blown up prop chip
Erik Friesen
Posts: 1,071
Any reason why a 14 volt input into a 20K resistor and a 3.3 volt zener to ground would fry the chip?· Or should I look to static electricity as the cause?· Anyone else have problems and fixes for static?
Comments
-Phil
Post Edited (Phil Pilgrim (PhiPi)) : 6/2/2007 4:34:14 PM GMT
-Phil
Graham
-Phil
-Phil
-Phil
-Phil
I would not use a zener and resistor if I were you. Regulators are not that expensive and much safer. You may be blowing out your voltage-drop resistor that you are using to get power through! I didn't see any power resistors on your board and your power input resistor is going to disspate some heat. Even though you are not drawing much current, you are dropping lots of volts, and that adds up fast. For example:
Let's say you have a 13.3 volt input. You are getting 3.3 volts out, so your input resistor is dropping 10 of the 13.3 volts. Now say you are drawing just 50 milliamps. That's 50 mA at 10 volts, which comes out to 500 mW (half a watt). Your surface mount resistors are probably rated for 1/8th watt. "Common" through-hole resistors are usually 1/4th watt. You need at least the big half watt kind, or even 1 watt, to be safe.
You didn't mention what the signal was coming from, if it's from the primary of the ignition coil for instance then you are going fry the chip.
There is no problem running a much higher value resistor of around 100K or even 1M in which case you will not need an input zener. If in doubt an NPN to buffer the signal is practical.
*Peter*
Seriously, look into optocouplers. It's not just the LED at left and phototransistor at right combinations, although those are typically the cheapest. You can get a 3.3V CMOS-level output, even with hysteresis. There are even analog (linear) optocouplers. (LOC110 comes to mind, although there are lots of others).
It also wouldn't hurt to have a real chassis ground. I know cars are weird sometimes where they connect the (-) onto the frame and have the current flow back to the battery that way. That is an awful noisy way to complete the circuit.
On your board itself at the power input connector, the Ground would typically be isolated from the Common (-) except for a ceramic cap connecting the two together. (A cap from (-) to (ground) near the board's power input and also a cap from (+) to (-). Any other plug into the board would also have this ground carried out through either a conductor in the cable or though the cable shield where appropriate. It's this Ground, and not the (-) that would locally connect to the frame.
The point is that the current flow from Power (+) should be about that of what's flowing out of Power (-), and Ground to the frame should not really be conducting any current except from static or surges. When driving inductors like solenoids, you actually want the surges to go out into the ground wire and into the chassis instead of the Common.
Have you considered in your circuit any input going below the 0V (Vss) level? Most chips can't handle even -0.5 volt. And remember it's the voltage at the pin, not the current.
Oh, good heavens! Now that I've reread the entire thread, the comedy of miscommunication is all too obvious. Of course it's an input! It says so right there in the first post. I just had a knee-jerk connection that zener = cheap voltage regulator and launched into a lecture that seemed to get nowhere — for good reason, it turns out. My apologies for leading everyone down the garden path...
D'oh!
-Phil
That's what happens when you over-clock your brain...
If I was trying to do as much as you, I'd be a blithering idiot by now...
Come to think of it...
Rich
Graham
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Paul Baker
Propeller Applications Engineer
Parallax, Inc.
To power a circuit takes two wires, a (+) and a (-). Think of a battery.
In a car,
(+) = ~14V = Positive = The red wire = "supply voltage"
(-) = 0V = Negative = The black wire = "common return"
It's convenient to use any sort of hunk of metal as a "Ground", in this case the vehicle's chassis. The connection from the (-) of the battery to this ground is made at the battery terminal. Given that you say your circuit is powered from the radio harness, it's pretty likely that the (-) makes its way through a real wire and not the chassis back to the negative of the battery. This is good.
If you have your project in a metal box, it would be a good thing to connect in some manner this metal box to the chassis of the vehicle. It would be a bad thing to connect this metal box directly to the (-) terminal of the power coming into your board, not saying that you did this... You could put a capacitor between (-) and this "ground" to pass any spikes off to ground. I'm not saying use a metal box, and I'm not saying you have to ground it. But you were worried about static and I was giving a way to solve the issue in my last post.
I am trying to explain in some manner or fashion, that by the time the (-) gets to your board, it can't be considered a "true ground". It is unfortunate that conventional digital electronics calls the common reference or 0V reference, a "ground". The two power leads, the red and black, by the time they get to your board have all sorts of interference and spikes induced on it. They have small but meaningful resistances. Hence voltage fluctuations.
Try this thought experiment: Imagine measuring voltage from one black (-) wire somewhere in the car, to some other black (-) wire in the car. It should be 0V, right? In the real world, it won't always be. Once you understand the issue, that all devices don't have the same exact 0V reference, things should become clearer.
Signal wires, through induction, also get fluctuations induced on them.
As for optocouplers...
Let's say your datasheet says the LED should have a minimum of 2.5 mA flowing through to be considered definitely on. The datasheet also says that the LED will have a forward voltage drop of 1.45V at this current.
If you want to detect a 6V minimum signal, the resistor would be R = (6 - 1.45) / .0025 = 1.8 K
From the datasheet:
The LED has an absolute maximum sustained rating of 20mA.
At this 20mA, the voltage across the LED is 1.55V
20mA through a 1.8K resistor gives a delta-v of 36 volts. 36 + 1.55 = 37.55 volts, which is a lot imho. Unfortunately, anything but a 1 Watt resistor is going to get very hot (I squared R). But the point being you can detect reliably a voltage from 6V to 37V or whatever without hurting the LED.
To get even better range, you can add another resistor across the LED in parallel, and adjust the first resistance, so that there is significant current flowing through the second resistor as well as the LED. Just watch out to not overheat the first resistor again.