thank you for this support service…it’s very useful!!
here my problem:
I need to drive a 12V DC automotive fan motor and the Timer mk 111 is one of the right choises I can make…
I want to know if its possible in your opinion to change the relay given in the kit with a 12v primary coil automotive one capable of higher loads than 3 amps… my motor need 3,6 amps unloaded…
I know I could drive another relay higher load capable with the timer as given… but I need to have small dimensions and the smallest number of pieces…
The relay will handle 3.6A, the problem is with the PCB tracks.
You might want to solder some extra wires between the relay connections and the screw connector.
Yes, this should be possible the way you propose it but make sure you have very good contact on the wires of the screw connector or you risk burnt connectors/cable caused by heat caused by bad contact in the screw connector.
Theoretically, you can drive an automotive relay of about 120 mA (as the used IC, 555 can source 200 mA) directly but I prefer to do such things always with a simple transistor buffer.
but I’m not so skilled and I have a pair of questions
I know how it works… but do you mean that I can use a transistor to protect the IC??
if possible, where do I have to put the transistor in order to obtain the desired effect?? and what kind of transistor??
The free end of the resistor goes to the output pin of the IC, take a 2k2 for the resistor instead of the indicated 1k. You can use a BC637/639 if that is easier to find (in Europe it is). And put a 1N4007 diode OVER the relay coil with the colored band to the positive side of the relay coil (antiparallel). This protects the transistor against inductive voltage peaks when the relay is switched off.
No, normally this won’t give any problem. The MK111 is built so straight forward that it does all you want ;-). Through its simplicity it is also very reliable.
assembled the original circuit on a breadboard and it works…I Only noticed that when the power is given for the first time is remains on for 9 sec while the following on modes it stills on for 6 sec…
then I tested the transistor mod… 2k2 resistor on a BC639 and a 1n4007 diode…
the circuit does not work correctly…when the power arrives for first time the circuit starts and the relay coil is switched on…when the circuit goes in pause mode the relay coil remains switched on…
I measured the tensions on the coil sides: led on 12v… led off 6,7 v… this tension does not allow the coil to switch off…
resistor connected on the original spot for the + side of relay coil… transistor base connected to resistor, emitter to ground and collector to negative coil side… diode between collector and positive side of the coil …
help needed if possible… is it normal that +6,7 v when circuit is switched off???
No this is not normal: a 555 gives almost 0 volts when its output is low. So on the output of the IC you also measure 6.7 Volts when not active? This is pin 3. This should be close to zero as the output of the 555 contains transistors for pulling to + and for pulling to 0.
Great! Was a possible cause I did not think about. Good you have found it!
No, changing the capacitor should cause not any problem in this circuit.
It is only C3 that influences the pulse and pause times. C2 is just to stabilize the power supply
This is indeed possible now you mention it: The 555 works with internal divider that divides the total voltage in thirds for its comparators. So internally the voltage on the capacitor C3 (the one for the timing) is compared to this “thirds” to decide what should happen.
Output (3) is high as long as voltage on C3 is lower than 2/3 of the supply voltage. If voltage on C3 increases above that, the output will go low and capacitor will be discharged through pin 7. Discharging (and low output) continues until voltage on C3 goes below 1/3 of the supply voltage. If that happens: output goes high again and capacitor is charged again to 2/3 of the voltage and so on …
As you see: during normal cycles: the capacitor is charged and discharged between 1/3 and 2/3 of the supply voltage. But the first time power is applied, capacitor does not charge from 1/3 of supply voltage to 2/3 but charges from 0 volts to 2/3. That is 1/3 extra that has to be charged the first time and that takes ROUGHLY double time.
if I change RV1 with a fixed resistor (for example 0,5M ) connecting the two tracks that are not common under RV1 place on pcb I will obtain a fixed pause rate…am I right??
or can I joint the tracks under RV1 and increase R2 value???
