My question is regarding the 200K potentiometer BradtheRad put in his drawing,
Could you give me resistor and capacitor values to close the relay for about 3 second (plus or minus half a second). Or should I just figure it out with the pot?
I don’t understand. Are you talking about the input impedance on your oscilloscope?
Or are you saying I can build the circuit with no resistor as pictured?
View attachment 112246
Regarding the ULN2803, it has several problems.Regarding the relay it should be able to drive .5 amp at 5V on the contact side I think maybe I’ll use a small reed relay like HE721A0510.
And I think I’ll use the ULN2803 instead of the 4069 hex inverter because the former has more gates per package( eight as opposed to six).
Yes, it's CMOS. As long as it can sorce or sink sufficient current for the relay coil.Does using the 4049 or 4069 hex inverters avoid the Darlington problem?
Since the relay coil requires even more current than the 1.35 mA I discussed earlier, it will require a gargantuan capacitor to do this directly. While theoretically possible, it is not cost- effective.Technically you don't need a logic gate at all, simply use the switch to charge up a capacitor, and then discharge that capacitor through the relay's coil. When the capacitor discharges through the coil, its voltage will drop and when the current drops below the holding current of the relay it will disengage. So aside from the switch, only a single capacitor is needed.
Why not look at the datasheet for the CD4069 hex inverters IC? When it has a 5V supply its minimum output current is only about 1.5mA. It can drive a ULN2803A. The input resistance of the CD4069 is infinite so you would use a high value resistor to slowly charge or discharge a fairly small capacitor.Does using the 4049 or 4069 hex inverters avoid the Darlington problem?
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