High Current Boost Circuit

Discussion in 'General Electronics Chat' started by JBmtk, Feb 17, 2011.

  1. JBmtk

    Thread Starter Member

    Jun 29, 2009
    Objective: High Voltage Output (45V @ 1A)

    I have created a basic boost converter as seen in below. The frequency is around 40KHz and pulse width around 20%. This should give Vout:=Vin/1-DutyCycle = 5/(1-.2) = 6.25V...ideally, but this is not the case. Why?


    With the 1k load resistor, I am only able to achieve around 10VDC. This is verified by the simulation and my voltmeter.


    The oscillations occur because I am only using a .33uF cap. If it is increased, the voltage will be smoothed out more obviously, but the time constant will increase.

    Without a 1k load resistor...say a 200k load, I am able to generate around 72VDC. This is verified by simulation and my wet fingers :)

    My question is how do I increase the current capabilities (assuming the input wattage is greater than that of output) ?

  2. marshallf3

    Well-Known Member

    Jul 26, 2010
    L1 probably doesn't like 40 KHz but forget that for now as a 1N4148 is far from correct either - it's a tiny signal diode - use at least a 3A fast, ultra fast or a Schottky switching rectifier, not a logic diode.

    C1 & C2 need to be far larger as well.

    Put a 4,700 uF and at least 0.47 uF for C1 close as batteries don't source current as you'd expect them to under high frequencies and at least a 1,000 uF for C2 however more is better.

    People forget to remember that batteries have an ESR as well, not to forget the wiring inbetween. I'll extopolte leter.

    Spend few $ @ http://www.goldmine-elec.com/ and you'll have a bonanza from the surprise boxes. 3 sizes and top quality stuff in there.
    Last edited: Feb 17, 2011
  3. Audioguru

    New Member

    Dec 20, 2007
    The old IRF540 needs a gate voltage of 10V to turn on pretty well. But you are feeding it only 5V so it does not turn on hard enough. Use a logic-level Mosfet and select a newer one with a very low on-resistance.
  4. Adjuster

    Well-Known Member

    Dec 26, 2010
    There is another basic issue here. Apart from sorting out drive levels to suit the FET (or vice-versa), and a getting a satisfactory rectifier, you need to be sure of the operating mode. You are assuming an averaged or continuous mode of operation, but at present your circuit operates in discontinuous conduction. In other words, the coil current falls to zero before the end of each cycle.

    This explains why the output voltage was larger than you expected. In this mode, the output voltage is strongly dependant on the load resistance, as you have seen. It may be possible for the voltage to rise to damaging levels with a sufficiently high resistance, so never operate a circuit of this kind with the load open-circuited.

    We can see that your circuit operates discontinuously with the present duty cycle, even with only 10V output. Applying 5V to a 220μH inductance for 5μs would raise the current by 114mA. If the inductor then discharges into (10V-5V), in about another 5μs the current would have dropped down by about the same amount. The transistor actually switches off for about 20μs, so there is plenty of time for the current to decay to nothing.

    To obtain more output current for a given voltage, the duty cycle would need to be increased. Above a certain value for a given load, the circuit would change from discontinuous conduction to continuous. This condition may facilitate higher output currents, but the output can be less easy to stabilise, and the continuous current places a greater demand on the rectifier switching speed: the diode current will not have fallen to zero before its voltage reverses.