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I have built a 3V to 5V boost converter with discrete components.

L = 220uH

Vf(Diode) = 0.25V

ZVN4210A MOSFET

C = 470uF

f = 47KHz

My load is a microcontroller board (M16C QSK62P Plus) that takes upto 90mA of current. I am using the PWM signal from the same microcontroller for the switching pulse for the converter. I tested my circuit for a constant load of R = 100Ohm

I measured average current flowing through the inductor and output voltage. From that I calculated powers Pin and Pout.

I am getting the efficiency of approx. 75%. I know that boost converters can be more efficient (90%). So, I need some suggestion about modifications in my circuit to improve the efficiency.

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    \$\begingroup\$ Before getting into it too much, how good are your measurements--what did you make them with? Most multimeters won't read 47 kHz signals (yours being a triangle wave has even higher frequency components) accurately. A better measure for Pin/Pout could be performed with resistive loads and shunts and very large capacitors. \$\endgroup\$
    – Nick T
    Commented Mar 1, 2011 at 6:10
  • \$\begingroup\$ lots of information on switching regulator design at the Open Circuits wiki. \$\endgroup\$
    – davidcary
    Commented Mar 2, 2011 at 4:20
  • \$\begingroup\$ @Nick T, I am measuring the frequency with an oscilloscope. So, it is accurate. But for measuring Pin and Pout I am using a DC multimeter to measure avg. current through the inductor (Pin = Vin * IL) and voltage output (Pout = V^2/R) \$\endgroup\$
    – Punit Soni
    Commented Mar 4, 2011 at 22:24
  • \$\begingroup\$ This is a too specific question. Please rephrase to be more general or it will likely be closed. \$\endgroup\$
    – Brian Carlton
    Commented Mar 9, 2011 at 1:48
  • \$\begingroup\$ How do you measure Pin from average inductor current? \$\endgroup\$
    – markrages
    Commented Mar 9, 2011 at 3:49

5 Answers 5

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  1. Your inductor probably have high serial resistence. You might want to take lower value with thicker wire. Increase frequency if needed (but not too high, switching losses increases).

  2. Use ulta-low ESR caps, tantalum ones. Add some 0.1 + 22-66uF of ceramic cap in parralel. If frequency is high enough just leave ceramic.

  3. Implement synchronious rectification to get rid of that 0.25 vDrop (and I am not sure it's really 0.25, should be higher)

  4. Use proper MOSFET. Why take 100V part? 1.5Ohm Rds? This is not suitable for efficient DCDC. You want some 12-24V logic-level MOSFET with some 0.1 - 0.01 Rds.

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  1. As noted - MOSFET is not very suitable. As well as poor Rdson it has marginal gate turnon voltage (Vgth or Vgsth) for use at 3V. It will work (and it does) but could be better. With your oscilloscope observe the voltage when turned on. If you can measure it its too large :-).

  2. As noted also, the inductor should have a small enough resistance to drop minimal voltage with FET on and it must have a core designed not to saturate at your operating current. You have not said what inductor you are using. Please advise.

  3. Inductor value is possibly a bit high.

Ignore probably: At, say, 3V available and assuming a current ramp to double I avg it will take 220 uH/3V x 0.1A x 2 =~~~ 14 uS to reach required current. with continuous mode operation 220 uH is OK.

  1. Diode drop is suspiciously low. What diode are you using?

At 0.25V, as you state, the diode loss at 5V is > 0.25V/5V = 5%. Done less than properly add MOSFET 5%, Inductor 5%, other 5% = 80% overall. Sound familiar :-)

Ensuring FET and inductor are OK will help. 90% is hard without synchronous rectification,which has its own issues. Well over 80% should be possible.

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Or you could buy a properly implemented ready-made boost converter with > 90% efficiency for under $5. If you make many of these, vendors like Recom or Murata may actually have pre-built modules. If you only need one or a few, Pololu has a nice 5V 200 mA boost converter with 90% efficiency for $4.95. Link, while it lasts: http://www.pololu.com/catalog/product/798

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Here is a trick I learned some time ago. The best way to improve efficiency is to have the field of the inductor collapse as quickly as possible. Adding the PNP/Signal Diode combination will improve efficiency a lot.

enter image description here

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You could design a Snubber Circuit for a boost converter, just follow this link:

Design of snubbers for power circuits

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    \$\begingroup\$ Could you summarize the content of the link? \$\endgroup\$
    – clabacchio
    Commented Jun 5, 2012 at 7:26

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