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Woooohooo! (Boost converter success)

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rupidust

Sun Apr 02 2006, 11:54PM

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Registered Member #110 Joined: Fri Feb 10 2006, 12:23AM
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Posts: 85

Like the others said, next step is to integrate feedback. Actually trying to up the watts power should be next. Once you varify its at least 70% efficienty then its time to move one. You measure the current in series with the IGBT at ground level. Take that current and multiply it by the voltage of the inductor supply to get Watts. Compare usage watts to charging watts to rate performance.

Here is a pic of an advanced booster operating at 30 Watts from 12v supply. This can charge caps or battery packs. Image is in battery charge mode. The difference is the voltage divider used to monitor the charge voltage: divide by 100 for caps and 10 for batteries. The heart of the booster is a PIC 16F819. Using 3 of the 5 analog to digital converting lines, I monitor the voltage of the battery supply, load voltage (cap or another battery), and current of the charger. The PIC also drives the LCD in 4-bit address mode. Boost signal is amplified by an 8 pin dual Mosfet driver, which relays the signal to an IGBT. To the left there are 3 buttons. One to decrease the HOLD/regulate voltage, one to increase this, and another to start/stop charging. Of course once the charge voltage reaches the HOLD voltage, the charger stops then begins regulation mode. If there are no bleeders across the cap bank, the charger stops for a long time. The heavier the bleeder, the more regulation pulses are sent to the cap to maintain charge level. There is always a bleeder because the feedback voltage divider across the cap is in essence a bleeder.

Link to Video of Cap mode http://geocities.com/rupidust1/booster-lcd1.wmv

More one the PIC. Ok, this is so far beyond beginner level but here goes anyway. The booster is running in an interrupt background, turning the IGBT on/off at programmed times and duty cycles. In the normal flow of code is everthing else. One A2D line voltage is compared with an A2D_level variable. When ever the A2D line voltate is >= the A2D_level variable, another variable is then toggled to stop the booster when the next interrupt service routine happends. The buttons are continuously monitored along with voltage. When a particular button is pressed, the code vectors to another sub routine to either, decrement the Hold voltage(A2D_level), increment it, or stop charging all together. For either increment or decrement, the LCD display of HOLD = XXXv is updated. Before the LCD is updated 3 math sub routines are used. First, multiply a 16bit value by another 16bit value resulting in a 32bit value: A2D_level * 20(mv) to create a calculated representation of load voltage. The second sub routine converts 16bit Hex values into a 5 digit BCD value Ten thousands, Thousands, Hundres, Tens, and Ones. The third sub routines converts decimal numbers into LCD ascii numbers by adding 0x30 to a number. Thanks be to internal PIC EEPROM, the last HOLD voltage is automatically remembered on powerup. Battery mode increments the HOLD voltage by 1 and Cap mode increments it by 10. Cap charge range is from 50v to 500v. If I need higher voltage, I just change the voltage divider ratio and modify the code.

Dr. Shark

Fri Apr 07 2006, 11:51AM

Registered Member #75 Joined: Thu Feb 09 2006, 09:30AM
Location: Montana, USA
Posts: 711

Aahz, inspired by your converter I decided to put one together to do some overvoltage stress testing on MMC capacitors I recently accquired. Shame was, it worked too well! By using a rather small inductor (output filter on the 5V line of a ATX PSU) of 100uH or so, I managed to overvolt a 600V IGBT in an instant. This was at 50% duty-cycle, discontinous mode of course. Just to let those nay-sayers know how much voltage these little things can pack!

Steve Ward

Fri Apr 07 2006, 07:43PM

Registered Member #146 Joined: Sun Feb 12 2006, 04:21AM
Location: Austin Tx
Posts: 1055

Just to let those nay-sayers know how much voltage these little things can pack!

This is easily predictable with V=-L*di/dt. You have to be careful about the inductance path between IGBT/diode and output capacitor. Anyway, with 100uH, and switching say 10Apk in 100nS, your max V would be about 10kV. This is why you must either 1) keep a load on the converter at all times, or 2) use voltage feedback. There are many design considerations with converters... far too much to get into right now, but if you dont want them to blow up, you better read up on them more first

IamSmooth

Wed Apr 26 2006, 09:54PM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
Location:
Posts: 1567

Regarding your schematic for the Boost Converter, why is R10 (1M ohm) resistor needed?
Is it to provide a discharge path in case there is no load on the output?

Second question: by changing the value of the inductor the voltage output goes up; so, what are the limitations as to how large an inductor one can use for a set input voltage? Is it the amout of current you want delivered?

Wilson

Thu Apr 27 2006, 02:14AM

Registered Member #78 Joined: Thu Feb 09 2006, 11:27AM
Location: Sydney, Australia
Posts: 133

I presume R10 is there as a bleeder resistor for any caps aahz might charge.
Using too large an inductor may limit your max output voltage, as di/dt (current rise time) won't be as high as with a smaller inductor.

Steve Ward

Thu Apr 27 2006, 09:30PM

Registered Member #146 Joined: Sun Feb 12 2006, 04:21AM
Location: Austin Tx
Posts: 1055

Using too large an inductor may limit your max output voltage, as di/dt (current rise time) won't be as high as with a smaller inductor.

Thats not how it works. The di/dt is determined by the switch (not the inductor). So in fact, a larger inductor would give a bigger inductive kick assuming the same current (built up at the time of switching) and switching speed.

IamSmooth

Fri Apr 28 2006, 12:45AM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
Location:
Posts: 1567

So, what is the difference in performance if one starts increasing the inductance? The voltage goes up as di/dt is fixed, but does the current go down? There has to be some trade-off if the output voltage continues to increase?

Steve Ward

Fri Apr 28 2006, 07:21PM

Registered Member #146 Joined: Sun Feb 12 2006, 04:21AM
Location: Austin Tx
Posts: 1055

So, what is the difference in performance if one starts increasing the inductance? The voltage goes up as di/dt is fixed, but does the current go down? There has to be some trade-off if the output voltage continues to increase?

You might be "missing the point" with boost converters. There are 2 basic operations going on in the conversion. 1) build up energy in the inductor, and 2) discharge energy into output storage device (capacitor). Then comes in the numerous factors that you must consider if you want to build the best converter given any constraints that you have. Basically, start with what output voltage and current that you want, and depending on the mode of operation (continuous vs. discontinuous) you can figure out the inductor current more or less. From here you have to juggle switching frequency and inductance. These 2 will be very closely related once you decide on the peak inductor current (and the change in current during the discharge period if running continuously). Because you must deliver some specific amount of energy per cycle, this determines the frequency you must operate at. Also, since the current can only build up at a specific rate (V=L*di/dt) this also puts constraints on L and freq. Anyway, im starting to ramble here, and the point is, you cant explain the whole converter design in a paragraph. If you are really curious, i suggest you search for "boost converter design" on google. This is how i learned what i know.

For you guys that are just charging capacitors and dont really care about getting high output current, you probably dont care as much about a proper design. Just keep in mind some way to protect the output voltage from going too high. A boost converter with no load (or regulation) is a very dangerous thing to itself. As i proved before, its easy to produce multi kV output from the inductor unless you have some capacitor to clamp this voltage as stored energy.

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