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.
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.
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.
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.”
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.
lutronjim:
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.
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.
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.
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.
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.
I’d need to know the relay coil data to be sure it’ll work. The TL431 can only sink 100 mA.
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.
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.
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.
That relay should be fine, the current needed is less than the recommended 100 mA (or 150 mA max). Yes, my schematic was showing the coil as it would be modeled by an EE. That is as an inductor (w/no resistance) in series with a resistor ... which for your relay would be a nominal 200 ohms.
lutronjim:
I assume pickup and dropout voltage problems just go away because voltage should be 0 or 13.5 or more.
The problem goes away because the TL431 has a precision comparator built into it. When the divided battery voltage at the ref input is a few mV above 2.5V, it switches to “zener state”. When the ref input is a few mV below, it’s back to “open circuit state”. Thus you’re not dependent on the precision of the relay characteristics.
lutronjim:
I’m not seeing a 1uF 14v dc cap. Will a higher voltage work?
Yup. In general it’s wise to use a 50% over rating. Your link was a good one.
Thanks for reply - been off grid for a few days. I ordered parts you suggested from SparkFun but changed cap to 10uF at 25v. Knowing enough to be very dangerous about circuits, I assumed the cap was to ‘smooth’ out voltage so I assumed more storage wouldn’t hurt.
lutronjim:
I assumed the cap was to ‘smooth’ out voltage so I assumed more storage wouldn’t hurt.
Correct. IIRC with what I expect will be potentiometer setting and the 1 uF, they made a low-pass filter with a bandwidth of ~100 Hz. That should have filtered out most noise and spikes. The new value will filter more. The "cost" is that when setting the pot for the trip point, you'll have to move it more slowly to let the ref voltage settle.
And that begs the question … how do you adjust the pot to get the desired trip point (voltage) ? In a perfect world you’d have some adjustable voltage that you can set to that trip voltage … and then you’d just tweak the pot until the bulb just comes on/goes off at that voltage. But I suspect you don’t have such an adjustable voltage supply. So there’s another method that should work, so long as you have a voltage above and a voltage below the trip voltage (which could be your battery w/the car off and with it running). And a voltmeter to measure those voltages and the voltage at the ref pin.
Parts came in today so I’ll be trying to solder up a board this week.
Don’t quite understand your method. Easy enough to measure on and off battery voltage - I have a Fluke DVOM. I guess I could measure the ref pin with car off and set pot so I could see 1.4 or so and just raise it a bit and try.
No doubt because I haven't detailed it yet. :mrgreen: Here it is, see if it makes sense.
Let’s say your voltage trip point is Vtp. When Vbatt <= Vtp, the light goes on. When Vbatt > Vtp, the light goes off. You’re going to set a pot to make this happen but you don’t have a battery where Vbatt = Vtp. How do you set the pot ?
OK let’s recall 1 simple working principle. That the pot “divides down” the Vbatt voltage that goes into the reference input. When that ref voltage = 2.5v, the switching occurs. So ;
Vref = 2.5v = k* Vtp, k = 2.5/ Vtp, k is the divisor value
As an example only : if Vtp = 13.5v, k = 2.5/13.5 = 0.185
So 1’st compute k for your Vtp.
Then with the car not running but the switch on, I’d expect Vbatt to be ~12.6v. Use the DVM to measure Vbatt and Vref and adjust it so Vref = kVbatt. Using the above numbers Vref would be 0.18512.6v = 2.33v. Thus the light should be on.
Lastly start the car and measure Vbatt and Vref. I’d expect Vbatt to be ~14.2v and Vref = ~2.63v. The light should be off. If Vref is a bit higher or lower than k*Vbatt, then slightly adjust the pot so as to “split the difference” with the ‘car off’ setting.
