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I just finished setting up an LM2577 as a boost converter, and have run into some unexpected issues.

For starters, here is the recommended "typical application circuit":

Application Circuit

My application circuit is slightly different. At the 5V input, I can actually have anything between a 8V to 14V input as part of a PMOS common source setup so I can turn off the current coming in to the circuit, like so:

Modified Circuit

However, the output voltage is nothing like what I expected. I had assumed that if I'm drawing at least a few dozen milli-amps, and have a +9V DC rail at the source of the PMOS, I should expect just under 9V at the drain of the PMOS when it's turned on, and that the output voltage of the boost regulator, Vcc, would be approximately 12V. My assumption was also that the output voltage is just a function of the external discrete components connected to the regulator, and not the input voltage (ie: 9V DC) provided the input voltage to the regulator is less than the output voltage it generates. Is this assumption correct? I want the regulator to always output a fixed voltage (ie: 12V) for a wide range of input voltages (ie: 7.5V to 10.0V).

Also, the output voltage I initially measured was much greater than expected, a whopping 41V. When I removed the 0.1uF capacitor that connects the input to GND, it went back down to +15VDC. How can this be happening? Is it because the input to the regulator is from a PMOS drain and not a supply rail?

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  • \$\begingroup\$ Try driving VIN directly from a supply rail (bench supply perhaps?) to verify the regulator is working. Measure the voltage across R11 (FBck), expect 1.23V if the feedback is working correctly. Confirm feedback network R10 and R11 are 1% tolerance. May need to add 220uF across C9 -- see data sheet figure 31. \$\endgroup\$
    – MarkU
    Commented Oct 9, 2014 at 6:50
  • \$\begingroup\$ did you actually build this? or is this all in simulation? Have you checked that it's not a hardware/soldering issue? \$\endgroup\$
    – KyranF
    Commented Oct 9, 2014 at 7:21
  • \$\begingroup\$ Most of the spec's for the LM2577 are for 100mA-800mA load currents. What happens to the output when you apply a more substantial output? \$\endgroup\$
    – helloworld922
    Commented Oct 9, 2014 at 7:47
  • \$\begingroup\$ @KyranF I actually built the circuit. I don't see any misplaced contacts (solderless BB). \$\endgroup\$
    – Cloud
    Commented Oct 9, 2014 at 13:15
  • \$\begingroup\$ seems like a strange behaviour indeed, It "should just work" unless something went wrong. The IC itself might actually already be damaged/broken so any further testing may be useless. Have you tried putting in a brand new LM2577? \$\endgroup\$
    – KyranF
    Commented Oct 9, 2014 at 13:28

1 Answer 1

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It turns out that I was using a partially correct application circuit. According to this link, it turns out that when using the fixed output version of this chip, I need to short the feedback (pin 2) to the output of the Schottky diode, and I'm all set.

The schematic is simple, easy to build and cost effective, producing 12V from a 5V unregulated supply with a maximum output current of 800mA. Pin numbers shown are for the TO-220 package (LM2577T-ADJ).

The design uses the adjustable version of LM2577, but the 12V fixed-voltage version (LM2577T-12) will also work if you remove R1 and R2 and connect the feedback pin directly to the regulator's output. The UC2577 - a pin to pin compatible replacement available from Texas Instruments can also be used.

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