# Notes - See [[ECE2k7 Lab1 Manual.pdf]] - Measure voltage in parallel w/ component (probe on each end of it) & current in series w/ component -> must "break" circuit first - In resistor, each color is a digit 1 - 9 & each stripe is a place value # I. DC voltage measurement with a Digital Multimeter > [!table] Measured voltages for positive and negative DC outputs | Set Voltage (V) | Measured Voltage (VDC) | %error | | --------------- | ---------------------- | ---------- | | +13 | 13.001 | 0.0000769% | | -13 | -13.004 | 0.0003076% | To get a 35V output of a DC supply, we connected two of the power supplies in series by connecting the ground of PSU1 (5V) to the power of PSU2 (30V), then using the power of PSU1 with the ground of PSU2 to measure the voltage using the multimeter --- >[!table] Measured voltage for series-connected power supplies, 35V > | Measured Voltage [VDC] | % Error | | ---------------------- | ------- | | 35.005 | 0.143% | --- > [!table] Voltage measurement comparison using PSU display versus DMM as reference > | **Set Voltage (V)** | **Measured Voltage (V)** | **%Error** | | ------------------- | ------------------------ | ---------- | | 5.125 | 5.1257 | 0.000137 | | 5.12435 | 5.125 | 0.000127 | The errors between using the PSU and the DMM displays (both approximately 0.013%) are very close, well within 5% of each other. This occurs because both instruments have high precision in this voltage range, with the small differences representing calibration differences. The DMM is generally considered more accurate for voltage measurements as it is specifically calibrated for measurement purposes, whereas the power supply's display is primarily for monitoring output. --- Current flowing through $220\ohm$ resistor w/ voltage of 5V given by: $I=\frac{V}{R}=\frac{5}{220}=0.023A$ --- When the PSU is adjusted to output 5V w/ 15mA current limit & a 220$\ohm$ resistor is placed as a load, the circuit will only generate a current of 0.015A, because while it’s capable of producing a current of 0.023A, it is limited by the given current limit, 0.015A. --- When the 220$\ohm$ resistor is attached to the outputs of the supply, the voltage displayed on the multimeter is less than that originally output by the power supply, 5V, to 3.142V. This is because the current is limited to 0.015A, so we get the equation $V=IR=.015(220)=3.3V$, which is in the range of error of the number displayed on our multimeter, $3.142V$. In other words, current limit is exceeded, so the voltage defaults to that defined by Ohm’s Law --- >[!table] Voltage and current measurements comparing power supply and DMM readings > | %%null%% | **Voltage [V]** | **Current [A]** | | ---------------- | ----------- | ----------- | | Power Supply | 5.000 | 0.015 | | DMM | 5.0004 | 0.0002 | To calculate the minimum amount of resistance needed s.t the current lmit is not exceeded, we use $V=IR\to R=\frac{V}{I}=333.333\ohm$. After using this value as a resistor in the circuit, we indeed get a voltage of 5.000V on the power supply and ~5.005V on the multimeter. We should rely on the DMM for measuring voltage more because the DMM is calibrated specifically for measurement while the power supply is used for monitoring. We should rely on the output of the power supply for current because the circuit is not set up in series to measure it. # II. Verification of Ohm's Law >[!table] Measured Voltage and Current Values for 1kΩ resistor, Varying Voltages >| Set Voltage [V] | Measured Current [mADC] | | --------------- | -------------------- | | 0 | 0 | | 0.5 | 0.4966 | | 1 | 0.9754 | | 1.5 | 1.4877 | | 2 | 1.9833 | | 2.5 | 2.4802 | | 3 | 2.9767 | | 3.5 | 3.4768 | | 4 | 3.9722 | | 4.5 | 4.4697 | | 5 | 4.9679 | ---- >[!figure] Measured Current due to Set Voltage w/ Line of Best **Fit** ![[f.png|center|500]] The value of the line of best fit is given by $y=0.9944x-0.55$ with an $R^2$ value of 1 (likely rounded from 0.9999), showing the very strong relationship between resistance, current, and voltage. The slope from the line of best fit represents the conductance of the resistor, which is the reciprocal of resistance. From Ohm's Law, $V=IR\to I=\frac{1}{R}V$, so the slope $m=\frac{1}{R}=\frac{0.9944mA}{V}=0.0009944S$. To find the resistance from the slope, $R=\frac{1}{m}=1005.6\Omega=1.006k\Omega$ . This value represents the resistance of the resistor, which closely matches the expected nominal value of $1k\ohm$, with only a 0.56% deviation. --- At V = 5.000, we get a measured resistance of $0.99396k\Omega$ , an expected resistance of $1k\Omega$, and a resistance of $1.006k\Omega$  according to the line of best fit. The largest error is between the expected resistance and measured resistance, with a percent error of 0.6%, which is within the reasonable range of error because the resistor has a tolerance band of 1%, meaning the resistor can vary in value up to 1% from its original value, greater than our calculated percent error. --- Based on the tolerance band of the resistor (brown = $\pm1\%$), we can determine the acceptable range of resistance values. The nominal resistance is $1000\Omega$, so the tolerance is $R_{tolerance}=10\ohm$, giving an acceptable range of $990\ohm<R<1010\ohm$. Both the line of best fit resistance and DMM measured resistance fall within this acceptable range, confirming that the resistor is operating within its specified tolerance.