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Trying to Increase the Voltage Output of a Boost Switch Mode Supply

Discussion in 'General Electronics Discussion' started by Eugbug, Jul 25, 2013.

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  1. Eugbug

    Eugbug

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    Nov 20, 2012
    I have built a simple boost switch mode supply. It runs off 6 volts and uses a Darlington arrangement for switching the inductor. I'm not sure what the value of the inductor is, it's a ferrite coil with several hundred turns of wire. I'm only getting about 24 volts output and trying to increase this. The output of a boost converter should be Vo = Vi(1/(1-D) where D is the duty cycle. I got better results when using a MOSFET, so I'm wondering is the limiting factor whether the transistor is being driven sufficiently into saturation. The duty cycle is about 0.9. Ltspice and the oscilloscope traces show the collector voltage is dropping to about 1 volt when the transistor turns on.
     

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  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    There are several reasons why this may be underperforming.

    Firstly, what is the resistance of the inductor? Several hundred turns on ferrite sounds like something designed for a radio, not a SMPS. It's also unlikely to be as low as 10uH if that's the case.

    Secondly, the switching speed of your transistors may be low enough that it limits the maximum voltage (the time it takes them to switch off is critical in this application). A mosfet may be better suited (it will also allow a greater voltage across the inductor).

    I would also consider taking the power for the IC from the top of the two 1N750's and to place a capacitor across them to minimise disturbance from the heavy currents flowing into the inductor.

    Once you've done all of that, you need to determine a safe ON time so that you don't saturate the inductor.
     
  3. Eugbug

    Eugbug

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    Nov 20, 2012
    I altered the circuit and am now using a 555 which gives squarer pulses, doesn't require the zener diodes for splitting the supply and the resistors produce a frequency of 110 khz. I'm using a MOSFET because it is easier to drive, has a faster switching time and I don't have to supply a large drive current as the hfe falls in saturation. I changed the inductor to one out of an old PC SMPS. Its series resistance is 0.06 ohms (still don't know the inductance). The Rds of the MOSFET is 0.025 ohms. I'm using several tantalums in parallel at the output to reduce the ESR.
    The 555 is running on 5.5 volts and I am managing to produce a 44 volt output into a 220 ohm load, that's 9 watt, efficiency works out at about 75% which is on the low side. The 10k resistor in the 555 oscillator circuit is the optimum value. Increasing or decreasing the resistance reduces the output voltage. Increasing the resistance increases the duty cycle but this doesn't increase the output voltage although it should do so (in theory anyway). I think that this could be because the output level from the 555 is borderline for gate drive or the power supply is limiting the current. I will try adding a large electrolytic on the power line. However does the ESR of an electrolytic limit the pulse capability?
    This isn't for any particular application but rather academic and I want to see how far I can push the voltage upwards from a low voltage supply, although I will probably modify the circuit to act as a switching current regulator for high power LEDs in a torch.
     

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    Last edited: Jul 26, 2013
  4. BobK

    BobK

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    Jan 5, 2010
    That is actually pretty good performance. If you have a scope I would look at the waveform on the inductor to see if it is saturating. If so, a higher frequency would improve things, if not a lower fequency might give you more output power.

    Bob
     
  5. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    The idea of placing a capacitor across the power supply is to reduce the impedance of the power supply (essentially ESR) to allow it to provide a higher peak current.

    A small resistor (perhaps 0.1 ohm) in series with the mosfet will allow you to see (on your scope) what is going on with the inductor and mosfet.

    What you want to see is a sawtooth waveform.

    If you see the current rise steadily and then shoot up, that means your inductor is saturating.

    If you see the falling edge of the sawtooth is slow (i.e. it appears as a slope) then you are not turning the mosfet off fast enough.

    For maximum power (and with a fixed resistive load, maximum voltage) the on time should be just short of that required to saturate the inductor, and the switch off should be as rapid as possible.

    Note that a 0.1 ohm resistor in series will have an effect on performance, but it should only be small. Also beware that if you adjust the 555 so the inductor is *just* short of saturating, removing the resistor will push it a little further toward saturation.
     
  6. Eugbug

    Eugbug

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    Nov 20, 2012
    Update!

    Pushing the supply voltage upto 6 volts improved the efficiency to over 80%.
    I added a large value electrolytic close to the power pin on the 555 which cleaned up the "dirty" supply rail.
    Adding a high value electrolytic capacitor close to the top of the inductor didn't appear to make any difference. On the scope waveform, the voltage wasn't dropping during the period when the MOSFET turned on.
    The Vds was getting close to the 100 volt rating for the MOSFET I am using so I didn't push it any further.
     
    Last edited: Jul 27, 2013
  7. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Yeah, either the capacitor increased the available current, or the higher voltage allowed you to turn the mosfet on more (or both)
     
  8. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Yeah, that mosfet requires about 5.5V Vgs before it turns properly on. And it's likely you're not getting there even yet.

    See the top left graph on page 3 of the datasheet.

    You would probably get higher efficiency from a higher input voltage (7V should show further improvement) and reducing your ON time to limit the voltage (or adding feedback)
     
  9. Eugbug

    Eugbug

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    Nov 20, 2012
    Thanks for the suggestions Steve, I'll try monitoring the current tomorrow if I can find a low value resistor.
     
  10. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    A few random thoughts.

    I've had good results with the UC3845 current-mode controller for the flyback topology. I've always used it with a transformer, but I'm sure it would work as a boost converter too. Its power supply voltage requirements may be too high for you though.

    Sensing the inductor current is easily done with a small resistor between the MOSFET's source and the 0V rail. This is needed for the UC3845 anyway, if you use it.

    You should be able to estimate the inductance by measuring the slope of the current increase. dI/dT = V/L where dI is the change in current, in amps; dT is the time interval over which that change occurs, in seconds; V is the voltage applied across the inductor, and L is the inductance in henrys. This rearranges to L = V dT / dI.

    Alternatively, if you have an oscilloscope, make a tuned circuit with the inductor and a capacitor and see what frequency it rings at. You can use the horizontal output (if the scope has one) to generate a pulse to start the circuit ringing.

    I always use a good solid ceramic capacitor around 1 uF at 50V (something like http://www.digikey.com/product-detail/en/C4532X7R1H105K160KA/445-3913-1-ND/1923420) across the power supply rails to give a nice low ESR on the rail into the inductor.

    As usual all of Steve's advice here is excellent.
     
  11. Eugbug

    Eugbug

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    Nov 20, 2012
    Thanks Kris, I'll try that. The best way to make a low value resistor is probably to wind some wire onto a former, drive a couple of amps through it and measure the voltage drop to establish the resistance. The inductance of the coil shouldn't be too high relative to the inductor in the circuit if it is air cored?
     
  12. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    I guess so, but low-value metal film resistors are readily available; I would use one of those. Something like 0.1 ohms, as Steve suggested, would be appropriate.
     
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