Re: transistor base input

lerameur wrote:
On Apr 30, 4:09 pm, John Popelish <jpopel...@xxxxxxxx> wrote:

Is the voltage swing labeled as "source, below also limited
to swing between +5 volts and ground? I guess I don't
understand why you need this circuit, at all. What prevents
you from connecting the "source" directly to the microprocessor?
(snip)
Well section 13.9 of the pic16F88 data*** is copied below:
From what I can read, if I supply 5v to the chip, my input to the
microcontroller has to be between 4.4v and 5.6v.
So the circuit is there to have a steady output of 5v into the
microcontroller for any voltage between 6 to 9 v off alarm system
(source)

I assume this is a PIC input configured as a digital input, based on your two transistor interface circuit.

A simplified circuit for an analog input is shown in
Figure 13-4. Since the analog pins are connected to a
digital output, they have reverse biased diodes to VDD
and VSS.

Why would you use the input configured as an analog input, instead of digital? Your circuit will output only two states-- high and low.

The analog input, therefore, must be between
VSS and VDD.

Yes.

If the input voltage deviates from this
range by more than 0.6V in either direction, one of the
diodes is forward biased and a latch-up condition may
occur.

Yes.

A maximum source impedance of 10 k? is rec-
ommended for the analog sources.

Only if you need the full analog to digital converter resolution over the full temperature range. What, exactly do you want the microprocessor to know about the input signal?

Any external com-
ponent connected to an analog input pin, such as a
capacitor or a Zener diode, should have very little
leakage current.

Of else, what?

--
Regards,

John Popelish
.

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