Power-On delay circuit

Thread Starter

tzitzikas

Joined Jun 3, 2007
41
i am looking for a circuit which has the follwing oparation:

when i give voltage to this circuit (12 v dc), after a time (approximately 40sec-1min) this circuit
to give 12 volts dc to a relay (12v dc/ 230 volts). i want this time to be variable with a variable resistor.
the circuit i am looking for is a power-on delay circuit.
i have listen many ideas like cd4060, CD4541 lm555 but i have not a circuit. please help me and post a circuit here.
thanks
 

lightingman

Joined Apr 19, 2007
374
Hi.......I have these little circuits as a small module that provide a delay on my dimmers. They use a 4047 and a 4022.I have all the pcb lay-outs and schematics.I can over the next 24 hrs, make PDF's of them if you wish, and post them here......Daniel.
 

bloguetronica

Joined Apr 27, 2007
1,541
You can use a simpler design than one using a LM555. Use a LM324 as a comparator, with a RC network attached to the plus input, and a RR network attached to the minus input. Knowing that the mid charge time of a capacitor is ln(2) * R * C, and having the RR network equal resistors, you could use this circuit with a relay.
 

lightingman

Joined Apr 19, 2007
374
I HATE 555's !!!!!I Think this comes from my teaching days, when allmost every project had a 555 in it.Also with my little "DIGITAL" circuit and a some transistors, you could put countdown LED's on it.......Sorry you 555 lovers....... I may have lost some friends here (if I ever had any !!!!!).....Daniel.
 

bloguetronica

Joined Apr 27, 2007
1,541
cumesoftware can you post the schematic of the circuit that you described above?
These solutions can be applied to the LM324 or the LM358 without modifications.

Here are the shematics:

View attachment 955
Direct output from the LM358

View attachment 956
Output using a TIP32 transistor

First, to choose the solution you have to take into account what type of relay are you using. If your relay demands more than 40mA (or 20mA if you want safety/reliability), you have to resort to the second solution, since the LM358 (or LM324, the quad version), can source only up to 40mA (and is capable of sourcing more current than it can sink).

The second step is to calculate the values of C, R1, R2 and R3. You can use R2 = R3 and resort to the formula of the mid-charge time of a capacitor through a resistor. Thus, given the resistor R1 and the capacitor C, T = ln (2) x R x C. But if you want to work with R3/R2 ratios different than 1, you can consider the formula T = ln ((R2 + R3) / R2) (deduced by myself, a generalization of the formula for the mid-charge:) ). Notice that precision will not be guaranteed if R3/R2 >> 1. I would advice 2.3 as the limit.

If you use the second solution you can calculate R4 so it will conduct up to 1:10 of the current demanded by the relay. Taking into account that the Vbe of the transistor is 0.7V (typically...did not confirmed that):
R4 = (Vcc - 0.7) / (0.1 * Irelay)

Notes:
- Never use the capacitor at Vcc. The capacitor will discharge much more quicky, but the discharge will be done throught the input transistor of the LM324/LM358 where the capacitor is connected and through the load, and that may damage the op-amp. The capacitor should be at ground, as sugested.
- The diode D is to protect the op-amp or transistor (in the second case) from the reverse current surge generated by the relay. Use of a 1N4001 rectifier diode will be suficient to protect the circuit from this high voltage surge (quite noticeably in the case of power relays).
- In automotive applications, the voltage (Vcc) may reach 18V. Also take into account the minimum voltage that can be reached to calculate R4. The important think is that the current through R4 cannot be bigger than 1:10 of the current flowing through the relay. See if that condition is satisfied for both extremes. If not, the transistor will over-saturate.
 

hgmjr

Joined Jan 28, 2005
9,027
Greetings tzitzikas,

I'm not sure what lightingman has in mind but I thought I would throw something out there for you to consider. It doesn't use a 555, or a 4060, or a 4541. It just uses discrete components and a few semiconductors to do the job.

Even if you elect to go the integrated circuit route to implement your delay timer it may be useful to see how it can be done fairly simply.

Admittedly, this approach cannot produce the degree of accuracy that can be achieved using a crystal controlled oscillator surrounded by several ICs but it is always important to be aware that there are less complicated ways to perform a simple delay function.

Questions and comments welcomed.

hgmjr
 

Attachments

Ron H

Joined Apr 14, 2005
7,063
These solutions can be applied to the LM324 or the LM358 without modifications.

Here are the shematics:

View attachment 955
Direct output from the LM358

View attachment 956
Output using a TIP32 transistor

First, to choose the solution you have to take into account what type of relay are you using. If your relay demands more than 40mA (or 20mA if you want safety/reliability), you have to resort to the second solution, since the LM358 (or LM324, the quad version), can source only up to 40mA (and is capable of sourcing more current than it can sink).

The second step is to calculate the values of C, R1, R2 and R3. You can use R2 = R3 and resort to the formula of the mid-charge time of a capacitor through a resistor. Thus, given the resistor R1 and the capacitor C, T = ln (2) x R x C. But if you want to work with R3/R2 ratios different than 1, you can consider the formula T = ln ((R2 + R3) / R2) (deduced by myself, a generalization of the formula for the mid-charge:) ). Notice that precision will not be guaranteed if R3/R2 >> 1. I would advice 2.3 as the limit.

If you use the second solution you can calculate R4 so it will conduct up to 1:10 of the current demanded by the relay. Taking into account that the Vbe of the transistor is 0.7V (typically...did not confirmed that):
R4 = (Vcc - 0.7) / (0.1 * Irelay)

Notes:
- Never use the capacitor at Vcc. The capacitor will discharge much more quicky, but the discharge will be done throught the input transistor of the LM324/LM358 where the capacitor is connected and through the load, and that may damage the op-amp. The capacitor should be at ground, as sugested.
- The diode D is to protect the op-amp or transistor (in the second case) from the reverse current surge generated by the relay. Use of a 1N4001 rectifier diode will be suficient to protect the circuit from this high voltage surge (quite noticeably in the case of power relays).
- In automotive applications, the voltage (Vcc) may reach 18V. Also take into account the minimum voltage that can be reached to calculate R4. The important think is that the current through R4 cannot be bigger than 1:10 of the current flowing through the relay. See if that condition is satisfied for both extremes. If not, the transistor will over-saturate.
This is a nice, simple circuit for power-on delay. There are a couple of issues I have with the parts chosen and the discussion, though:

1. According to the datasheet, LM324/358 output will not go close enough to the positive rail to reliably turn off the PNP driver. I would use LM339 or LM393 (true comparators), with a pullup resistor, if you use the PNP driver stage. You might be able to add a pullup to the LM324/358, but I haven't tried it, and it doesn't work well in simulation.
2.
The important thing is that the current through R4 cannot be bigger than 1:10 of the current flowing through the relay. See if that condition is satisfied for both extremes. If not, the transistor will over-saturate.
This is not a problem except in very high speed switching circuits, although it is generally a good rule of thumb.
 

bloguetronica

Joined Apr 27, 2007
1,541
1. According to the datasheet, LM324/358 output will not go close enough to the positive rail to reliably turn off the PNP driver. I would use LM339 or LM393 (true comparators), with a pullup resistor, if you use the PNP driver stage. You might be able to add a pullup to the LM324/358, but I haven't tried it, and it doesn't work well in simulation.
although it is generally a good rule of thumb.
Yes, this is true. The LM324/LM358 will go only close to the positive rail. In that case, in the second circuit you would swap the + and - inputs, and use the op-amp to sink current. You could use a NPN transistor like TIP31.

Nevertheless, in the first situation it is apropriate to use as it is (sourcing current), since when the output is zero volts, is is really zero volts. Also, 10V (for example) from the output is enought to latch a 12V relay.

Resuming, the second situation would have the inputs being used as in the first. Good catch! Have to take notes on that! (Actually that same problem caugth my attention when projecting the LED strob, now in the projects collection, but that didn't constitued a problem for a supply of approx. 5V).
 

lightingman

Joined Apr 19, 2007
374
First of all, I am very sorry for the "delay" (excuse the pun) in posting this..

The timing is set using the pre-set.... you can play around with the timing capacitor (470nF) but do not use electrolytics....

On switch on the 4022 is reset at pin 15 by the capacitor (100nF) and the resistor (470K), the diode just dicharges the capacitor if the unit is switched off and on very quickly....

The 4022 counts to 8, and then is inhibited by the counter output holding pin 9 of the 4047 high.....

The LED flashes 8 times during the timing cycle...

Hope this helps....

Daniel.
 

Attachments

lightingman

Joined Apr 19, 2007
374
Just a note about the power on delay that I posted e few days ago.....You could substitute the 4022 (not pin for pin) to say a 4040, and by selecting the outputs, you could make longer delays.......Daniel.
 

Thread Starter

tzitzikas

Joined Jun 3, 2007
41
lightingman i would like to ask you a few quastions about the circuit that you have posted above.
the fet bs107 as 3 or 4 pins? can i use another fet like bs170. i have some bs170.
please give me the range of the time that i can adjust
thanks
 

lightingman

Joined Apr 19, 2007
374
Hi...Yes any small N channel MOSFET will do, or an NPN transistor with a base resistor of around 5 to 10K.....Range of timing:- depends on the speed of the oscillator (4047), the 4022 is a 1 of 8 counter. Change it for a 4017 (1 of 10 counter).Or a 4040 will give very large delays, as the 4040 is a binary counter, the delays will be double at each output I.E. 1, 2, 4, 8, 16, 32 clocks and so on....... 4040 reset = pin 11, clock = pin 10.....Daniel.
 
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