Need to Sense DC current draw. 3

[skiier]

Unusual question here.
I want to sense when a car lighter is on. A car lighter is a self contained simple unit. When you press the lighter "in" you "load" a spring while engaging the contact points of the lighter resistance element to 12VDC and in turn this causes the lighter element to heat up. This heat causes expansion and the unit moves to a point where the loaded spring releases and pops the lighter out. Easy enough right?
There is no way to sense a voltage change at the lighter as the lighter is self contained. The +12V and -12V leads to the lighter never change state.
I also think that if you placed anything in series with the lighter - a resistance of some sort say to create a voltage divider - you might reduce the voltage to the lighter to such a magnitude that the lighter will not function properly mechanically. Power developed in the lighter element being the inverse square of voltage. Hence a 30% or 3.6V reduction in voltage to the lighter element means 1/2 the original power to the lighter element.
Also keep in mind the lighter resistance is about 1.0 ohm. Any sensing element placed in series would have to be much smaller than 1.0 ohm for the lighter to function properly.
So basically I don't want to mess with the lighter properties I just want to sense current flow.
So what I was thinking was using a power transistor as a switch. The transistor will reduce the voltage at the lighter slightly but I am hoping to use the base of the transistor as a switch while offering the lighter a low resistance path thru the collector in this application. If you think of the collector driving the base instead of the base driving the collector you will understand what it is I am doing.
I connect the 2 leads of my lighter to the emitter of my transistor cct. and to GND (-12V).I connect the collector directly to +12V. I know the original current draw of the lighter is 8A@12V so I can calculate the lighter resistance R=12/8= 1.5ohms.
Now I assume a certain votage drop at the lighter that I hope will allow the lighter to work properly. I need this voltage to overcome the base voltage drop and allow me some level of voltage in the base cct that I can use to drive some sort of relay. Let's say my lighter will work properly at 10V. So my emitter current is now 10/1.5=6.66A. If my transistor had a gain of 100 then my base current would be about 6.66/100=6.6mA. A resistance in the base cct would have a voltage drop of 12-0.7-10=1.3V.
So what I need is a relay that operates at or around 1.3V and draws close to 6mA with an impendance of 1.3/0.006=216ohms. Any thing commercially available?
Yea right! This is where I become stumped. Is there an easier way to do this. I want simple thank you. Just simple electronic components. No complex ccts needed or required thanks. You get the idea.
Any way to make the base drive something else?
Any help would be appreciated. I do not want to modify the lighter in ANY WAY. I mean any way. The lighter has to operate as normal. A replacement lighter must also work in the same receptacle.

 

[buzzp]

There are many devices that measure DC current that look alot like a standard current transformer torroid. The conductor simply passes through the CT window. Try

http://www.crmagnetics.com

,

http://www.rbeelectronics.com

,

http://www.sypris.com

, NK technologies or for more search for DC current sensor on google. All will have options for setting a trip level and will have a solid state or relay output to trigger when you reach the current level where you want to take action (lighter being used). Very easy to hook up. Good luck.

 

[nbucska]

Modify the lighter : add an LED in series with a 1K resistor.


<nbucska@pcperipherals.com>

 

[logbook]

Nah, all too complicated.

You have 12 amps of current?
Wrap two turns of the lead wires around a reed switch, as used in reed-relays. These will operate with 10 to 20 Ampere-turns.

Job done

 

[dhwilliams]

Disconnect the negative supply wire from the lighter, reconnect via a 0.1 ohm 10W resistor. Connect an NPN transistor's emitter to the negative supply wire and connect the base via a 1K resistor to the other end of the 0.1 ohm resistor. Connect a 12V relay coil from the positive to the collector and fit a diode across the coil with its cathode to the positive end.
When the lighter is on, 8A will flow through it and the 0.1 ohm resistor, there will be a voltage drop of 0.8A across the resistor which will turn the transistor on and the relay will operate. Just make sure the transistor can handle the current through the relay coil.

Dave

 

[skiier]

Thanks Buzzp but I did say simple. I have looked at these components. I think I should be able to make this work with junk from cct boards in my shop. I hopefully don't have to buy anything.
Thanks Logbook. But wouldn't the reed switch coil have to pass the lighter current thru it? This is a DC cct and I don't understand how wrapping wires is of any consequence. Can you point me to a link. What kind of internal resistance in the coil are we talking about?
Would have to be 0.1ohm or less. Is this possible?

 

[dhwilliams]

logbook,

You must have been writing your reply while I was writing mine - I should have remembered the reed relay trick! (I'm old enough to have used them...)

However, skiier said that the current was 8A so he may need more turns and he should only use one of the lead wires - unless he coils them in opposite directions.

Dave

 

[xnuke]

skiier,

I think the point of the reed switch idea is that it uses the magnetic field set up by the DC current flowing to the lighter through the wraps you put around it. You're essentially making an electromagnet with a coil of wire, and putting the reed switch (which is activated magnetically) inside it. There is no "reed switch coil" for current to flow through. Connect the two leads coming out of the switch to an indicator of some sort, and you're off and running.

xnuke

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[skiier]

I see. You make the lead wire the "coil" in a reed relay by wrapping the lead wire around the "armature" of the relay.
How simple. Who would a thunk it?
Thanks. This solution is too easy to believe. I will find some reed relays and try it out.
I did rethink my cct as well. I decided I could use a 2V power zener in series with the lighter. The zener uses less power than a resistor. I don't have to worry so much about warm components. I use the 2V to bias the base on a transistor. I found some readily available 12V relays that only use about 1W@12VDC. The transistor drives the relay.
I guess I just needed to review my electronics.
DHwilliams' idea of the 0.1OHM resistor led me to think of the zener. Thanks

 

[logbook]

Thanks to Dave and xnuke for clarifying my posting. I of course meant to only wrap one of the wires around the reed switch.

Skiier: to try this you are better to use a bare "reed switch" rather than a "reed relay". Let me clarify. A reed switch is the glass envelope containing the switch contacts. A reed relay is a package containing one or more reed switches with an activating coil. The reed switch contacts will only be able to handle around 30VA to 50VA load current though.

If you can’t easily find a bare reed switch you could just buy a reed relay and wrap four turns of one of your lighter power leads around it. There will be plenty of magnetic field to activate the reed.

As Dave noted this is an old automotive trick. I recall someone using it to sense if one of their tail lights was blown. The power leads to each of the tail lights were wrapped around the reed in opposite directions. If one bulb was blown and the other was on, there would be a full strength field, activating the reed. If neither bulb was on, or if both bulbs were on, the field would be zero and the switch would be off.

 

[ScottyUK]

Another possibility is to use a diff. amp or a high-gain comparator with one input connected to the main fuseboard and one at the lighter itself, using the volt-drop in the panel wiring itself as a shunt. Do you need to measure the current, or just need an on/off indication?



-----------------------------------

Start each new day with a smile.

Get it over with.

 

[OperaHouse]

Just to add to the fray, this is an easy way to do it without breaking the wire. Get a flat steel washer and cut a slice in it to the center. Slot should be wide enough to accept a hall effect sensor. Wire goes through center of washer to create magnetic field They make these devices linear and on/off. Check your junk box for an old floppy drive, should be one mounted around the flywheel motor. I still have a box of them but those reed switches are becoming harder to find.

 

[skiier]

Thanks ScottyUK. Great idea. I thought about using a current shunt but decided against it because of cost. An OP AMP comparator is a great idea as current shunt. So cheap and no interconnection with the load.
Operahouse thanks as well. I should have thought of a Hall Effect. Thanks for pointing it out.
Lots of ideas. I'll dig thru my box of goodies again and see what I can come up with. Thanks to all.
I'll let you know what I use.

 

[busbar]

If you don't need galvanic isolation, a shunt made of 1/16-1/8" 308-alloy TIG fill rod can work well. 0.025 ohms = 0.25V/10A. This could trigger the comparator.

 

[jghrist]

Quit smoking and throw away the lighter. ;-)

 

[panelman]

reed switches (in little, but easily removable cases) are available for pennies from your local burglar alarm shop and are by far the easiest way to do what you want

 

[skiier]

jghrist. I live in Canada. You can't buy a Honda here with a lighter or an ashtray. I have a Honda. Thanks for your concern about the cigs though.
I do have a friend who has a lighter in his car and he wishes to install an "Anti-theft" cct of some sort. To simply slow down or discourage a would be thief. I suggested he use the momentary operation of the lighter to interlock his starter.
There are 1 problems with this idea though. If you own a standard you can still push your car to start it.

 

[panelman]

Skiier

A cheap and easy anti-theft device is a 12V flashing led wired across the ign switch, it does nothing to stop people stealing it but it makes it look like it has an alarm and so encourages them to go steal someone else’s car.

Wire the led in parallel with the ign position on the switch and mount the led somewhere conspicuous on the dashboard

 

[dhwilliams]

Some years ago I fitted an anti-theft device that worked as follows:

When I turned the ignition on I then operated a hidden momentary action switch. This operated a double-pole relay which latched using one set of contacts. The other contacts were used to enable the 12V ignition coil circuit. When the ignition key was turned off the relay de-energised and disabled the ignition (so you can't forget to use it!).

You would have to check the cars ignition wiring diagram to find a suitable place to enable/disable it. If it's a diesel you could wire it up to the fuel cut-off solenoid valve. Take care when building and installing something like this, you don't want your engine to cut-out while you are overtaking!

This method would stop someone push-starting the vehicle.

I used a reed switch operated by a magnet glued into a rotating ashtray to operate mine but I had to remember to close the ashtray each time.

Keep It Simple!

Dave

 

[circuit1952]

The March 2004 edition of Circuit Cellar magazine had the article FROM THE BENCH Intelligent Current Sensing: Harness the Power of the ACS750 Hall Effect Current Sensor . The Allegro ACS750 looks like a neat sensor.
I recall a project in the late 60's in a magazine like Popular Science in which a silicon diode was placed in series with a DC load (headlight?) and the base-emitter junction of a germamium transistor was placed in parallel with the diode.

|
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|_______
| |
| |
| | / R
| | /
| | /
| | |/
_|_ |_______|
D \ / | Q
---- | | |


Silicon diodes have a forward voltage drop of about 0.6 v, while germanium diodes have a forward drop of about 0.2 volt. The transistor can switch a relay for higher cuurent loads.