Using the DRV8662 chip to create 105V DC from battery voltage.

This is how I generated a little over 100V of DC using a 3.95 V power supply with about $10 of parts.

Summary

Using a DRV8662 chip we can create a battery powered board that will generate a 105V DC power rail using around $10 worth of parts. Some surface mount soldering is needed though. If you have an input signal below about 500Hz, you can use the chip to amplify this to have a peak to peak output of up to around 200V.

Why?

I need a 100+ V DC rail as an input to a circuit to drive some piezoelectric crystals. The crystals I’ve been tasked to use resonate at 40KHz.  How should I go about this? First off, the crystals need a signal with an amplitude of around 100+V. How hard could that be. Errrr….

How?

There are a couple of standard methods to create a boost converter:

Transformer

For instance flyback transformers as used in ‘old school’ television sets. The type with tubes. Remember those? No? I’m old…

Inductor switching circuit

Switching a voltage into an inductor ‘bounces’ the output voltage up. One interesting design based on this idea can be found here. The switching frequency needs to be controlled to maintain a steady output voltage, ideally using some kind of a controller. It is a beautiful piece of electronic design which I think I would enjoy, but it’s all about time.

Have somebody else make it

Of course. The simplest and fastest way. Stand on the shoulders of giants. Now, how many battery powered 100V DC-DC converters can I find on eBay. None. However, after some searching, I did find an integrated chip that does most of the work for me made by Texas Instruments.

Enter the DRV8662

Texas Instrument’s DRV8662 chip is designed to drive a haptic feedback piezoelectric transducer at up to about 500HZ. If we look at the functional block diagram in the data sheet, which I copied below, we can see that an external inductor (L1) is used by the IC to generate a voltage rail at up to about 105 V. This voltage rail is called VBST. This is used to power an internal operational amplifier. This op-amp can be used to amplify an input signal from IN+ and IN-. The output can be applied to a piezoelectric transducer, shown acroos OUT- and OUT+. 

So we have an inductor switching circuit to generate the DC voltage rails and a high voltage op-amp with a gain-width bandwidth somewhere around 500V which can be used to drive a piezo actuator. The chip is designed to be used inside of e.g. mobile phones, to make a piezoelectric crystal vibrate so that your phone shakes. Now you know where that comes from.

The simplified schematic below the functional block diagram is basically the same, but the outline of the chip is dotted in. The components outside of the dotted line are the ones that we need to add to make the circuit work. What values to use? I copied the ones from the demo board that Texas Instruments make.Then started playing.

I used R1=768K and R2=16K gives me a 105V DC output. According to my calcs, R2 = 10K should give me a 105V output. I think the extra inductance from using flying leads on the inductor affected this. On my first board with the inductor soldered close to the IC, using R2=20K gave me about a 52V VBST. On board two, with the inductor on flying leads, I got VBST as 84V.

On a PCB I would place 10K to get the 105V output. Or maybe start with 12K and resolder it if the rail was too low. I used L1 = 3.3uH and Rext = 7.5K, using a recommended inductor from the data sheet. GAIN0, GAIN1 and EN are tied high using 1K resistors. Don’t forget that EN needs to be high. Or the chip no worky.

However, the chip does allow me to tap the boost voltage that it creates from the VBST or PVDD pins. I can use this high voltage with some extra circuitry to generate a high voltage oscillating signal at 40 KHz to make my crystals shake. Watch this space.

‘But, but but, you’re using a fraction of the IC’s capability to just generate the 100 V rail, wouldn’t it be cheaper to make your own voltage boost circuit?’. The DRV8662 costs $3 in low quantities. Even Farnell in the UK only charges £3 each. Plus VAT. Plus postage. Even so, no, it wouldn’t be cheaper to spin my own circuit using transformers or a switched inductor, especially when you factor in my valuable time at $1 an hour. I know, I earn the Big Bucks.

So I bought a few DRV8662s from Farnell and soldered one down onto a QFN20 to DIL converter board. Which in plain speak is a little circuit board that converts a stupid small chip with no legs into something that I can plug into some breadboard to play with. The soldering is not too tricky if you use a decent small bit in your soldering iron, plenty of liquid solder flux and have a fan to blow the toxic fumes away. I use a USB powered fan designed to go inside computer cases with a power brick for this. eBay.

The first board I made is on the picture below on the left. Note that I soldered the inductor directly to this board. It worked. Lovely. Then I shorted the VBST pin to ground with a crocodile clip. Careless. Not that I’m bitter. Still, a bit of care would have avoided this. I learned my lesson.

Why is there a bit of brown tape on the second board? I put a corresponding bit of brown tape on one edge of my breadboard to remind me which way around to stick the module into my breadboard as it looks pretty much the say both ways around, but only works one way around and may never work again after being put in the wrong way around. I don’t want to find out if the last statement is true or not.

With the second board, I soldered the inductor onto some leads and stuck this into the breadboard. I found that the output voltage was higher than when the inductor was stuck directly to the surface mount to DIL converter board. Why? I suspect that the extra leads and path to and from the inductor increases the overall inductance.

Why did I use that particular inductor? Because the data sheet said that make and model would work. That’s why.

VBST -||–||- GND

Because the output on VBST is 105V and each of the two capacitors is only rated to 63V. So I daisy-chained two of them. There is probably not much risk of a 63V rated capacitor cooking off at 105V, but if it were to short, then the DRV8662 could die. As I found out when I shorted VBST to ground with a crocodile clip. Not that I’m bitter about this. Not at all.

How smooth is the 105V voltage rail? No way to tell using breadboard. If you want a smooth rail, don’t use breadboard.

I tested using an input from a signal generator. Each output pin generates a signal in anti-phase to the other, so the peak to peak difference is about twice the VBST voltage.

Please find a couple of photos showing this testing below. For these tests I only had a 52V VBST. Then I learned that adjusting R1 in the simplified schematic above changed the output voltage. Reading the data sheet always helps. I started out using an old-school ‘scope as I like them. Then switched to a new-school ‘scope because I could and nobody stopped me.

Useful extra stuff

How do we know if we have soldered down the chip correctly? Please find some impedances that I measured between various pins for a working board below.