Post by thread:Ohmimeter

That is easy. 5V | | |___[res]__| ADC Pin | Read the adc pin. then calculate the voltage drop. Like this: drop = 5 - adcvalue; Then figure out what value resistor drops x amount of voltage. IT may NOT WORK... Hope this helps, Andrew On Mon, 7 Feb 2000 10:50:56 -0200 "Cassiano J. Pereira" <cassianoKILLspamEXATAS.UNISINOS.BR> writes: {Quote hidden}> Hi all! > > How can I to build an ohmimeter using a Pic16c711? > If somebody to know of something, answers me. > Thanks! > > -- > +-----/-------------------------\-----+ > | \ CASSIANO J. PEREIRA / | > | \ Eng. Eletrtnico / | > |-------\---------------------/-------| > || -> LAPRO - Centro 6 - UNISINOS <- || > || emails: .....casspereiraKILLspam.....sinos.net || > || EraseMEcassianospam_OUTTakeThisOuTexatas.unisinos.br || > +-------------------------------------+ ________________________________________________________________ YOU'RE PAYING TOO MUCH FOR THE INTERNET! Juno now offers FREE Internet Access! Try it today - there's no risk! For your FREE software, visit: dl.http://www.juno.com/get/tagj.

part 0 2653 bytes 5V |                   | |___[res]__| ADC Pin             | Read the adc pin. then calculate the voltage drop. Like this:     drop = 5 - adcvalue; Then figure out what value resistor drops x amount of voltage. IT may NOT WORK... Hope this helps, Andrew </UL> I can be a little more definate.  It won't work!  The input impedance of the adc is pretty high and worse, is not precisley defined, so unless the the resistor you are measuring has a very high value, you will only drop a very small amount of voltage.  Remember a resistor will only drop voltage when a current is flowing through it. The normal way to measure a resistor is to put a known current through it and measure the voltage drop.  You extend the range of an instrument, you would have several ranges that would put different amounts of current through the resistor.  You simply apply ohms law R=V/I. The tricky(er) bit is making a precision current source which isn't my speciality, but I'm sure someone on the list will oblige. Mike </BODY> </HTML> </x-html>

On Mon, 7 Feb 2000 08:22:29 -0500 Andrew T Kelley <k_andrewspam_OUTJUNO.COM> writes: > That is easy. > > 5V > | | > |___[res]__| ADC Pin > | > > Read the adc pin. > then calculate the voltage drop. > Like this: > drop = 5 - adcvalue; > Then figure out what value resistor drops x amount of voltage. > > > IT may NOT WORK... > > Right, it probably won't. You can either use a pull-up resistor to +5V (assuming the A/D is using +5V as its reference) and have the unknown to ground (which will give a nonlinear A/D versus resistance curve) or use a current source as the pull-up, which will give a linear curve. The current source could be made using any of several techniques. Could be a current mirror, a PNP transistor with zener, a constant current diode, or an op-amp constant current source. If you don't mind floating the unknown resistor, you can use an op-amp to do both the constant current and current to voltage conversion. If, for example, you ground the noninverting input, connect the inverting input through a 1K to -1V, put the unknown resistor between the inverting input and the output, the output voltage (in volts) will represent the unknown resistance in k ohm. You can scale resistors and voltages to whatever you need. Harold FCC Rules Online at http://hallikainen.com/FccRules Lighting control for theatre and television at http://www.dovesystems.com ________________________________________________________________ YOU'RE PAYING TOO MUCH FOR THE INTERNET! Juno now offers FREE Internet Access! Try it today - there's no risk! For your FREE software, visit: dl.http://www.juno.com/get/tagj.

Any electronic instrumentation measures the physical variable element by comparison, or to another known part or to a known internal reference. All the other kind of magic only with "Magic Lights and Sound" in Holywood, or using a RC timming constant :) You can choose any of the following 3 options. Pick any TWO from "Fast", "Good", "Low Cost". OPTION #1: (Not easy, Not simple) The most common way to produce a ohm-meter is via a constant current source. .----------o---> ADC Input | | / \ | +10Vcc-----/ \ '---> -10Vss----/_____\ UNKNOWN R (UR) + - .---> | | | VRef o----' o--------o 0.1V | | === C R 100 Ohms (RF) | | Gnd Gnd The above circuit will try to supply 1mA through UR doesn't matter what. High values of UR will generate high voltage to ADC input, zero ohms at UR, will generate 100mV at ADC input. As the current is constant 1mA (with some UR), there is a 100mV over RF always, so your software should reduce it from the actual ADC output. RF = 100 Ohms, and UR from zero to 4900 Ohms will generate ADC input voltages from 100mV up to 5000mV. So, RF would be your scale range. RF = 10 Ohms, UR range up to 490 Ohms. RF = 100 Ohms, UR range up to 4900 Ohms. RF = 1000 Ohms, UR range up to 49k Ohms. RF = 10k Ohms, UR range up to 490k Ohms. ADC input overflows above 5V means the op-amp output swings toward +10V (>5Vdc) when UR is above RF range. UR = infinite (open), op-amp output goes close to +10V. You could also control VREF to change range (restrictions). OPTION #2: (Simple, but not easy) Using a discrete ADC unit that can accept a VREF from 0.1 to 4Vdc, you could build this circuit: +4Vdc | | R1 RANGE (=R2) | | o-----> ADC VREF | R2 (RANGE) | o-----> ADC INPUT | R3 (UNKNOWN) | | Gnd Vref = 4 * (R2+R3) / (R1+R2+R3) Input = 4 * R3 / (R1+R2+R3) Suppose ADC is 12 Bits unipolar, so full scale (FS) = 4095 If R2 = 10k and R3 = Unknown R3 = R2 / ( FS / ADC_OUT - 1 ) ------------------------------ Examples: ADC output = 50 ADC output = FS / ( 1 + R2 / R3 ) 4095 / 50 = 1 + R2 / R3 81.9 - 1 = R2 / R3 R3 = R2 / 80.9 = 10000 / 80.9 = 123.60 Ohms ADC output = 300 4095 / 300 = 1 + R2 / R3 13.65 - 1 = R2 / R3 R3 = R2 / 12.65 = 10000 / 12.65 = 790.51 Ohms ADC output = 1000 4095 / 1000 = 1 + R2 / R3 4.095 - 1 = R2 / R3 R3 = R2 / 3.095 = 10000 / 3.95 = 3231 Ohms ADC output = 2047 4095 / 2047 = 1 + R2 / R3 2.000 - 1 = R2 / R3 R3 = R2 / 1.000 = 10000 / 1.000 = 10k Ohms ADC output = 2356 4095 / 2356 = 1 + R2 / R3 1.738 - 1 = R2 / R3 R3 = R2 / 0.738 = 10000 / 0.738 = 13550 Ohms ADC output = 3000 4095 / 3000 = 1 + R2 / R3 1.365 - 1 = R2 / R3 R3 = R2 / 0.365 = 10000 / 0.365 = 27397 Ohms ADC output = 4094 4095 / 4094 = 1 + R2 / R3 1.0002 - 1 = R2 / R3 R3 = R2 / 0.0002 = 10000 / 0.0002 = 50 MOhms ADC output = 4095 4095 / 4095 = 1 + R2 / R3 1.000 - 1 = R2 / R3 R3 = R2 / 0.000 = 10000 / 0.000 = Open Circuit In any case, VREF would swing between Vcc/2 to Vcc, since R1=R2. R1=R2 defines the mid range of the curve. OPTION #3: Using a discrete ADC unit with diferential VREF and Input pins. +4Vdc | | o-----> ADC VREF + | R1 (RANGE) | o-----> ADC VREF - | o-----> ADC INPUT + | R2 (UNKNOWN) | o-----> ADC INPUT - | Gnd Vref = 4 * R1 / (R1+R2) Input = 4 * R2 / (R1+R2) R2 = ADC_OUT * R1 / ADC_FS If ADC_FS (full scale) = 4095, ADC_OUT = 1000 R1 = 100k R2 = 1000 * 100k / 4095 = 24420 Ohms R1 will be your full scale range. OPTION #4: (Easy, but not simple) PIC port pin A output-----. | R Unknown | PIC port pin B input------o | === C | Gnd Flip A to + and measure time it takes to charge C. Long time means High Value R. Create a linearization formula to process the measured time, or a PWM output at A (increasing the + pulse width along with the time), so the ramp on C will be linear.