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Re: Just to verify my reset circuit

Subject: Re: Just to verify my reset circuit
From: "Chris"
Date: 10 Apr 2006 20:00:35 -0700
Newsgroups: sci.electronics.basics
crazy frog wrote:
> you suck
>
> "John Fields" <jfields@xxxxxxxxxxxxxxxxxxxxx> wrote in message
> news:75rk32df0le5crjma7lag80k4tobtps36n@xxxxxxxxxx
> > On 8 Apr 2006 05:07:45 -0700, "Ant_Magma" <vcteo1@xxxxxxxxx> wrote:
> >
> > >My reset pin is active low, this would work right?
> > >
> > >Vcc
> > >|-------
> > >|      |
> > >|      -
> > >R10k 1N4148
> > >|      +
> > >|      |
> > >|-------To reset pin
> > >|
> > >C10uF
> > >|
> > >|
> > >GND
> >
> > ---
> > Yes, but why do you need such a large capacitor?
> >
> > --
> > John Fields
> > Professional Circuit Designer

Actually, pin 46 is a CMOS input, so it neither sources (blows) or
sinks (sucks) current.

By the way, if you want a simple transistor circuit for your one minute
delay, and you happen to have a 10,000 uF cap and a 10K pot handy, you
might want to try this (view in fixed font or M$ Notepad):

|
|      SW1
| VCC-o
|     __--o-o-------o-------.
| GND-o     |       |       |
|           |       |       |
|           |      .-.      |
|           |  .-->| |10K   |
|           |  |   | |      |
|         D -  |   '-'      |
|           ^  |    |       |
|           |  '----o       |
|           |       |       |
|           |      .-.      |
|           |    1K| |      |
|           |      | |  .---o
|           |      '-'  |   |
|           |       | |/    |
|           '-------o-|     |
|                  +| |>    |
|           10,000 ---  | |/
|             uF   ---  '-| TIP101
|                   |     |>
|                   |       |
|                  ===   .--o
|                  GND   |  |
|                       D|  C|
|                        -  C|RY1
|                        ^  C|
|                        |  |
|                       ======
|                       GNDGND
|
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

Simpler, and fewer components.  This relies on the fact that the
guaranteed minimum turn-on voltage for a 12V relay is almost always
less than 9.1V  Even if Vcc goes down to 10.5V, this circuit should
still light up the relay.  This circuit doesn't discharge the 10,000uF
cap through a base junction on turn-off (eeeww), or rely on two diodes
to limit Vbe to 1.4V (?ouch!) on turn-on.  You don't have to worry
about exceeding maximum wiper current if you turn the pot to min.  You
don't have a 100uF cap which performs no useful function at all.  Nor
do you have a second transistor which really doesn't do much except
drop voltage.  You will also be able to drive a wider range of relay
coil currents reliably (can drive a 12 ohm coil, but over 200mA timing
becomes beta dependent -- and be sure to use a heat sink!)  This
circuit has provision for quick reset on switching.  However, just like
your circuit, the time delay on this circuit will be power supply
voltage dependent.  Use 1N4002 diodes for D.

Call it a day.  Mr. Fields may have been a little brusque, but you're
embarrassing yourself.  Big time.

Good luck
Chris


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