# Timer for Relay - Referenced (Can a capacitor hold open a relay ? post from 2008)

#### AJKBajwa

Joined Feb 20, 2018
8
The moderator wendy in the referenced article made a circuit that looks like the one I have attached.
I wanted to know if I could use this circuit for the following problem.

When the voltage is supplied, the relay turns on and a capacitor starts charging.
After a certain period of time the capacitor will be fully charged and it should stop the relay.
Now if the power supply is removed the capacitor should discharge itself.

Can it be achieved?
Also tell me how to calculate the charging time.

#### Attachments

• 1.6 KB Views: 21

#### LesJones

Joined Jan 8, 2017
4,008
Add a resistor between Vcc and ground on your circuit to provide a path for the capacitor to discharge. It is also a good idea to add a diode in parallel with R1 to avoid exceeding the reverse vbe rating of the transistor. The cathode of the diode (Band end) to the base of the transistor. The value of the added resistor needs to be low enough to discharge the capacitor in the time the poweris removed.

Les.

#### AJKBajwa

Joined Feb 20, 2018
8
Is after adding the things you mentioned. Would it be possible to achieve the result I want?

#### Tonyr1084

Joined Sep 24, 2015
7,358
You want the relay to come on with the power. After "n" time you want the relay to shut down. Correct?

How long is "n"?

What voltage are we talking about?

How quickly do you want to discharge the capacitor once power is removed?

As for how to calculate the numbers, I'm sure I could figure it out. I've done it before, but I don't remember the formula. I'm sure it's something like "T=RC" or something like that. But I'm not sure I could describe it. Let alone be confident I've got the formula correct. That can actually be found in the education section of this forum. OR, I'm sure Wendy already covered that subject. Just do a little more reading and you'll probably figure these things out for yourself.

#### ian field

Joined Oct 27, 2012
6,536
The moderator wendy in the referenced article made a circuit that looks like the one I have attached.
I wanted to know if I could use this circuit for the following problem.

When the voltage is supplied, the relay turns on and a capacitor starts charging.
After a certain period of time the capacitor will be fully charged and it should stop the relay.
Now if the power supply is removed the capacitor should discharge itself.

Can it be achieved?
Also tell me how to calculate the charging time.
A 555 is easiest if only because the application note gives you a ready to use C/R timing formula.

#### ebp

Joined Feb 8, 2018
2,332
I would change the circuit a bit.
Connect the emitter of the transistor directly to ground.
Put a resistor in series with the base of the transistor.
You don't really need R1 if you add the diode as Les suggests.

As the circuit stands right now, unless R1 is a low resistance, most of the current through the capacitor will flow into the base of the transistor. If the resistance of R2 is low, the current will be quite high. If R2 isn't quite low resistance the voltage drop across it may be enough to prevent reliable pull-in of the relay.

If you connect the emitter to ground, you don't have the voltage drop across R1 to worry about, but now without a resistor in the base the current into the base at turn-on would be more or less unlimited. With the resistor in the base and no R1, the charging of the capacitor is entirely through the base-emitter junction, so you can accomplish a longer relay hold time for the same capacitance.

Calculating the time that the relay is pulled isn't easy because of unknown parameters, and you must expect a lot of variation from one unit to another. The coil voltage at which a relay will drop out is usually not well-specified for any particular relay type. The current gain of the transistor is likely to vary by a factor of at least 2 from one transistor to another of the same type (there are a few transistors with reasonably low gain variation actually specified), and that isn't including the effect of temperature variation on gain. If you know the drop-out voltage of the relay coil you can calculate the corresponding drop-out current. Divide the drop-out current by the current gain of the transistor to determine the base current for the transistor. Now you can use the standard formulas for capacitor charging to determine how long it takes from the time Vcc is applied until the current falls to that level.

If you need a reasonably precise time you'll need a more elaborate circuit, such as something based on a 555 timer.

Also be aware that this circuit will cause the relay contact to open more slowly than usual. This often is not any problem at all, but if you are switching sufficiently high voltage and current with the relay it may have implications for contact arcing.

#### sghioto

Joined Dec 31, 2017
4,221
If you add another transistor to make a darlington you can get a delay of several minutes as in the circuit below. As ebp said above charging time is difficult to calculate. Best method is to just measure.
SG

#### AJKBajwa

Joined Feb 20, 2018
8
@ian field
I cannot use 555 timer as the circuit requires to be made up of discrete components.

@Tonyr1084
I need it to turn off after 15-30 minutes.
and discharge in about 1-2 minutes.
The voltage level is 230 V.

#### AJKBajwa

Joined Feb 20, 2018
8
@ebp
Thanks a lot. I will try and implement it.
Be sure to tell me if you have any other suggestion for the circuit.

#### AJKBajwa

Joined Feb 20, 2018
8
@sghioto
how will the calculations for timing change for darlington pair then?

#### AJKBajwa

Joined Feb 20, 2018
8
Also to make it clear. I have calculated the drop out voltage, now I just need to apply Ohm's law and calculate the drop out current. Am I correct?

#### AJKBajwa

Joined Feb 20, 2018
8
This is the circuit I have made according to your instructions.

Please note: I have yet to change the values of the resistors and capacitor , after I calculate the drop out voltage.

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• 20.1 KB Views: 6

#### AJKBajwa

Joined Feb 20, 2018
8
Also, I would add a darlington pair if you suggest it as I need to operate high voltages of 230 V and need the charging time of capacitor to be between 15-30 mins.

#### sghioto

Joined Dec 31, 2017
4,221
Also, I would add a darlington pair if you suggest it as I need to operate high voltages of 230 V and need the charging time of capacitor to be between 15-30 mins.
230 volts! H...S... Batman are you serious? Or do you mean the relay will be switching 230 volts. What voltage level is the circuit using? 15 to 30 minutes is not practical with this design.
SG