-2
\$\begingroup\$

I'm designing the boost converter with 3 mH inductor and 680 uF capacitor 20 kHz fequency. When I try with 12 V, 5 A power supply in the input and fixed 50% duty cycle, I can reach the 24 V output voltage since I put the load resistance greater than 50 ohm. When I use the resistance less than 50 ohm the output voltage will decrease, and the lower the resistance, the lower the output voltage until it is similar with the input (about 12 V). In this condition my MOSFET IRFP460 is very hot. When I try to take out the resistance, the output voltage is increasing continuously until more than 80 V with the same duty cycle at 50%.

Anyone can explain me what happen in my circuit when working at the resistance less than 50 ohm and when no load condition please.

\$\endgroup\$
9
  • 1
    \$\begingroup\$ Inductors work that way. When an inductor kicks it is using the energy in the collapsing magnetic field to generate whatever voltage is required to push current through itself. What that voltage is is a voltage that is higher than the voltage in the outcap cap. \$\endgroup\$ Commented Aug 2, 2021 at 15:46
  • 1
    \$\begingroup\$ "when working at a resistance less than 50 ohm" I think you mean greater than 50 Ohms since A no-load condition is like infinite Ohms. But a LARGE load means low resistance (because it will draw more current) while a small load means high resistance (since it will draw less current). So it is opposite in English when talking about the value of load resistance and the "size" of the load. \$\endgroup\$ Commented Aug 2, 2021 at 15:48
  • \$\begingroup\$ Is a regulator chip used? If so, add it to the question please. \$\endgroup\$ Commented Aug 2, 2021 at 16:08
  • 2
    \$\begingroup\$ @DKNguyen: OP does appear to be using the terms correctly, "less than 50 ohms" for higher load current and "more than 50 ohms" for lower load current. \$\endgroup\$ Commented Aug 2, 2021 at 16:31
  • \$\begingroup\$ @BenVoigt I guess it depends on what they are actually asking. To me I thought they were just rewording things. \$\endgroup\$ Commented Aug 2, 2021 at 18:32

2 Answers 2

Reset to default
3
\$\begingroup\$

A boost converter will only maintain a constant output voltage with constant duty cycle if it is in "continuous conduction mode" (CCM). This means (among other things) that current must be allowed to flow both ways through the coil when the output is lightly loaded. This usually means that synchronous rectification is used (MOSFETs instead of diodes).

Without this, the fixed amount of energy stored in the coil on each switching cycle has nowhere to go except to the output, driving it ever higher until something breaks down.

\$\endgroup\$
2
  • \$\begingroup\$ so when no load condition, with 12V input voltage and fixed 50% duty cycle, cant i keep the output voltage to 24V? because the output voltage will be more higher and higher as you said. i just want to test my circuit if it works with fixed duty cycle in no load condition because if i put the load, the circuit works fine. \$\endgroup\$ Commented Aug 3, 2021 at 3:42
  • \$\begingroup\$ We can't answer questions about your circuit unless you show it to us. Edit your question to add a schematic diagram, at least. \$\endgroup\$ Commented Aug 3, 2021 at 10:05
1
\$\begingroup\$

When eliminating the load current, the PFM must also decrease at some duty cycle to just overcome the losses. Or you decrease both d.c and PFM rate.

DCM has unstable characteristics over a full to null load.

\$\endgroup\$

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.