PIC speark driver

You can effectively double the voltage by using one pin on each speaker lead. You drive one pin high and drive the opposite pin low at the same time.

Doesn't this just yield the difference between a high pin output (~Vdd-0.7) and a low pin output (~0)? It would only yield double if a low pin output was the negative of a high. I tried it and the output level was lower than what I'm currently getting (probably from the ~.7V loss since it's currently connected between Vdd and a low pin output).

Would've thought you'll get 5V driving speaker current in one direction, then 5V driving speaker current in the other direction. Speaker cone movement 'll be full out, back to neutral, then full in. I.e twice the original distance. (and 4 times the power!)

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The peak current with a 5 volt possible swing is half the swing divided by the resistance. So in this case

2.5V/8ohms=.31amperes. Obviously your PIC cannot deliver .3 amps in either direction, so almost all of the 2.5 volt swing is dropped across the output stage of the PIC. Even 5 outputs in parallel are limited by the total port current spec for the chip and will make it hot. There are two simple ways to get more volume to your speaker. One way is to use a transformer to get a less terrible impedance match between the output and the speaker. Something like this: at about $10 each. If you look through the surplus suppliers you may find one for a dollar or two.

The other approach is to add a current booster stage to the output. The simplest would be an NPN and PNP transistor connected as a complementary emitter follower. This can boost the available peak current by something like 20 to 50 times higher, and will keep the output current at the PIC within specs.

This requires just the two transistors (e.g. 2N3904 and

2N3906 or 2N4401 and 2N4403). Both bases connect to the PIC output, both emitters connect to the coupling capacitor to the speaker. The NPN collector connects to the 5 volt supply, and the PNP collector connects to ground.

You may get some nasty (current) spikes while switching. I'd advise a small resistor between the collector of the NPN and the Vcc and a decoupling capacitor near the transistor circuit.

petrus bitbyter

I agree that a bypass capacitor from collector of the NPN to the ground return point for the speaker is a good idea. But there is so little extra voltage available, that I would probably try a ferrite bead on a lead between the collector of the NPN and the 5 volt supply, to retain as much voltage as possible for the transistor driver. If the supply cannot handle the current drawn by this driver, then it can't handle this volume from an 8 ohm speaker.

I see your point, but I tried it again and the results were the same. My understanding of analog electronics is limited, so maybe someone can explain why what you stated doesn't result (or what I might be doing wrong).

My guess (I'm a hobbyist, nothing more) is that this is due to the driver impedance. Figure about 120 ohms source and 80 ohms sink. When you were driving just one pin, you only got one of those inserted. With two pins, you've always got both reducing current.

Jon

your speaker will pull a lot more power that the PIC can source (or sink) directly... put a transistor in the way (same as you would drive a relay)... much better.