On May 25, 2005, at 9:45 PM, Bernd Felsche wrote:
> On Thursday 26 May 2005 10:16, Mike Murphree wrote:
>
>> On May 25, 2005, at 8:58 PM, Ralph Hilton wrote:
>
>>> Relative to pin1, pin2 goes from +5 volt to - 5 volt which is 10
>>> volts peak to peak.
>>>
>
> Not really. It's just how it appears (memntarily) at the buzzer.
>
>
>> This technique was used in car stereo amplifiers before internal
>> power inverters became commonplace. If you connect an oscilloscope
>> (with differential inputs) across the buzzer, it will appear to a 10
>> volt peak-to-peak waveform even though only 5 volts is applied in
>> each half-cycle.
>>
>
> To answer Peter's question; the "storage" device is the buzzer.
> It's mostly a capacitor. The gates on the uC act as diodes (when
> they go "high").
C'mon guys this is basic AC theory... nothing is being stored.
A 10 volt peak-to-peak square wave is (assuming symmetrical) +5V
applied to a load during one half-cycle followed by -5V applied to
the load during the next half-cycle. The CMOS outputs of the AVR can
either source or sink current depending on their state, so by using
two outputs, you can effectively double the voltage excursion applied
to the load. Peak voltage for an AC waveform is half of the peak-to-
peak value and since in this case we are talking about a square wave,
it is same as the AC RMS value. So with one output you can achieve
2.5 Vrms and with two outputs 5.0 Vrms.
Back to my car amplifier example, before they learned to bridge the
amplifiers (use 2 amps, one 180 degrees out of phase with the other),
about 4 watts was all you expect out of a direct-coupled solid-state
amplifier. With bridging this jumped up to 18 watts (20W at 10% THD)
and now nearly all of them include a power inverter to remove the
dependency on the supply voltage entirely.
Mike