You need to measure the actual output voltage of a nominal 120V DC power supply
ID: 2248604 • Letter: Y
Question
You need to measure the actual output voltage of a nominal 120V DC power supply that has an effective internal resistance of 5k2. The voltage from the power supply is expected to lie in the range 114 126V. The "meter" will be a data acquisition system [DAQ] that has a maximum allowable input voltage of 5VDC. Exceeding this input voltage by even a small amount will cause damage. This "meter" has a very high input impedance and the current through the meter can be neglected. Specify the values of the resistances needed in a simple voltage divider to safely enable this measurement with a loading error that is 0.1% or less. Explain your choices and state the "gain" of your device. Specify resistors with one significant digit only (eg. if you need 53,612(2 you can only use 50,000 or 60,000) 1.Explanation / Answer
To Keep loading error 0.1% or less total resistance of two resistor of volatge resistor should be at least 1000 times the internal resistance of power supply.
Rps = intenal resistance of power supply = 5 K ohm
Rt = Total resistance of voltage divider circuit = 5 *1000 K ohm = 5 Mega Ohm
Rt = R1 + R2
R2 is the resistor across which output will be taken and connected to DAQ and max value allowed is 5 V.
Now maximum power supply voltage can reach is = 126 V
Therefore
5 V / R2 = 216 V / ( Rps + Rt )
R2 = ( Rps + Rt )*5 / 256
R2 =( 5 K + 1000 K )*5 / 256
R2 = 19.628 K
R2 = 20 K ...........1 ( resistance with one signifacant digit only )
Rt = R1 + R2
R1 = Rt - R2
R1 = 1000 -20 K
R1 = 980 K
R1 = 1000 K ..............2 ( resistance with one signifacant digit only )
Gain of the device K = R2/R1 as signal is attenuated in ratio R1/R2
K = 1000 / 20 = 50 ..........3
This is the required gain of the device
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