Re: 555 Timer Question

On Fri, 23 Dec 2005 22:07:31 GMT, "DSallee"
<audiopulse@xxxxxxxxxxxxx> wrote:

>
>"John Fields" <jfields@xxxxxxxxxxxxxxxxxxxxx> wrote in message
>news:gopoq1dmcedirsot70lf2gjaqr96aeb354@xxxxxxxxxx
>> On Fri, 23 Dec 2005 20:44:58 GMT, "DSallee"
>> <audiopulse@xxxxxxxxxxxxx> wrote:
>>
>>>
>>>> Why make it so hard?
>>>>
>>>> Just use the relay connected to the transistor connected to Q14 of
>>>> the 4060:
>>>>
>>>>
>>>> +V
>>>> | +9V -9V
>>>> +--------+ | |
>>>> |A | O--> /<--+
>>>> [DIODE] [COIL] - - - -/
>>>> | | O
>>>> +--------+ |
>>>> | +----------> +/-V OUT
>>>> C
>>>> 4060 Q14>--[R]--B
>>>> E
>>>> |
>>>> +--------------------> GND
>>>>
>>>>
>>>> --
>>>> John Fields
>>>> Professional Circuit Designer
>>>
>>>Hey John,
>>>
>>>Thanks for the input as I am also learning... :)
>>>
>>>I'm a little confused on how this would work? With how you have it,
>>>wouldn't
>>>you have to supply a -9v so it would switch between the +9v and -9v with
>>>just the one relay??
>>>
>>>Sorry if this is elementary but my "little" brain only came up with what I
>>>had with the relays... LOL Can't seem to comprehend what you are
>>>describing...
>>>
>>>The top 3 voltages ( +V , +9 and -9) in your drawing above, are those the
>>>"supplied" voltages??
>>
>> ---
>> Yes. If the + and -9V were batteries they'd would look like this:
>> (View in Courier) +V is just the voltage to operate the relay, and
>> if it were a 9V relay the + 9V could be used to energize it during
>> the times +9V was supposed to be the circuit's output.
>>
>>
>> +V
>> | +9V -9V
>> +--------+ | |
>> |A | O--> /<--O
>> [DIODE] [COIL] -|- - -/ |
>> | | | O |
>> +--------+ | | |
>> | | +-----|----> +/-V OUT
>> C |+ |
>> 4060 Q14>--[R]--B [BT1] [BT2]
>> E | |+
>> | | |
>> +----+----------+----> GND
>>
>> --
>> John Fields
>> Professional Circuit Designer
>
>
>Thanks for clearing that up John... I was taking it (what they was wanting)
>as using just one supplied voltage of +9v... the relays would then do the
>switching from neg. to pos. ...
>
>Either one would work I guess, yours is much simpler but requires 2 separate
>supplied voltages...

---
It depends on what he's going to do with the output signal. Your
way requires a floating load, (that is, the load can't be grounded)
while using two supplies lets it be grounded if necessary.

BTW, I went over to that page again (it was up this time) and it
struck me that by using a DPDT relay as the relay being switched by
the transistor and your circuitry to do the polarity reversal into
the load, both of the other relays could be eliminated, like this:



+9V
|
+----+-------------------------+
| | |
+--------+ | |
|A | O--> /<--O-+-O--> /<--O
[DIODE] [COIL] - - - -/- - - | - - - /
| | O | O
+--------+ | | |
| | | +--------->OUT
C | |
4060 Q14>--[R]--B +------------------------>OUT
E |
| |
+-----------------+
|
GND

:-)


--
John Fields
Professional Circuit Designer
.

Relevant Pages

  • Re: Remote Turn-On
    ... >the device type used for driving the relay? ... order to decrease its potential leakage current. ... John Fields ...
    (sci.electronics.design)
  • Re: relays
    ... John Larkin wrote: ... There's no limit on the power gain if you're patient ... The relay requires so many watts to activate. ... The total energy delivered to a load increases with time. ...
    (sci.electronics.design)
  • Re: 555 Timer Question
    ... >>>just the one relay?? ... >> John Fields ... >switching from neg. ... need diodes, if you're using another relay to switch them, then ...
    (sci.electronics.basics)
  • Re: T25 Headlights
    ... back into leather armchair) I suggest you trade the car in on a fucker that ... > John. ... >> work and I can still flash full beam, but no dipped lights! ... could it be switch or relay? ...
    (uk.rec.cars.vw.aircooled)
  • Re: relays
    ... John Larkin wrote: ... There's no limit on the power gain if you're patient ... The relay requires so many watts to activate. ... The ratio of these two powers is a constant. ...
    (sci.electronics.design)