 1, 2014 at 17:05
0
\$\begingroup\$

The LTC3063 effectively has a single MOSFET from Vin to Vout.
In a buck converter this MOSFET is turned on for part of a switching cycle and then turned off when Vout reaches the required level OR if certain other conditions are met.
If the feedback path is not acting as expected and the Vfb pin does not rise to its trigger voltage (indicating that Vout is at the required voltage) then the MOSFET may stay on and Vout will appear on Vin.

When 1st turning on it is VERY advisable to current limit the supply to a (probably) safe value and/or include a series resistor in the supply line that will drop the voltage if troubles occur.

The IC may have been damaged but there is a reasonable chance that it is OK.
If your circuit is anything like the above then ensure that the values of the feedback divider - shown here as 475k and 105k are correct and connected as expected. The 475k open circuit or too large a ratio of these two resistors would be good reasons not to operate.

Data sheet example circuit:

enter image description here

LTC3603 data sheet

\$\endgroup\$
2
  • \$\begingroup\$ In the design I have (without Cbst :-( ) I have resistors 165K and 22.6K I'm a bit confused by this design as the voltage divider's voltage to Vfb is dictated by the Output Voltage. In my case the output is 11.7 so Vfb is 1.4 Volts. The reason for my slight confusion is that Vout is set by Vfb but Vfb is set by Vout via the voltage divider. Anyhow from the datasheet Vout = 0.6v(1+R2/R1) which I would expect to be 4.98Volts. Hopefully my problem is just Cbst. \$\endgroup\$
    – jwhitmore
    Commented Oct 1, 2014 at 13:09
  • \$\begingroup\$ NO! - Vref is fixed at 0.6V. The device switches when Vfb from external sources exceeds 0.6V. or when Vout = Divide ratio x 0.6V. For 165K up and 22.6k down Vout ~= 5V as you say. The 1.4 V is what appears on the fb pin - not the V the device switches at. With no Cfb it cannot switch properly. \$\endgroup\$
    – Russell McMahon
    Commented Oct 1, 2014 at 17:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.