Can someone please confirm if my thinking is correct here ?

I have a small battery powered 3 LED light which I want to convert to my 12V power supply.

The 3 batteries ( AAA size ) are 1.5V each in series, so 4.5V total.

The insides of the light contain an 11ohm resistor, and then the 3 LEDs ( quite bright, 5mm diameter, 8 mm height ) in parrallel.

I placed some card between the first battery positive terminal and the contact, then measures with an ampmeter and it is drawing 40 mA from the batteries.

So my thinking is that if I have a 12VDC supply, I should place a resistor to suit 4.5V 40mA, and power via the ( now removed ) battery terminals.

The page :
http://led.linear1.org/1led.wiz?VS=12;VF=4.5;ID=40
says I need a 220 ohm 1 W resistor.

Since the best I have available is 220 ohm 0,5W I think I have 2 options available :

option 1 :
As far as I can understand, resistors placed in parallel will halve the resistance, while in series will add the resistance.
So I split the Vcc line to 2 x 220 ohm ( 0,5W ) in parallel, effectively using 110 ohm resistance from each, but halving the energy loss factor, and then after each of the 2 resistors, a second 220 ohm in series, to add the rest of the required resistance, and manage the rest of the heat loss.

so :
              / 220 ---- 220 \
Vcc -------                     --------------- LEDs
              \ 220 ---- 220 /

option 2 :
I have a few L7808 linear voltage regs and small heatsinks in the hobby box, so reduce the power from 12V to 8V, and then use a 100 ohm ( 0,25W ) resistor to the LEDs.

I think that as I have all the components for the above 2 options, option 1 looks the better and simpler choice.  Or am I missing something ?

C109 - that's an SCR.  Are you suing that as a place keeper?  Or what did you intend there?
You have no base resistors.
Is the drawing intentionally short on details as a preliminary sketch-up?

Sorry, it should have been C106.

My thought was to have the transistor control the ground / low side of the motor, and the C106 gate, once activated, the high side. Once power is removed from the transistor, current will stop flowing and the C106 would close ( not certain of the correct term ).

There is no base resistor as the 3V would be supplied by the board that this is being 'added' on to. I really like your idea of using the FWB, but I was hoping not to have to order more components and use what I have in my stash.

Regards and sincere thanks for the input.

Use a full-wave bridge.
Put either of the FWB's AC terminals to either of the motor terminals.
Connect the FWB positive terminal to +Vsupply.
Connect the FWB negative terminal to Gnd.

if I think I understand your idea, it is using a conventional bridge rectifier, that usually converts AC to DC, in reverse operation, with the motor connected to the conventional AC input, and 9VDC supply on the conventional DC output side ?

I assume then that I would require 2 rectifiers ( 1 for each motor direction ) and that the diodes would protect the battery pack from any surges ?

Consider adding a diode from positive to negative right in the same connections on you motor. This will make the voltage flow in one direction and would prevent the electronics from getting damaged when the motor if off but still spining.

Am moving back to this project to start clearing my hobby workbench.

I understand the need for the diode, but how can it possibly be added to this circuit ?  The 'Signal A' / 'Signal B' options allow for the motor direction to be controlled by changing the polarity ( correct term ? ) of the power going to the motor. So the positive supply to the motor could be on either pin ?

The only possibility that I can imagine, is if the diode is placed at the source of the 9V, but that would be before the 2N3904 transistor and the C106 Thyristor, thereby placing them at risk ?

Feedback :

I changed the power for the strike lock to a wall wart ( 16VAC ) as discussed, with one wire direct to a striker pin, and the other wire from the wall wart to a relay 'com', and the relay NO to the other striker pin.  The relay is controlled by the Arduino ( well, via a transistor on the relay board, to be more accurate ).

Although the wall wart is only rated 1A  ( the striker was drawing 3.3A on 12VDC ) it is working perfectly.

The only difference is that, on DC there was only a slight 'click' of the coil when it was activated, the AC powered coil makes quite a buzzing noise.  Googling has told me that this is quite normal for an AC powered coil.

I then replaced the wall wart with a 9VAC 1A unit.  ( the striker was rated 8 - 14 V, so I was a little concerned about the 16VAC wall wart output ) and everything still working perfectly.

Thanks to all for the education.

Had the opportunity to put my new knowledge to use again almost immediately.  It struck me that, as the AC makes the striker coil buzz, why was the second gate not buzzing ? ( much heavier duty Cisa Lock ). The Cisa lock is drawing around 2A when activated.

Googled and found that the Cisa lock is designed for 12VAC, and they also make an optional extra bolt-on  'Booster' that can accept 12VDC, 24VDC and 24VAC. It also uses much less current than the 3A it says in the specs for the 12VAC, so I assume that it has some sort of capacitor that stores the charge until it is required. This Booster is said to work well with power from intercom systems, which is around 150mA.

So why is my Cisa lock drawing 2A from my 12VDC supply ?  Quite simple - installed by an 'expert' it didn't have the booster added, and the 12VDC connected directly to the 12VAC coil.

This would explain the lack of "AC on the coil" noise / buzz, and also why the current draw is so high.

To make matters worse, there is no diode over the coil contacts. I am guessing that this would have also added additional strain to my old power supply.

That's the last time any 'expert' is going to install anything like this at my house again !!  If he can't show me that he has a minimum of 200 posts on this forum, and evidence that he actually understand what he is talking about, he can't be trusted.  I am just dumb-struck by the ( non ) expertise of the local installation techies - and that's with the company who has the best reputation in town.

Tomorrow I will disconnect the power to that gate lock until I can source and install the correct booster.

Many Thanks Papa G

I really have learnt SO MUCH from this forum.   I am amazed at the number of regular replies by members whose level of knowledge I could only dream of reaching.

My project is moving forward and, strangely, I look forward to the next problem challenge

Can you post a link to the data sheet for this striker that you are using?

Unfortunately not. This is Africa.

All I can tell you is the specs we have, and 'Made in Spain'

Brand name is not relevant, as I can probably buy the exact same thing with 3 or more different names.

Thanks Papa G

I suppose that the "problem", or non-problem as you may be suggesting, is one of personal perception.

From my point of view, I have a striker rated 350mA for use on AC or DC, which I think should use around 350mA. The fact that my estimation of the size of power supply takes this into account, and I purchase the power supply, and it keeps failing because of the current being far in excess of the 'advertised' rating, becomes a problem for me - it's a matter of perception, you see.

Now that I think I understand the cause of the 'problem'  ( now re-named to 'challenge' to suit various possible perceptions ), I can change things around to suit - will be running the striker from the AC adaptor.

I think that I really have to sincerely Thank everyone that has contributed in any way to this post - I have again learnt so much from this. I really appreciate all the time that you have all taken to reply.

My last question would be :

The wall fitting AC adapter ( 230V to 16VAC 1A ) has 3 terminals. 2 would be the live / neutral. The third is connected directly to the earth pin that plugs into the wall.

Would it be recommended to connect spare wires in the cable to that earth terminal, and then at the gate location where the strike is, connect the same wires to the metal gate frame that the strike lock is inserted into ?

You likely chose to use DC for your strikes so that the system could be operated by battery backup.

correct

if you want to maintain the battery backup capability, one solution is to use a power supply that can supply enough current. As suggested in reply #16.

I think that this problem should not be 'solved' ( masked ) with this solution.  With the very low resistance on the coil, I think that it would continue to drain current in an ever increasing manner, until either the power supply peak is exceeded ( like happening now ) or the 2 second that is coded for the coil to be energized has elapsed.  Only difference with the old supply, was that it was capable of managing the short for the 2 seconds and didn't quite reach its peak.

If I am understanding this problem correctly, I think that the stress placed on the components in the old power supply by this action was a major contributor to its' demise.

You are almost exactly drawing 10X the rated current.  Sounds to me like a decimal error in the reading.

I doubt this is the case - we actually tested with 2 different amp meters and both gave almost identical readings.

In addition, if it really was drawing 1/10th of the 3.3A on the meter, it wouldn't be causing a system power supply reset at the 5A peak.

OK. Update time.

Took the lock down to the auto electrician.

Connected the striker directly to a 12V battery, and energised as it should.

Also tried on a stand alone standard 12VDC alarm backup battery ( 7Ah ) and also operated fine.

Added an amp meter on the positive line in series, and registered 3.3A. So it looks like my home power supply and home system wiring is OK, and the hall effect sensor is registering the right current range that is actually being used. ( I was starting to wonder if I had some other serious error in my complete setup )

Went back to the security supply store. Chap there says that they usually connect these strikers to a wall plug-in AC adapter - input 230 VAC, output 16 VAC, 16 VA, 1A. He had a box full of these and says they never had any problems using them. He can't understand why it is drawing 3.3A. The rating is 350mA, 8-14 Vac / dc.

To solve the problem of not having a gate lock : if I want to connect the striker to one of these AC adapters, would it be safe to connect 1 of the AC lines directly and permanently to the striker ( like I currently have the 12VDC Ground connected ) and pass the other line through my relay ( com to no pins, controlled by the Arduino ) ?

Does it matter which line I use for the permanent connected line ?

Should the striker still have a diode over the contacts, or is that not applicable if I am using an AC supply ?

Is it possible that a coil that draws 3.3 A on a 12VDC circuit, would draw less than 1A on 16VAC ?


ps .. just found this statement on google :
DC coils need enough resistance in the windings to limit the current, and AC ones limit it with inductance.

So does this mean that the coil ( most likely designed for AC usage ), if used with AC would draw the rated 350mA, but has little resistance, so if used on a DC circuit, would be the same as a 'short' ( as suggested by the low 12 ohm reading between Ground and the positive side of the coil when in the off position ) and would rely on the DC power supply having a high enough peak rated to be able to 'over-power' the short for the duration that the coil is energised.  If this is the case, wouldn't a resistor in series on the 12VDC supply to the coil limit the current, thereby preventing the total and system fatal short ?  I think that if this is the case, the supplier should clearly indicate that although the coil is rated for AC and DC, the use of DC does require additional circuitry and / or a much more powerful ( and hardy ) power supply.

Just been looking at the board, and measured between Ground and the relay's NO pin, and it shows 17 ohms.  But if the disconnected coil was at 24 ohms by itself -- shouldn't the reading be higher -- 24 ohm coil + 100ft cable ?

Is this evidence of a short waiting to happen when the relay connects the 12V on com to the NO pin ?

or is this caused by the diode on the coil contacts ?


ps .. re-measured with multimeter black lead on Ground = 12.1 ohms, and with red lead on Ground = 19.4 ohms. I assume this indicates that the diode is in place ?

This was a comment I made early in the discussion, clearly wrong as written. After having read all that follows. First a fuse unless otherwise marked/made as being a fast acting fuse requires about 10 seconds to blow.. at 200% of rated current. Fuses are solely for the prevention of fire.. So the problem with the strike coil seems to point to a defective coil... IMO something within the coil moves when activated (I used to use solenoids for all my work, they controlled the irrigation valves that I was making controllers for). Occasionally one wouldn't work except from a battery and grew noticeably warm after 30 seconds or so of being connected to a hard 12V supply (A fresh 12V 7A SLA battery) and these solenoids usually had shorted turns inside the coil and they would move when the coil was activated drawing more current than a static test with an ohm-meter would indicate possible.
In other words a defective striker coil that fails when activated. I might add that all my solenoids were activated by dumping a charged 4700 uF cap into the coil with a relay for polarity and a Mosfet for the switch that grounded one end of the coil.

I wonder if there isn't a back emf diode in each striker solenoid and that they are reversed or damaged. An inductor can be expected to draw a heavy current for several hundred uS but not continually as Op's explanation would seem to indicate. OTOH a reversed diodeor damaged would do just exactly that and with any appreciable length of wire attached could well draw the currents mentioned by the OP because if the resistance of the wiring connecting striker coil to the controller... A quick test would be to clip one end of the diode and see if it changes and proof would be to measure the current drawn by the coil directly from the battery.

Doc

Thanks for the reply doc.

And for the info about the fuse.

My original suspicion was a faulty coil ( in the strike lock ) so I changed the complete unit with a new one - same result.

This did give me an old lock to strip down, and all I found was the 2 contacts connected to a small coil, with a metal piston through the center, which did the unlock action.  There are no other components what-so-ever in the lock unit or coil.

As far as time is concerned, the lock / coil is energized for 2 seconds ( in the code ) and resets after at least a full second has passed.

I have removed the lock and will be visiting the auto electrician in the morning.

You are drawing 4+ amps and NOT blowing the fuse on the power board even though that is a 25% over current?

That's what is so damn confusing. The sensor says 5A, then the power supply resets. And the strike is powered after the fuse and diode.

The power supply is a 5A Peak Current rating, so understandable that it resets at 5A.

OK. Here's the diagram. Apologies if it is a bit crude - I have tried to keep it as simple as possible.

I have drawn the positive lines in red, and ground lines in blue.  Added  A, B, C .. in purple just in case anyone wants to reference them.