Re: transistor base input

On Apr 30, 4:09 pm, John Popelish <jpopel...@xxxxxxxx> wrote:
lerameur wrote:
ok good thanks for the link. below is my circuit . I have a 7805 that
regulates the 5v. The ground is common to all. The source is anywhere
between 6 to 9 volts. The source as mentionned in the circuit is the
output of the analog alarm system going to the siren. I will hooking
up the 5v regulator most probably on the batteries (12v) actin g as
backup and being recharge by the analog circuit.
Hope that helps you help me.

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?





VCC 5V
+
VCC 5v |
+ |
| .-.
| | |
.-. | |
| | '-'
| | | ___
'-' -------|___|-- to
microcontroller
| |
| ___ |/
|----|___|--|
| |>
___ |/ |
Source--|___|- --------| |
|> |
|
| GND
|
GND

Once I understand exactly what the purpose of this circuit
is, I might have more suggestions.

--
Regards,

John Popelish

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)

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. The analog input, therefore, must be between
VSS and VDD. 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. A maximum source impedance of 10 k? is rec-
ommended for the analog sources. Any external com-
ponent connected to an analog input pin, such as a
capacitor or a Zener diode, should have very little
leakage current.
.

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