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TLC2543 question (odd) — Parallax Forums

TLC2543 question (odd)

JonathanJonathan Posts: 1,023
edited 2007-07-21 13:42 in General Discussion
Hi All,

I have used the TLC2543 many times, and one thing I have found is that if the input voltage on a channel exceeds VDD, the chip will go into a lockup state where the chip gets hot. This of course is not good for the chip.

I just made a new battery pack and got a new charger for my 'bot. I should have looked at the output of the charger on a 'scope, but I just measured it with a meter. It is a 7.6 volt battery pack, and the charger read 9.2V while charging.

However, the voltage is actually spiking above 10V when you look at it on the scope. One of the channels of the ADC wired to monitor the battery voltage. So, when charging, I blew the ADC [noparse]:([/noparse] .

So, my question is this: I would like to keep a 2:1 voltage divider to check battery voltage. If I add a 100kOhm resistor in series with the input, I would *imagine* this would limit the current enought to prevent destroying the chip when the voltage spikes above 10V with the charger.

Any thoughts on this? I'd like to find out the easy way, not the $8 buy another chip way. [noparse]:)[/noparse]

Jonathan

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Comments

  • NewzedNewzed Posts: 2,503
    edited 2007-07-13 19:48
    Jonathan, the current at any input is about 2.5ua.· A 100K resistor would only drop about .25V at that current.· I would think an easier approach would be to use a 3:1 or maybe a 3:2 divider.

    Sid

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    ·
  • Tracy AllenTracy Allen Posts: 6,667
    edited 2007-07-14 17:15
    Hi Jonathan,

    You should put resistors in series with all the ADC inputs that go to the outside world, just as you would for Stamp pins. The ADC has protection substrate diodes, just like the Stamp. I use 330 ohms. That is sufficient to limit the fault current to less than 20 milliamps even if you accidentally hook the input directly to the 10 volt power.

    For monitoring the power voltage, a voltage divider will be fine. 100k:100 or 200k:100k with a 0.1uF capacitor at the ADC input.

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  • JonathanJonathan Posts: 1,023
    edited 2007-07-14 17:39
    Gentlemen,

    Thanks for the help. I do put a 330 ohm on every input, but I didn't on the voltage divider (duh). Now I have put a 100k in series after the voltage divider of 10k/10k, and that seems to prevent the problem. The ADC is still working, just off a little. I'm hoping that the new chip will be ok with this setup.

    I had thought the 10k/10k divider would have limited the current enough to prevent the problem, but I guess not.

    Thanks again for the help!

    Jonathan

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  • Tracy AllenTracy Allen Posts: 6,667
    edited 2007-07-14 19:20
    If you had a 10k:10k divider, that should indeed have prevented the problem. From your original post, I thought the 10 volts was accidentally hooked directly to the input pin, with no resistor at all. pfoot! That may be what happened anyway, that the 10k resistor you thought was there became all too un-briefly short circuited.

    Put a 0.1uf capacitor at the ADC input and that should restore the accuracy. The drop in accuracy comes from the ADC sampling current into the 100k resistor. With a capacitor, it will hold the voltage so long as you don't read the channel too often.

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    Tracy Allen
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  • JonathanJonathan Posts: 1,023
    edited 2007-07-15 14:29
    Tracy,

    The ADC is reading off on all channels, not just the one with the battery input. It got hot [noparse]:([/noparse]. I realize the drop across the 100k will call for a little "fudge" to read correctly.

    I would have thought that a cap on the input would extend the length of time that the input is exposed to + 10, meaning that when the charger spikes above 10V that the cap will hold this voltage level on the input briefly. I'll try it with this alreadt damaged ADC and see if it gets hot.

    As to the 10k/10k divider, we damaged a TLC2543 on the hydrogen fuel cell robot the same way. I should have realized that the voltage from the charger might be exceeding 10V and looked at it on a scope, as it is some sort of switching type charger. It suprises me too that the divider doesn't limit the current enough, but this is the second time I have seen it happen.

    Thanks!

    Jonathan

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  • NewzedNewzed Posts: 2,503
    edited 2007-07-15 14:36
    Jonathan, you might try putting a large capacitor on the charger output, say 100uf.· This might absorb the spike.

    Sid

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    ·
  • Tracy AllenTracy Allen Posts: 6,667
    edited 2007-07-15 19:14
    Jonathan,

    I just want to be sure we're on the same page. Here is a schematic. Are you doing something like this?
    attachment.php?attachmentid=48204
    There is no reason for the ADC to see any voltage greater than Vdd. Give your divider plenty of leaway, as 2:1 in the attached diagram, or even go to 3:1 or 4:1 so the input voltage can go up to 20 ro 25 volts. With the capacitor, there should not be significant error in the measurement, given that you are reading that channel less than once per second. It has plenty of time to settle, and it will settle to a voltage between Vss and Vdd. There no reason for erroneous readings on other channels much less for the ADC to get hot.

    It is characteristic that the readings on an ADC will be in error on all channels if any of the substrate diodes are forward biased, even if it is not to the point of damaging the chip. On my data loggers, I also have an active clamp circuit that prevents the inputs at the ADC from going much above Vdd or below Vss.

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  • JonathanJonathan Posts: 1,023
    edited 2007-07-16 14:46
    Sid,

    I do have a 100uF cap on the vreg for the ADC, which is dedicated to the ADC. I'll try adding a larger cap and looking at the scope.

    Tracy,

    No, that isn't quite what I have. I have a 10k/10k divider instead of the 200k/100k. I am using a 100k series resistor, and have no cap in between the divider and the series resistor. I was hoping to be lazy and get away with just adding the series resistor. I guess for safety's sake I better unsolder the whole mess and do as you and Sid reccomend and go for a 2:1 divider.

    Thanks for the help!!!

    Jonathan

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  • Tracy AllenTracy Allen Posts: 6,667
    edited 2007-07-16 18:49
    Jonathan,

    What you have with 10k:10k should work for inputs up to 10 volts, but I'd go with higher values and a capacitor. The 330 ohm resistor is there only for protection, in case the resistor to the higher voltage gets short circuited, and it should be a relatively low value between the capacitor and the pin.

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    Tracy Allen
    www.emesystems.com
  • KatyBriKatyBri Posts: 171
    edited 2007-07-17 04:00
    Tracy,

    Could you share a schematic for your comment " I also have an active clamp circuit that prevents the inputs at the ADC from going much above Vdd or below Vss."? Thanks.
  • PJAllenPJAllen Banned Posts: 5,065
    edited 2007-07-18 16:13
    I/O clamp

    Update -- I chose/choose Schottky diodes for their lower forward voltage; germaniums would be better, but they're harder to find.· The idea being to clamp as close to VSS any input·voltage less than VSS and as close to VDD any voltage greater than VDD, as possible.· Schottky "leakage" is·primarily a concern·when near·reverse breakdown and more so at high temperatures.

    Post Edited (PJ Allen) : 7/19/2007 5:02:45 AM GMT
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  • Jason-WIJason-WI Posts: 35
    edited 2007-07-18 18:24
    Schottky diodes are inherently leaky. I use 1N914 or 1N4148 diodes for clamping ADC inputs.

    Jason

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  • Tracy AllenTracy Allen Posts: 6,667
    edited 2007-07-19 06:18
    I use the SP720 (Littlefuse). It is a 16 pin chip that contains an array of 14 protection networks similar to the circuit that PJ posted.

    attachment.php?attachmentid=48255

    The "diodes" are actually a couple of transistors wired like an SCR, so the response is sharp and fast with low leakage. One pin on the array is common to Vss and one pin is shared to Vdd. However, instead of wiring to Vdd, I wire the high shared pin to a transistor circuit that shunts the fault current to ground instead of to Vdd. The reason for that is kind of subtle. The regulator that supplies Vdd can only source current, not sink current, and when operating at micropower, you don't want a high fault voltage to raise Vdd itself to a high and perhaps damaging level. The shunt transistor is a power superbeta. The squares in the diagram represent a low ohms resistor to the ADC input pin and a second low ohms resistor to the outside input. The resistors assure that the fault currents will be carried predominately by the SP720 rather than by the ADC substrate diodes.

    The switching threshold of the SP720 is kind of high, higher than a shottky diode, so even though it works well for protection, it does not do so well for preventing an overrange signal on one channel from leaking into a neighboring channel.

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  • JonathanJonathan Posts: 1,023
    edited 2007-07-21 13:42
    Tracy,

    Hooked it up as you suggested, and it is of course working fine. Thanks!

    Jonathan

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