i need a timer circuit that when it has no power applied to it a relay has a secondary circuit open, when power is applied to the timer circuit i want the relay to keep open until it reaches the preset amount of time and then i want the relay to close and stay closed . does anyone have an idea of how i can do this ??
thanks

 

Hi themikestar,

play with the component values,
till you're happy with the delay period.

Best of luck with it, John

Attached Images

Easy_delay_cct.jpg (7.0 KB, 492 views)

 

i can get the capacitor to gradually charge and output 10 volts, bu i cant get it to switch a voltage through the transistor, i get the same reading from the transistor that i get from the capacitor (gradually increases from 0 - 10 volts). What am i doing wrong ??

 

can you post a pic of your assembly ?

 

That sounds about right actually,
the relay should jump in at about 8 or 9 volts,
assuming its a 12 volt relay.

 

when i connect the relay to the transistor it draws so much power it drops the volts back down to 1v (and the relay wont close), so i guessing the transistor is doing nothing and that the realy is drawing power direct from the cap and the cap cant keep up. I know i have done something wrong with the transistor, what is it ??

the diagram of what i have at the moment is below

thanks

Attached Images

trans_timer.JPG (19.9 KB, 468 views)

 

Hi themikestar,

from the values you've put there, you are aiming at quite an
interval, maybe six seconds or so ?

That transistor is capable of one tenth of an amp, maybe your
relay needs a little more ?

If you're happy that the Tr should operate that relay ok,
then you will have to reduce the resistor until the relay
operates,

then increase the cap to the interval you want.

(if the cap becomes larger than you would like, you could
'darlington-up' with another transistor, then the resistor
and cap would be different values, smaller cap, and higher
resistor.)

John

 

Hi themikestar,

If the circuit is as you have shown,
then you have not done anything 'wrong' with the transistor.

Its use as an emitter follower might just be new to you,
as a normal amplifier the collector would not be connected
directly to the supply.

It is possible that this is why you feel something is wrong,
but its ok.

John

 

i will try it again with a smaller relay. I'm am not really familiar with transistors, what is the purpose of the transistor in this circuit, my understanding is that when the base voltage reaches the same as the collector voltage then power passes from the collector to the emitter, is that correct ??
thanks for your help john1

 

i have found that it runs an led (with a 470 res) perfectly and if i tap the wire from the relay coil onto the emitter leg the relay cuts in perfectly, but if i start the circuit with the relay connected i get nothing. i dont know what to try next, i cant find a smaller relay until tommorow

 

do you have a working multi-meter ?

 

yes i have a working multimeter. the circuit works fine when i remove the resistor and charge the cap directly, bu there is no time delay. i tried with a smal relay even a 5v one and it still didn't help.

 

came across this on the net, i tried this new circuit and it works perfectly, it also explains why the other circuit didn't work

" The following circuit uses two 1N4148 diodes to protect parallel port against higher than +5V signals and also against wrong polarity signals (power on the circuit is accidentally at wrong polarity.


Vcc
|
+------+
| __|__
Relay /^\ Diode 1N4002
Coil /---\
| |
+------+
|
Diode | /
1N4148 4.7K B |/ C
parallel >-|>|-+--\/\/\/--| NPN Transistor: BC547A or 2N2222A
port data | |\ E
pin +-|<|-+ | V
1N4148 | |
parallel >-----------+------+
port ground |
Ground


Adding even more safety idea: Repalce the 1N4148 diode connected to ground with 5.1V zener diode. That diode will then protect against overvoltage spikes and negative voltage at the same time.

Bad circuit example
I don't know WHY I see newbies who don't THINK electronics very well yet always putting the relay "AFTER" the transistor, as if that was something important. Well it's NOT, and in fact its a BAD PRACTICE if you want the parallel port to work well! This type of bad circuit designs have been posted to the usenet electronics newsgroups very often. The following circuit is example of this type of bad circuit design (do not try to build it):


Vcc
|
| /
4.7K B |/ C
parallel port---\/\/\/\/---| NPN Transistor: BC547A or 2N2222A
|\ E
| V
|
+------+
| __|__
Relay /^\ Diode 1N4002
Coil /---\
| |
+------+
|
Ground

NOTE: This is a bad design. Do not build or use this circuit.

The problem of the circuit is that the voltage which goes to the relay is always limited to less than 4.5V even if you use higher Vcc supply. The circuit acts like an emitter follower, which causes that the voltage on the emitter is always at value base voltage - base to emitter voltage (0.6..0.7V). This means that with maximum of 5.1V control voltage you will get maximum of 4.5V out no matter what is the supply voltage (when it higher than 5V and below transistor breakdown voltage).
Other problem is that in some cases this type of circuit can start to oscillate if the base resistor is right on the edge. "

 

stupids pics stuffed up

 

You could post the addresses instead.