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hey i need to make a multi meter on a bread board without ic or timers. which measures current, voltage, resistance and also inductance . can any show me a circuit which will work . i tried some but all have ic or timers . some circuits i tried http://homemadecircuitsandschematics.blogspot.in/2011/12/make-workbench-multimeter-with-ic-741.html |
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Given an ammeter with full scale current I and internal resistance R. VOLTAGE
So - \$V_{fs}=I_{fs}\times R_{total} = I_{fs} \times R_{meter}+R_{series}\$ or, rearranging: \$R_{series}=\dfrac{V_{fs}}{I_{fs}}-R_{meter}\$ eg given a 50 uA meter with a 1000 Ohm resistance, make a 20 VDC full scale meter. \$R_{series}=\dfrac{V_{fs}}{I_{fs}}-R_{meter}=\dfrac{20}{50\times 10^{-6}}-1000 = 400,000 - 1000 = 399,000Ω = 399KΩ\$ In this case the meter resistance is irrelevant as it will have minimal effect on accuracy. Note that for 1 Volt, \$R_{fs}=\dfrac{V_{fs}}{I_{fs}}=\dfrac{1}{50 \times 10^{-6}}=20,000\$. SO a 50 µA ammeter produces what was called when such things were common, a "20,000 Ω per Volt" Volt meter. To make a multimeter, just add \$20,000\times V_{fullscale} Ω\$ for each range. eg for ranges of 1 V, 10 V, 100 V the series resistors are 20 KΩ, 200 KΩ, 2 MΩ. CURRENT If we use the same 50 µA, 1000 Ω meter we can divert current around it so that more current must flow to make the meter read full scale. If we place an \$R_{sh}\$ resistor in parallel with the meter, then at full scale, if 50 µA flows through the 1000 Ω meter then \$\dfrac{1000}{R_{sh}}\times 50 µA\$ will flow through the shunt resistor. So total current = \$I_{fs}=I_{meter}+\dfrac{R_{meter}}{R_{sh}} \times I_{meter}=I_{meter}\times \left( 1+\dfrac{R_{meter}}{R_{sh}}\right)\$ \$I_{fs}=I_{meter}\times \dfrac{R_{sh} + R_{meter}}{R_{sh}}\$ or rearranging: \$R_{sh}=\dfrac{R_{meter}\times I_{meter}}{I_{fs}-I_{meter}}\$ So eg to make a 100 mA meter with our 50 uA meter we see \$R_{sh}=\dfrac{R_{meter} \times I_{meter}}{I_{fs}-I_{meter}}=\dfrac{1000\times (50\times 10^{-6})}{0.100 - 0.000050}=\dfrac{0.050}{0.099950}=0.500250Ω\$ or close enough to 0.500 Ω. In the above when \$I_{fs}\$ >> \$I_{meter}\$ the \$(I_{fs}-I_{meter})\$ term can be simplified to \$I_{fs}\$ so \$R_{sh}=\dfrac{R_{meter}\times I_{meter}}{I_{fs}}\$ which makes a large amount of sense if you look at it long enough. So in th above case \$R_{sh}=1000\times \dfrac{50 µA}{100 mA}=1000\times \dfrac{1}{2000}=0.5Ω\$ as expected. So a Multimeter has ranges which switch shunts across the meter which are 1 \$N^{th}\$ of the meter resistance for \$I_{fs}=N\times I_{meter}\$. RESISTANCE: Look at the resistanmce scales on a non electronic analog ohmmeter. The Ohmmeter is an ammeter with the scale calibrated to suit. INDUCTANCE Harder. Can be done BUT electronic make it FAR easier. |
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If you were given the challenge of designing a device that could precisely measure voltage, current, and resistance, and what you had were a calibrated voltage source (say 10.00 volts), a calibrated resistor (say 10.00 ohm) and a mechanically-calibrated 10-turn pot (approximately 1.000K), and various multi-turn pots which need not be calibrated, and one or two sensitive but non-calibrated bidirectional current-sensing "meters" (e.g. a galvanometer), you could design an accurate and precise, albeit not terribly convenient, apparatus for measuring resistances, steady-state voltage, or steady-state current. For the simplest case, and assume the goal is to measure a voltage 0 to 10 volts. Simply tie the supply across the potentiometer, tie the negative of the supply to the voltage under test, and connect the meter between the positive of the voltage under test and the potentiometer (possibly with a resistor around 1K or so in series to protect it). Turn the pot until the meter reads zero. The voltage on the pot will be 10 volts times the resistance radio, which may be read out from the pot's mechanical position. Note that variable voltage dividers are called "potentiometers" because one of their first uses was the measurement of voltage in this fashion. If one needs to measure voltages 0 to 1 volt, it would be helpful to scale things so full scale on the pot was 1.000 volts rather than 10.00 volts. To do that, one could connect a resistor in series with the pot whose value was precisely 9 times the pot resistance. If one didn't have such a resistor, one can produce one by using the calibrated pot to set the wiper on a non-calibrated pot so as to generate 1.000 volts, and then put a variable resistor in series with the upper leg of the calibrated pot and adjust it so that the top of the calibrated pot matched that 1.000 volts. If one needs to measure voltages of up to 100 volts, one could use the calibrated pot to adjust a 10:1 voltage divider which could then be fed with the input voltage and scale it to the range 0 to 10 volts, which could then be measured as above. To measure current, one would use an uncalibrated variable resistor feed voltage through a precision resistor so as to balance out the meter, and would then use the above measurement setup to measure the voltage across that precision resistor. To measure resistance, one would put a precision resistor and adjustable resistor in series with it, vary the adjustable resistor until the precision resistor had a certain voltage across it (probably some power of ten), and then measure the voltage across the resistor to be tested. Using this style of apparatus, depending upon the sensitivity of one's galvanometer, one could produce reasonably effective measurements of a wide range of voltages, currents, and resistances; the basic approach would be to use the calibrated voltage, resistance, and potentiometer to produce additional calibrated voltages and resistances, and then use those as required to take actual measurements. Obviously nowhere near as convenient as simply using a calibrated meter, but potentially more precise. |
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