Loading and Loading Errors
Resistive Loading
Consider the following circuit. VS can be considered to be a voltage source from a sensor measuring for
example temperature where 100 mV = 100° C. The sensor will be considered to have a linear output
voltage from 0 to 200° C. The voltage source has a source resistance specified as RS and in connected to
some other circuit (shown as a box) that could be an amplifier or display device. Let’s assume it is a
display device. This device has an input resistance RIN that can be viewed as a load RL for the voltage
source. The concern for this circuit is - what effect does RIN have on the value of the input voltage VIN
connected to the display device? The relationship between RS and RIN determines this effect.
Examples:
1. If RS = 100 Ω and RIN = 1000 Ω, the value of VIN can be determined by voltage divider.
VIN = RIN/(RIN + RS) x 100 mV = 1000/(1000 + 100) x 100 mV = 90.9 mV
So the actual input voltage to the display device now represents a temperature of 90.9° C – this is an
error in the temperature of 9.1%. The display will actually read 90.9° C.
Is this a practical concern? The answer to that will depend on exactly what temperature is being
measured. Is it a household room temperature controlled by a thermostat or a critical industrial process
temperature?
2. If the calculation is repeated with If RS = 100 Ω and RIN = 10 kΩ, then the value of VIN is
VIN = RIN/(RIN + RS) x 100 mV = 10 kΩ /(10 kΩ + 100) x 100 mV = 99.0 mV
This represents an error of about 1% in the temperature reading.
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Conclusion
There will be a minimal error in the source voltage when the input resistance RIN is much larger than the
source resistance RS .
For minimal error in the source voltage
RIN >> RS
Rule of Thumb
The error in the source voltage caused by loading can be estimated from
If RIN = 10 x RS then the error is about 10%
If RIN = 100 x RS then the error is about 1%
What is the error if RIN = 50 x RS ?
What must the RIN/RS ratio be for 0.1 % error?
Example
This example illustrates the loading effects that occur with a Non Inverting Op Amp buffer amplifier.
Previously it was determined that the voltage gain of this amplifier was 1, RIN = 200 GΩ, and ROUT,INV =
0.75 mΩ.
Loading Effect at VIN
The ratio of RIN / RSource = 200 GΩ/50Ω is a huge number (4 x 109:1) so there is no loading effect.
Loading Effect at VOUT
The ratio of RLoad / ROut = 200/0.8 mΩ = 250,000:1. Once again there is no loading effect.
This amplifier acts as a perfect buffer amp – the input and output
voltages are the same because RIN is very large and ROUT is very small.
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AC Capacitive Loading Errors
AC loading becomes a factor when a circuit such as an amplifier or display device has an input
capacitance CIN as well as the input resistance RIN. RIN and CIN are of course connected in parallel.
You should recall that the capacitive reactance XC associated with CIN is inversely proportional to
frequency – as frequency increases the capacitive reactance XC decreases.
Note: The capacitive reactance XC = 1/(2πfC)
Examples
3. Assume that RS = 100 Ω, RIN = 10 KΩ, C = 0.004 µF and the frequency is 100 Hz. The
capacitive reactance XC is 398 kΩ.
The input impedance of the circuit ZIN = RIN || XC,IN
For this example the capacitive reactance XC,IN is so much larger than RIN so the quantity ZIN = RIN || XC,IN
can be effectively approximated by RIN.
The loading error in this case is the same as the example 2 ( 1%) seen above
4. Assume that RS = 100 Ω, RIN = 10 KΩ, C = 0.1 µF and the frequency is 10 kHz. The capacitive
reactance XC is 159 Ω.
The input impedance of the circuit ZIN = RIN || XC,IN is 159 Ω at a phase angle of -89°
This impedance forms a voltage divider with RS = 100 Ω. The value of VIN can be calculated as 84.5 mV.
This represents a 15% error.
calculations not shown
Capacitive loading of an AC circuit becomes more problematic
at higher frequencies and with larger capacitances.
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Practical Concern – A Digital Voltmeter
A DVM has an RIN = 10 MΩ in parallel with a CIN ≈ 60 - 80 pF. This meter will have substantial errors in
the measured values if used to measure AC voltages at high frequencies.
Example
What is the input impedance of a DVM with RIN = 10 MΩ and CIN = 80 pF at 50 kHz?
The capacitive reactance of C is
XC = 1/(2πfC) = 39.8 kΩ
The input impedance ZIN (ZIN = XC || 10 MΩ) = 39.8 kΩ at a phase angle of -90°. Calculation details
not shown. The input impedance is essential the capacitive reactance of C. The 10 M Ω input resistance
is too large to be a factor (“something large in parallel with something small is the something small”).
Conclusion: The large input resistance of the DVM present for DC
measurements can become a small input impedance (ZIN = XC ||
10 MΩ) when measuring AC voltages at higher frequencies.
Application Note
Written by David Lloyd
Computer Engineering Program
Humber College
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