Re: 0.20 delay in 6v control circuit
From: Robert Monsen (rcsurname_at_comcast.net)
Date: 11/10/04
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Date: Wed, 10 Nov 2004 00:06:46 GMT
Bob wrote:
> On Mon, 08 Nov 2004 22:10:19 GMT, Robert Monsen
> <rcsurname@comcast.net> wrote:
>
>
>>(fixed top posting below. Please post replys after the main body of
>>text. It's just convention here)
>>
>>Bob wrote:
>>
>>>On Mon, 08 Nov 2004 07:37:01 GMT, Robert Monsen
>>><rcsurname@comcast.net> wrote:
>>>
>>>
>>>
>>>>Here is a slightly better circuit for you needs:
>>>>
>>>> 6V
>>>> ----------------------------o--o
>>>> | | |
>>>> | - C|
>>>> | o ^ C| Relay Coil
>>>>|=| switch | C|
>>>> | o | |
>>>> | '--o
>>>> | |
>>>> | 10k |/
>>>> o---/\/\/--o-----o-----------|
>>>> | | | |>
>>>> | | | |
>>>> o----|<----| | + |
>>>> | * --- 220uF |
>>>> / --- |
>>>> \ | |
>>>> / 1k | |
>>>> \ | |
>>>> | | |
>>>> | | |
>>>> '----------------o-------------'
>>>> GND
>>>>
>>>> * Schottky Diode
>>>>
>>>>created by Andy´s ASCII-Circuit v1.25.250804 www.tech-chat.de
>>>>
>>>>The base will drop immediately from 850mV to about 200mV when the switch
>>>>is opened, due to the 1k resistor and the schottky diode. When the
>>>>switch is closed again, the base will come up from 200mV to about 850mV
>>>>before the transistor conducts, so
>>>>
>>>>V(t) = V ( 1 - exp(-t/RC) )
>>>>
>>>>thus
>>>>
>>>>t = -ln(1 - .65/6) * 10k * 220u
>>>> = .25s += 0.05s
>>>>
>>>>Once the cap is charged up, the current will be about 500uA, so the
>>>>current through the transistor can go up to 25mA (assuming a beta of 50).
>>>
>>>
>>>Thanks again Robert,
>>>
>>>Do you have any idea on the specs for the diodes needed. I went to
>>>pick up the Schottky diode and the one in parallel with the relay but
>>>was not able to choose from the variety available. Thanks for the
>>>design.
>>>
>>>Bob
>>>
>>>
>>
>>You can use any of 1N4002-1N4007 diodes for both of them. A Schottky
>>diode would be nice where the * is (it doesn't have to carry any
>>appreciable current, so any of them will do) but isn't necessary. It
>>just makes the recovery time a bit better, because it has a lower
>>forward voltage drop.
>
>
> As you can probably tell, this is the first time I ever posted to a
> news group. Anyway I built the circuit as you designed. What
> actually occurs is the relay is energized for only .20 seconds instead
> of a delay of .20 seconds then energized until the switch opens. (Did
> you expect this? I checked the path of the circuit 3 times)
>
> Bob
When you close the switch, it should take 0.2 seconds to charge up the
capacitor enough to turn on the transistor. Thus, assuming the relay is
open when you start, the following should happen:
1) relay open, switch open
2) switch closed
0.2 seconds
3) relay closed
...
4) switch open, relay opens immediately.
It sounds like you have the 10k resistor and 220uF capacitor swapped.
That would cause it to
1) relay open, switch open
2) switch closed, relay immediately on
0.2 seconds (or less!)
3) relay open
...
4) switch open, relay still open
From your reply, I can't quite figure out whether that is what you are
seeing, though.
(You probably know this, but closed means passing current, open means
unable to pass current.)
--
Regards,
Robert Monsen
"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.
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