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RC Series Circuit
2

RC Series Circuit

RC Series Circuit

(OP)
Now this should be elementary, but I have an operational question regarding a series RC circuit.  I have a small 12V Potter/Brumfield relay that I would like to make slow release by about 5 seconds.  I figured I could use a simple RC circuit across the coils.  I chose a 1Mb resistor and a 5uF capacitor (electrolytic), and am using a simple bench power supply for the 12 volts.  However, when I denergized the coil it immediately drops out.  I replaced the cap with a 10uF and a 100uF with no difference.

Am I missing something here?  Is it because I am using an electrolytic?

————————————————————————————————
James W. Hudgins, III
Union Switch & Signal
 

RE: RC Series Circuit

Analyze it as an LRC circuit and see where you get.

RE: RC Series Circuit

There isn't, perhaps, a built-in freewheeling diode, is there?

Ed

RE: RC Series Circuit

assume 200 mA holding current

0.2A*5s/12V = 83 mF, that's MILLI, not MICRO.  And since you can't actually allow the voltage to drop to zero since there's resistance in the relay, you actually need something more like 830 mF, so a 1F capacitor might do the job pretty well.  

TTFN

FAQ731-376: Eng-Tips.com Forum Policies

RE: RC Series Circuit

You should use a RC network, but not the reason you thought.  The resistor should be about 15 ohms to limit charging current to the capacitor.  This resistor protects switches or other circuity from damage due to initial turn on surge.  Figure the resistance of the relay  for the RC time constant. Relay should drop out at about 40% of voltage.

RE: RC Series Circuit

Do you also want it to engage slowly?

RE: RC Series Circuit

(OP)
Ok, here is some more information...

Relay: http://relays.tycoelectronics.com/datasheets/KHA_DS.pdf
Potter/Brumfield KHAU-17D12-12
12Vdc, 160ohm, 0.9W
(no diodes)
_____________________________

Addressing IRStuff:

Question...Since "holding current" is not listed in the spec. sheet, is it safe to assume that the holding current can be estimated at 20-50% of pick-up current?  For now I will assume 50% for now.

Calculating pick-up current at 75mA with an estimated holding current at 38ma (~50%).  Therefore, using your formula for capacitance (0.038A*5s/12V = 15 mF)
_____________________________

Addressing OperaHouse:

Ok, so using my 160ohm coil as my R value and a desired time constant of 5 seconds, I get the following result: 5s/160ohm = 31.25mF (not micro).  So using three 10000uF caps in parallel in conjunction with my 15ohm resistor would get it close.  Am I following you right?
_____________________________

Keep in mind that I don't need it to slop pick, just slow drop.  I really appreciate all the feedback.  I will play with this tonight and post all results.

-j

RE: RC Series Circuit

Everyone wants a two lead solution but this is very bad form. So bad, I will not even answer your question.

If you have power available use a FET or a 339 op amp to drive the relay and then you can get the performance you want without a monster cap.   

RE: RC Series Circuit

(OP)
I understand that the use of an RC circuit probably is not the best solution, but I was trying to use parts that I have laying around versus something more elaborate.  I usually follow the rule of perfection being attained not by nothing more to add, but nothing more to take away.  However, if a simple solution is not feasible, then so be it.  I am willing to entertain a better mouse trap, and do have a few op-amps so I will see what I can come up with.

Thanks again for your input.

RE: RC Series Circuit

I found a similar relay and ran a simple test. See attached recording with comments.

See macgyvers2000 (Electrical) first response. The 1 megohm resistor is the same as a circuit interruption. Use a much smaller current limiting resistor in series with the capacitor. A few ohms are adequate for 12 V coils.

Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

RE: RC Series Circuit

This question is very similar to a previous question in another forum: "How can I add soft on/off functionality in simple lamp"

http://www.eng-tips.com/viewthread.cfm?qid=261642&page=1

It's worth pointing out the obvious:

If the power-off transition completely removes all power sources from the circuit, then an active circuit is going to do nothing in terms of holding the coil on for an additional five seconds in the absence of power.

If you want to coil to remain on for five seconds, then you need to supply current to the coil for those five seconds. A huge value capacitor, a battery, an alternate power supply, a flywheel motor-generator, a trained mouse...
 

RE: RC Series Circuit

I forgot to mention that you will need a capacitor in the 22 - 33 000 microfarad range. Use a 16 or 25 V capacitor. Not 'huge' in any respect. Cheap, too.

Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

RE: RC Series Circuit

You'll actually need something about 6 times larger.  The current is being generated by the voltage on the capacitor, so you'll need to keep the voltage droop from eating into your holding current.  

My example was simply to show that your initial capacitor choices were off by several orders of magnitude.

TTFN

FAQ731-376: Eng-Tips.com Forum Policies

RE: RC Series Circuit

IR did you see my IRL  ( smile ) test? 22 000 - 33 000 uF does the job.

Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

RE: RC Series Circuit

'Huge' is relative. winky smile

I wonder what would happen if the drop out voltage of a particular relay was higher than 2.34 volts?

 

RE: RC Series Circuit

You will get shorter delay, you knew that.

Typical drop out voltages for relays with DC coils is in the 10 - 25 % of rated voltage. So any precision cannot be had with a simple parallel capacitor. Also, the tolerance of an electrolytic capacitor can be from -50 % to +100 %. If all possible variations like drop out voltage, capacitance tolerance and vibration are taken into account, you will see variations in drop out time that are really 'huge' smile

Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

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