Jump to content
Electronics-Lab.com Community

Supa Dupa Basic Timed Latched Relay


Recommended Posts

Hi All,
Just want to say im really new here. My question is that I need to make a 12volt, 40amp auto relay latch on for about 10 seconds. I however only want to use a capacitor and a resistor to do it. Can this be done?? If so what values should I use and where should I put the capacitor and resistor??? I seen basic capacitor articles where it just shows a capacitor can make a relay stay on for a time, I just need to know if it can be done. I do not want to put transistors in the circuit. All comments and answers are highly appreciated. Thank you

Link to comment
Share on other sites

Without a transistor, your circuit won't be very supa dupa.
You will need about 330,000uF across the 90 ohm relay coil. Thats 330, 1000uF capacitors, all in parallel. Or use a big, expensive cap that is used for high-power car audio. :(

With a transistor, a resistor and a diode, you need only one cheap, little 3300uF cap. A little bit of technology makes a big difference. ;D

I'll let Ante figure out a few transistors without even using a relay!

Link to comment
Share on other sites

Laet, audioguru,

In fact there is a way without semiconductors involved! My first question was to establish if a cap over coil solution was right for this. If there is high current it will arc when the contacts open slowly as they do when the cap discharges. Maybe not to the eye but slow enough to burn the relay. My suggestion is to avoid semiconductors, find a small relay with high resistance coil and let the small one control the big one. This way the high amp relay will open at normal speed and no unnecessary arcing will occur. And no enormous cap is necessary. ;D

Ante ::)

Link to comment
Share on other sites

A diode across a relay coil does slow its release, because it causes the current to continue during the collapse of the magnetic field.

Since the enormous cap would be in series with the small switch contacts, without any current-limiting, then the charging current will be huge, more than enough to fry those little contacts.

Link to comment
Share on other sites

Hey Guys, Thanks heaps for your replies and interest here :) Okay, Ill try to be more specific. - I have a few ther questions, but I think Its first best to outline the application.
I need to hook up daytime running lights on a car (DRL's). However, they need to be automatically switched on once the car is started and once the alternator is charging the battery. As the application is for a fleet, just hooking the headlight relay to ign is no good, because the batteries will die as these cars are getting started on and off all day.
So, what i need really is a 12volt 40amp mini auto relay to switch the low beam on only when the alternator is charging. Now I dont want to go off the alternator dynamo because its really hard to get to the alternator on these things. Second, I only want to use realy simple stuff if i can get away with it and I mean basic. If I have to use a transistor I will but not too many. So breaking down the problem again, i need the realay to close when i get about 13volts. Oh yeah the resistance of the auto relay is about 6.7ohms between 86 and 85. I was thinking along the lines of a voltage regulator on the negative side of the relay, also thinkg about a zener put on the postive side in series with the coil. Also maybe a voltage divider with some resistors, but I m not sure how to apply these.

Anyway, have a think and let me know, id really appreciate it. It might be to hard without transistors etc, but let me know, thanks

Link to comment
Share on other sites

Hi all,
Thanks for the info MP, very good. Well i decided to set up a little project in multisim to try and work my problem out. A pdf of my circuit is attatched. Now your eyes are probably going to pop out of your head and i bet you all think im mad, but im only learning ::)

I have refined my relay values a bit so they make better sense, the coil resistance should be about 75ohms and the realy coil should draw 0.160 amps at 12 volts. Relay switches on at about 8volts.

Okay, so you see how i set it up, its working but I think I need to change values or do something. I used the BD139 because its a high output transistor, I somhow thought this might be good for driving the relay, but was not sure. Now I did not really do much math here just played in multisim untill i got it half right.

So my aim: To make the relay switch on at about 13+ volts, but not 12-12.9 volts, or there abouts (ie at least when alternator charging).

So if this circuit is a super pooper let me know how I can improve it, should i use a different transistor, should I have a stabalisation resistor somewhere, and should i have a set up that determines when the transistor turns on and off better. The simulation in multisim is really fine with the resistor value, it will work sometimes and not others, althoug im not getting an error message.

You comments and thoughts, most helpful
Laet :)


Link to comment
Share on other sites

Your circuit will not work with the relay at the emitter of the transistor because the circuit is like a see-saw: As the current in the relay coil (transistor's emitter current) increases, then the current in the 25K resistor (transistor's base current) decreases:
1) The moment that you close the switch, the 25K resistor will have about 11.8V across it (12.5 minus 0.7 Vbe), so the transistor's base current is 0.47mA.
2) Using a BD139 with a minimum hFE of 63, its emitter current will be only about 30mA, so the voltage across the 75 ohm relay will try to rise to only 2.25V.
3) Even if your BD139 has its maximum hFE of 250, then the voltage across the relay will try to rise to 8.9V, but cannot because the voltage across the 25K resistor will be only 2.9V and the transistor's base current will be only 0.12mA, so the relay current will be only 30mA, which again is only 2.25V across the relay coil. But then the base current will be higher, so the voltage across the relay coil will end up to be somewhere between only 2.25V and 8.9V, but much less if the transistor has less hFE.
4) To get more base current you could choose a much lower value for the resistor, such as only 1000 ohms. If the BD139 has its minimum hFE, then the voltage across the relay coil will be about 9.7V, and higher if the transistor has more hFE.

This change of the base current and range of hFE is why Ante suggests having the relay at the collector of the transistor.

Link to comment
Share on other sites

  • 4 weeks later...

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.


  • Create New...