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Voltage Range of Analog Inputs
Hi All,
What is the voltage range of analog inputs? In other words, a byte of 255 translates into how many volts? Oh, and can anyone give the schematics for the current monitoring circuit? -Thanks |
i *think* the voltage range is 0-5V, but don't quote me on this, i am only going from memory... anyone have an affirmative answer?
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Yep....+0Vdc to +5Vdc...hence the +5Vdc on some of the analog port pins....
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Does anybody know what is the maximum current that can go through the controller without burning it up? We are trying to figure out how to put a current monitoring circuit together, if anyone could give us hand we would greatly apreciate it. It is pretty hard to come up with one without any experience or engineers.
Eugene :confused: |
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I would love to help, but I might need a little more information on what your thought are. Actually, FIRST sent out a couple of good hints in the third update. (I think) -Quentin |
To Quentin
Forgive me if I'm blind, but the only reference to the electronics in the 3rd update that I saw was about the fuses, please correct me if I'm wrong.
The reason I'm asking this is because I wanted to build a custom circuit board that would monitor the current, and using the analog inputs on the controller input the data. Problems with this idea: 1. Is it legal to have the circuit board connected in a serial circuit to the motors, or the main braker? The current has to be measured in a serial circuit, right? 2. Which transitors would have the correct rating to take the currents that might go as high as couple hundred AMPs? 3. What is the maximum current that can go through an analog input on the controller? And here is a unrelated question, do i need to have a power source when i connect a potentiometer (it is just a resitor right)? Thank you all for helping us out!!! We don't have any proffesionals on outr team. Eugene |
I think the analog connector can only source something in the range of 50mA to the pins when the resistor is at a dead short (0 ohms) so I would guess this is a safe max amount. (The max rating is listed in the Innovation First instruction manual
Matt |
To measure current from the battery (or to the motors for that matter), you should put a very low resistance shunt in series with the power lead. Technically it doesn't matter which one but we use the positive lead. You then have a low voltage (in the millivolt range) that can be measured across the shunt and amplified (with a simple op-amp circuit) to accomodate the 0-5v of the A/D converter on the controller. We (Team 802) are measuring total battery current and that's how we are doing it.
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series vs parallel
Correct me if I'm wrong. It's been a while since I've had electronics. I do believe that if you hook an amp meter in series, you will here a loud and distinct pop of a fuse (a breaker if it's a good one). I think you measure amperage in parallel and voltage in series. Don't forget the difference between the two, it's an important one....
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Here's a circuit I banged out that might be able to help you.
It has a small current sensing resistor in series with the battery and circuit breaker. The voltage across the resistor, which is proportional to the battery current, is determined and amplified by a difference amplifier circuit that has a gain of 25, which allows the 1 milliohm resistor to fill the available range on the A/D input. (200A through the circuit corresponds to 255 A/D counts). And the wheel on the bus goes round and round. You might want to throw on an output filter to remove the PWM ripple, and perhaps another gain stage to fine tune the circuit's range, but it should work reasonably well now. Please note that Digikey doesn't have any 1 milliohm resistors that can handle enough power to work here - so you have to 'make' your 1 milliohm resistor from five 5mOhm precision resistors, or ten 10mOhm precision resistors (in parallel). Five of these would be able to handle 24W continuously, which is short of the 40W required for the anticipated 200A spikes, but I think they would hold up in practice, because the average power dissapated would be well below 24W. If you have any questions, let me know. I know that I didn't document the image too well, but it shouldn't be too hard to get the general idea. |
Re: series vs parallel
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Like you said, it's important to remember the difference between current and voltage... CURRENT is a measurement of the number of electrons flowing past a certian point in a given time frame. I've always pictured it as: you break the circiut, forcing all of the electrons to go through the meter, "counting" them. VOLTAGE on the other hand is a measurement of potential. "Pressure," if you will. When you measure voltage, you've to keep track of what point you are using as refrence. As such, you *could* wire a volt meter in series... but you'd read 0V. The difference in potential accross a perfect conductor is 0. (Try measuring the voltage drop accross a resistor, and then measure the voltage at a point with respect to ground...) Of course, keep Mr. Ohm in mind! Voltage = Current * Resistance Hope that helps! |
Thanks A lot guys!
Thanks, this was really helpful, without you guys I would have been lost.
Eugene |
Inferring current MAY be easier...
This all sounds pretty tricky to an escapee from M.E. land...
For my money, I would rather infer current. You know battery voltage. Assuming you have consistant batteries (big assumption I know), your battery has an internal resistance that will give you some information about battery current. AND... you can refine your estimates by doing a test. It would be possible for example to stall a motor of know resistance for a very short time (say by moving an arm against the stop), when you first install your battery or upon first power up. If you watched battery voltage during this test, you could calibate your sensor (i.e. the built in battery voltage sensor) to various current levels. It would be also possible to give the motor several different voltages in order to get several data points. For my money, this seems the better path than putting resistors in your battery path. Joe J. P.S. Of course, there are other ways to infer current as well, but that is a message for another day -- perhaps for another year. |
A lead-acid battery's internal resistance actually depends upon the state of the battery's charge, and the temperature. While you could use the drop in the battery's voltage to determine the current that it's sourcing, the calibration could vary greatly over the course of an event, and even over the course of a match.
Of course, now you're going to come back and tell me that that's what you do, and that it works fine, aren't you? :) What we need to do is petition Innovation First to include a current monitoring output on their Victor speed controllers. |
Many of the ideas presented here are good ones but there are some potential hazards.
1. (My personal favorite) Measuring battery current on a working robot is best done with a current meter like an AmpProbe or other similar device. These are clamp on meters that will not put your controller (or you) in the presence of the several hundred amps that may be periodically delivered by your battery. 2. I would frown on any resistance you add between the battery and controller. The controller resets at about 8 volts and a resistor will get you close under extreme loads. Feed the controller from the battery side of the resistance. 3. If you are not sure how to do it, don't connect circuitry to the robot controller inputs except as listed in the manual. A bad hit on the analog input is a fatal (i.e. permanent, non warranty) failure. 4. Opening the circuit and inserting a voltmeter would prevent anything from working and should read near the battery terminal voltages. (Open circuit voltage) 5. Much truth about the variable battery internal resistance, it also varies from unit to unit due to production variances. Ironically, it is the internal resistance that ends up dropping the available voltage to the robot during large current draws. When the voltage is so low that it drops below 8 volts (controller dropout) it is because four volts is being dropped across the internal resistance. Good Luck All, Al |
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