I have a TPS61222 Buck boost voltage converter, in a circuit intended to boost a 3.7V battery source to 5V. The output voltage however, is consistently less than 5V - scoping shows a 4V signal with spikes up to 6v at about 90kHz, low duty cycle.

The current draw is 56mA nominal (into the converter). The converter powers a charge pump inverter, to generate -5V, along with a number of other devices. The feedback pin voltage should be o.5V, with about 13mV tolerance (as per data sheet) - the value it is showing is more like 0.35-0.42V.

Full schematic and PCB layout are here

  • \$\begingroup\$ You need to do and present more measurements, like actual waveforms, precise dc voltage measurements etc. \$\endgroup\$ – PlasmaHH Jun 27 '16 at 15:36
  • \$\begingroup\$ @PlasmaHH While i take your point, I cant get properly characteristic waveforms, as they change constantly, its only the vague shape that is consistent. Similarly for the DC voltages - they drift over time, so there is no precise measurement to be made \$\endgroup\$ – Mauvai Jun 27 '16 at 15:38
  • \$\begingroup\$ exactly that is the point to show us. If e.g. the reference voltage is drifting, then no wonder everything else is following. \$\endgroup\$ – PlasmaHH Jun 27 '16 at 15:40
  • \$\begingroup\$ How much current load is on the Buck converter? Too much or too little current draw can cause it to misbehave. \$\endgroup\$ – Brendan Simpson Jun 27 '16 at 16:30
  • \$\begingroup\$ @Brendan Simpson the load is approx 40-50 mA depending on the connected sensors. At the supply voltage I'm. Using (3.7) I should he able to draw 200 mA \$\endgroup\$ – Mauvai Jun 27 '16 at 16:58

You are trying to run a charge pump from another switching device. In general, it is best not to run this type of device from the output of this type of converter, when the capacities and switching frequencies are close. The TPS61222 relies on voltage feedback to regulate, and your LM2663 is providing a load which is changing at its switching frequency.

I would try to run the charge pump generating your negative voltage directly off VBatt as a first choice. If this is completely unacceptable, you could add a cap to the feedback point of the boost converter, although this will slow down the converter's response to changes in input voltage and output load, possibly causing overshoot and undershoot.

| improve this answer | |
  • \$\begingroup\$ This makes a lot of senses, thanks. I can't run the charge pump from the battery, as I need -5V, 3.7 won't do. Are you suggesting a cap between the feedback pin and the output voltage, or feedback pin and ground? \$\endgroup\$ – Mauvai Jun 28 '16 at 12:21
  • \$\begingroup\$ Feedback to ground with a time constant much longer than the loop time constant. Your boost converter reacts to increases in load by increasing the PWM duty cycle, which will increase the output voltage. Avoid a second switcher running on a first one because the input of the second switcher has a negative dynamic impedance. When increasing the voltage on the input, the input current does not increase as it would with a passive load; it actually decreases. I can see it is late in your design, but I would consider flyback topology for a switcher when you need multiple polarities and outputs. \$\endgroup\$ – John Birckhead Jun 28 '16 at 14:11
  • \$\begingroup\$ For the sake of an update - i put a 100nF capacitor across the 110k resistor (RDC2), and while it brought the average voltage up to ~5V, the output was still spiking at about 90kHz. Adding a 47uF Cap across the output (for a total of 57uF, well over the recommended 10uF), retuces the volatge variation enormously (though not completely). It did however raise the average output voltage to 5.3V, which was undesireable. The voltage on the feedback pin is now a very steady 513mV \$\endgroup\$ – Mauvai Jul 5 '16 at 11:12
  • \$\begingroup\$ You might be able to make a voltage doubler from your charge pump and get away with running directly from VBatt. I would be worried that the large capacitor in the feedback will effect load regulation. \$\endgroup\$ – John Birckhead Jul 5 '16 at 15:23
  • \$\begingroup\$ Fortunately I am powering op-amps almost exclusively, so power regulation is not an issue. I'm pretty much where I need to be I think. \$\endgroup\$ – Mauvai Jul 5 '16 at 15:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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