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By lutronjim
#172290
I want to control a warning light from a one wire alternator. I have a 12 vdc relay that I want to pick when the voltage goes from static ( about 12.6 volts or lower) to high (13.5 or more). The light would be wired from the battery+ and the relay would ground the circuit form the light to be on when the battery voltage is low. When the voltage goes high, the zener would pick the relay causing the light to go out.

I got a cheap 13v zener diode and hooked to the battery in series with a dvom. In one direction, I get a very low voltage(.1 or .2) in the other I see the 12.6.

I suspect that the zener is not accurate enough for my needs.

Does Sparkfun have a more accurate zener?

Thanks from a newbie.
By Mee_n_Mac
#172291
lutronjim wrote:I got a cheap 13v zener diode and hooked to the battery in series with a dvom.
What ? Can you draw a simple schematic of what you're trying to do.
User avatar
By Ross Robotics
#172292
He wants to turn a light on when a car voltage goes above 13.5V. I would think he needs a circuit to sense the voltage.

http://lmgtfy.com/?q=voltage+sense+circuit
By lyndon
#172293
I saw your response to my question in the other forum. You are misunderstanding how a zener diode works. What you're seeing is probably due to the leakage current through the diode.
Image

The schematic above should do what you want (no idea where all the extra space is coming from). Replace the zener diode with the one of your choice and the LED should turn on just around it's breakdown voltage. If you need better accuracy, then a comparator circuit would be preferable.
By lyndon
#172294
Sorry. Replying to my own post: I don't know what your electronics level is. Can you read schematics? Also, you asked about detecting a high voltage, but your post title says "High Temp."
By lutronjim
#172313
I can read diagrams somewhat.

I am sure I don't understand zeners. I thought they were non conductive in one direction like a normal diode but would conduct in the other direction when the voltage went beyond their rated level.

Background - 70 chev truck had warning light when alternator wasn't charging battery. A light got power form ignition and 'saw' ground through alternator field. When alternator staring charging, the voltage equalized across bulb and it went out.

Current - installing a LC9 engine and transmission. The new engine has a one wire alternator to battery and no way to 'drive' bulb.

Being newbie, I don't now how to insert diagram but have attached one.

The battery (a) has a hot to ignition switch (b). When the switch is on, it connects to existing bulb (e) through nc contacts of a relay (d)and then goes to ground and lights bulb.

When the switch is on it also goes to a zener diode (c) and then to the coil of relay. Since the battery voltage when the alternator is not charging is below the voltage needed to 'turn on' the zener, the relay doesn't 'pick'

When the alternator starts charging the battery, it raises voltage above 'turn on' for zener and it picks relay away from NC and turns out light.

Hope this makes sense. Apologize for drawing, the only tool I have is Open Office.

Couldn't attach file, forum wont let me add pdf nor odg files - hope description is enough.

Thanks
User avatar
By Ross Robotics
#172316
lutronjim wrote:I thought they were non conductive in one direction like a normal diode but would conduct in the other direction when the voltage went beyond their rated level.
They do to a point. But they what they call "leakage" which just means it will let some voltage through but not all of it. Different Diodes have different leakage values.

If your circuit doesn't work then obviously your theory is incorrect. You should try something else, like the circuit posted.
Last edited by Ross Robotics on Mon Jun 30, 2014 10:09 am, edited 1 time in total.
By Mee_n_Mac
#172325
I think I understand what you were doing. If you're using a relay you should be aware there's a fairly wide range of voltage (for a given coil voltage) that it will "pull in". And then there's a wider and lower range of voltage before the coil will "drop out". We'd need to know the relay you're using. And even then, while I think you could adjust (using a pot) the voltage where the relay would close and turn off the light, the light will stay off until the voltage drops by (perhaps quite) a bit. What is your desired range of bulb turn-on and turn-off voltages ?

The transistor circuit above will also work to provide a ground for the existing bulb, just substitute the bulb for the LED and R3, but it's logic is reversed from what you want (you want "high" voltage = bulb off). That's fixable. You should tell us what kind of bulb it is so we can know what transistor is appropriate for the current flow. That or you can use the DMM to measure the current flow.
By lyndon
#172327
I think what you are trying to do should work... with caveats. By selecting an appropriate zener, you should be able to get the light to go on/off depending on voltage. The caveat is that it can be fooled by battery voltage since you no longer have access to the alternator field. What you really want to do is to detect whether or not current is flowing into the battery. The circuit above will work, but by the time the LED/bulb comes on, your battery is probably dying.

I think the original method was really an "is alternator turning at the right speed" detector. From your description, even if the battery was disconnected, the light would stay out, but if it was dead or dying, the bulb would come on. It's a clever idea.

The simplest way I can think of to measure current flow out of the alternator and into the battery is more complex and would need either a current shunt or a Hall-effect current detector.
By Mee_n_Mac
#172336
Here's a schematic of your circuit, as I might modify it. Your problem is that the relay and zener both need to be rated at less that the 12 - 14v in the circuit. What happens is when Vin is turned on (let's say it's just the battery voltage of ~12.6v), current will flow through the relay, the zener and the potentiometer (aka a "pot") that I'm using as a variable resistor. After a short time the current stabilizes and, depending on the relay coil resistance, the pot setting and the input and zener voltage, relay may or may not "energize". In your usage you wanted to use the normally closed contacts to send Vin to the bulb (yellow in the schematic below). So long as the voltage across the relay (shown in red) is less than this particular relay's pick-up voltage, the relay is not energized and the contacts remain closed. Then as/if the voltage increases the current increases and the voltage across the relay does as well. At some point the voltage across the relay is high enough that it does energize and thus the contacts open, removing power from the bulb.

What you have to realize is that the voltage across all the components is determined by the current flowing. And the same current is flowing through all the components. That current is (once stabilized) :

I = (Vin-Vzener)/(Rrelay + Rpot)

and the voltage across the relay is :

Vrelay = I*Rrelay

The values shown below are just some I picked that seemed about typical for a 5v relay and a ~5v zener. The problem w/the circuit below is that the drop-out voltage can be very small. That's the voltage at which the relay is guaranteed to de-energize. And it always takes less voltage to hold the energized relay than it does to initially energize it. So you may get the circuit adjusted to energize and turn off the bulb at 13.5v but, once energized, the battery voltage may have to drop below even 12v before the circuit turns on the bulb. And that value can't be precisely known ahead of time. Perhaps you'll get lucky ... or maybe not.
relay_comparator.jpg
So you may want to rethink how you're doing this.
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By lutronjim
#172344
Hadn't thought about less current to hold than to pick. Maybe I should look into a solid state 'relay'.
By Mee_n_Mac
#172356
If I've understood the device above properly and if you have a compatible 12v relay, then the below should work as a more precise comparator, switching the bulb on/off at a voltage you set via the 10k ohm potentiometer. Basically the TL431 acts like a switchable 2.5v Zener diode. When the voltage present at it's reference input is 2.5v above the voltage at the anode pin (a), then the cathode pin (k) is 2.5v above the anode pin ... just like a 2.5v Zener. When the ref voltage is less than 2.5v different, the cathode will not sink any current ... it acts like an open circuit. So you can use this to control the relay coil akin to what you were trying to do before, except that you now need a 10k ohm pot to set the reference voltage input. Bascially if you want the switch point to be 13.5v at the battery, you set the pot's wiper to be 2.5v with 13.5v across the pot.
TL431_relay_ckt.jpg
I'd need to know the relay coil data to be sure it'll work. The TL431 can only sink 100 mA.
To understand how the pot works, read this ...
https://learn.sparkfun.com/tutorials/voltage-dividers
Due to the ugly nature of a car's voltage, you may want to add some additional protection circuitry to the reference pin, perhaps a 1 uF cap ... maybe some transient suppressor though given the expected voltage divider ratio it may not be needed.
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By lutronjim
#172386
Relay - nominal coil current 60ma - coil resistance 200 ohms +_ 10% - pickup votage 9.6 - dropout 0.6.

Not quite sure I understand Lrelay and Rrelay. Are you trying to show a resistor in series with coil? If 220 ohms is needed, that is what the coil has.

I assume pickup and dropout voltage problems just go away because voltage should be 0 or 13.5 or more.

I'm not seeing a 1uF 14v dc cap. Will a higher voltage work? Like http://www.galco.com/buy/Mallory/M30U105M5.

Thanks for info. I think my idea would work but trying to find a zener for the close voltage difference seems very difficult.