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I have been experimenting with DC-DC boost converter design. After I made some calculations in order to obtain a certain output based on certain parameters, I tested the SMPS and saw that it's not working as I designed it to.

schematic

simulate this circuit – Schematic created using CircuitLab

Parameters:
\$V_{in}\$=2.4V
\$V_{out}\$=5.1V
\$I_{out(min)}\$=1A
\$I_{out(max)}\$=1.5A
D=0.629 or 62.9%
L=94 uH
C=1517 uF
f=6239.54 Hz

The batteries are rechargeable (2 X 1.2V) NiMH 800 mAh. I checked them after I used the converter last time and they were almost fully charged (2.3V). The PWM part (the 555 and resistors + capacitor) are powered by a 9V battery.

I used this site for most of the variables, and this site only for C1. There are two supply rails in the schematic (one for PWM and the other for the actual boost converter) and only one ground.

By checking the boost converter with a multimeter, I found that the output was 20V and 80mA, which is far from what I wanted. I noticed that once I connect the probes it takes some time (but it's still short) before the output reaches 20V.

It sounds like I traded voltage for current without knowing. The diode I used is not great, but I would not consider it to be the cause. According to the second site, output current is influenced by IC current and this can be a possible cause, but I'm not sure of it. Reducing the value of the inductor doesn't seem to help in increasing the current.

What do I have to change in my circuit, in order to get the desired results? How do I increase output current?

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    \$\begingroup\$ What do you mean by '20V and 80mA'? What happens if you draw more or less current? Without regulation the output voltage will vary with loading, and when unloaded it will go as high as it can (limited only by parasitic elements in the flyback circuit). 1N4007 is not good at high frequencies - with a better diode the output voltage might go even higher! \$\endgroup\$
    – Bruce Abbott
    Commented Oct 22, 2016 at 14:33
  • \$\begingroup\$ removed comment... \$\endgroup\$
    – Nils Pipenbrinck
    Commented Oct 22, 2016 at 15:45
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    \$\begingroup\$ Try swapping the 1N4007 with a 1N5819. It's a Schottky diode equivalent to what you're using and is going to have a more responsive switching time. The responsiveness of the diode is important in a boost converter topology. \$\endgroup\$
    – ev3670
    Commented Oct 22, 2016 at 16:08
  • \$\begingroup\$ @Bruce 20V and 80mA are present at no load(I connected the multimeter directly at the output).I thought that it should draw more current than that(at least closer to 1A).But as you can see it's far away from that value. \$\endgroup\$
    – Daniel Tork
    Commented Oct 22, 2016 at 17:47
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    \$\begingroup\$ your C1 & battery both have ESR, you are open loop, your L can saturate , trigger goes to threshold and nobody seriously uses a 555 for a boost regulator \$\endgroup\$
    – D.A.S.
    Commented Oct 22, 2016 at 20:48

2 Answers 2

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A boost converter stores energy (in the first half of the switching cycle) in an inductor and then releases that energy to the load in the second half of the switching cycle. The energy it stores (and how often it repeats this process every second) HAS to match the power required by the load in order to sustain the correct output voltage.

In other words, in its basic form, it's a power regulator and not a voltage regulator.

So, a boost converter (without a sophisticated feedback system to control duty cycle) is a power regulator AND, without a load it will continue to dump that inductor energy into the output capacitor until that output capacitor gets so charged up it breaks or the output transistor starts to break down.

You need to have the correct load to match the duty cycle to maintain the output voltage at your desired level. It's not a voltage regulator until you apply a level of sophistication and negative feedback.

The diode is very important - for instance the 1N400x series has an appalling switch off time in several tens of micro seconds and if your operating frequency is tens of kHz it will behave very badly.

C1 at 1,500 uF sounds very much too high unless your switching frequency is a few hundred Hz then, 94 uH sounds too low.

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  • \$\begingroup\$ So without the feedback network,the converter is not useful.Doesn't continous mode mean independent output(not dependent on load)?That's how I tried to set it. \$\endgroup\$
    – Daniel Tork
    Commented Oct 22, 2016 at 17:05
  • \$\begingroup\$ Continuous conduction mode means that the inductor current never falls to zero i.e. it lways remains conducting. Discontinuous mode means the inductor current will fall to zero and stay at zero until the next charge cycle begins. It's as useful as an op-amp without negative feedback. \$\endgroup\$
    – Andy aka
    Commented Oct 22, 2016 at 17:32
  • \$\begingroup\$ Ok,so it's always dependent on the load.I hope you saw the links I gave in the question.I understand that all those formulas help calculate variables for a certain load.As I set it up,the converter will supply 5.1V and 1A to a load that requires 5.1V and 1A.If I change this load,duty cycle and frequency must change too.Or perhaps the rest of the components,too? \$\endgroup\$
    – Daniel Tork
    Commented Oct 22, 2016 at 17:54
  • \$\begingroup\$ Only duty cycle needs to change but, lowering the frequency (energy passed per second) also lowers the output voltage. I didn't follow the links BTW. \$\endgroup\$
    – Andy aka
    Commented Oct 22, 2016 at 18:20
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This circuit does not have any feedback to stabilize the voltage output, so it just pumps voltage higher and higher until some parasitics stops it.

If you are serious, you should use any of industry offerings of boost converters, with internal voltage reference etc. There are plenty at Digi-Key.

If you are more serious, you better use one of the special power-management IC that manages both the charging your NiMH and upconverting it to 5V.

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