Potentiometer
Potentiometer
[1] A potentiometer (POT) is a 3-terminal variable resistor (made of 3 terminals) and a rotating analog device.
Embedded Files
The POT symbol is a resistor with an arrow.
Note: POT symbol is different in Europe (right).
Usages
Usages
It controls/divides output voltage from a fixed input, in a circuit by varying resistance (called variable resistance).
output voltage: amount of produced power
It controls electrical functions (volume, tone, and brightness) and lets a person control manually.
Fine-tuning and calibration: It adjusts devices precisely (tuning audio tools, set reference voltages, adjust sensor sensitivity.)
calibrate: to adjust a device to make accurate results
Position sensing: It measures rotation in joysticks, servos, and robotics by converting mechanical moves into electrical signals.
Diagram
Diagram
A POT's 2 outer terminals are outside/fixed/end terminals, which make a fixed resistor a certain value (here it's 10 kΩ).
Its middle terminal is the wiper arm/terminal.
One of its 2 end terminals connect to its power supply and the other to the ground.
An end terminal may connect to the input voltage (e.g. power) and the other, to ground. Wiper arm then adjusts voltage output between the end terminals by varying resistance.
diagram1
diagram2
diagram3
A POT's top voltage is the voltage applied to the 1st outer terminal (connected to the input voltage/power source; diagram3).
Its bottom voltage is the second outer terminal's voltage (connected to ground/reference point; diagram3).
Changing a POT, changes resistance, which affects voltage drop across the POT's 2 sides.
A POT has a resistive element, a strip/track made of carbon/metal oxide.
The wiper is a movable contact sliding along the resistive element to vary resistance.
A wiper divides resistive element into 2 parts, making a variable resistance, which adjusts output voltage
Resistance between the wiper and each end terminal varies as a wiper moves.
Voltmeter
Voltmeter
TBA
Schematics
Schematics
This schematic is an equivalent circuit, both being a POT (resistor with an arrow) at 50% of its max, for equal resistance on either side.
These 2 are the same circuits.
Equations
Equations
Voltage per section correlates to each's ratio to the POT's total value.
Voltage across voltage drops = input voltage
Voltage drop across potentiometer/voltage divider equation: Vout = Vin*[R2/(R1 + R2)]
Rbottom = Volmeter voltage/battery voltage = Pot resistance - Rtop
Rtop = Pot resistance - Rbottom
Vout = voltage drop/output voltage
Vin = input voltage
R1 = resistance of wiper to POT's one end
R2 = resistance of wiper to POT's other end
[Q2] Examples
[Q2] Examples
1# The voltage drops across R1 and R2 is each section's ration to the POT's total value: VR1 : VR2.
E[R1/(R1 + R2)] = E[R1/(RT)] = VR1
E[R2/(R1 + R2)] = E[R2/(RT)] = VR2
Calculate the voltage drop of the resistors if:
R1 = 8 kΩ
R2 = 12 kΩ
Answer:
Voltage across each section correlates to the ratio per section to the POT's total value.
2# This circuit has a 10 kΩ POT at 60%. What's its top and bottom voltages?
top = 10 k * 60% = 6 k
bottom = 10 k - 6 k = 4 k
Vtop= (4k/10k)12 = 4.8 V
Vbottom = (6k/10k)12 = 7.2 V
2.1# If the percentage is now 75%, what's the top and bottom resistance?
Rtop : 10 k * 0.75 = 7.5 kΩ
Rbottom: 10 - 7.5 = 2.5 kΩ = RT - top resistance
2.2# If the percentage changes to 75%, what's the top and bottom voltages, and the voltage drop?
Vtop = 2.5k/10k * 12 = 3 V
Vbottom= 7.5k/10k * 12 = 9 V
Voltage drops must add to input voltage: 3 + 9 = 12 V
3# If the voltmeter is 8 V, what can be found from the circuit?
Vtop = 4 V
Resistance is found with the ratios:
8/12 = Rbottom/10 k = (8 x 10k)/12 = 6.667 kΩ
4/12 = Rtop/10 k = (4 x 10k)/12 = 3.33 kΩ
Or
VT = 12 V; RT = 10 k Ω, so IT = 12 mA.
Rbottom = 8 V → R = 8/1.2 m = 6.667 kΩ
Rtop = 4 V → 4/1.2 m = 3.333 kΩ
4# E.g. If the voltmeter is 3.84 V, what can we find from the circuit?
VTop = 24 V - 3.84 V = 20.16 V
Rbottom = 3.84/24 = Rbottom/(2k + 500) = (3.84 x 2.5k)/24 = 400 Ω
Rtop = 2000 – 400 = 1.6 kΩ
Vtop/24 = Rtop/2.5 k = (1600x24)/2500 = 15.36 V
VRS = 24 – 15.36 – 3.84 = 4.87 V or Vrs/24 = 500/2.5 k = (500x24)/2.5 k = 4.8 V
Wheatstone bridge circuits
Wheatstone bridge circuits
It works by comparing the ratio of R1/R3 to R2/R4.
If the ratio is the same, it's a balance bridge circuit and voltage at Vab is 0 V.
If a value other than 0 volts is found at this point it is called an unbalanced bridge. This means there Would now be a voltage at Vab.
Same circuits drawn differently:
[Q2] Examples
[Q2] Examples
1# What's this circuit's value of R1 for a balanced bridge?
Often one resistor will receive an analog input from a temperature sensor.
R1 =
2# Each branch has 30 V. For this circuit, what's Vab?
Voltage divider rule:
(Va) V60 = (60/150)30 = 12 V
(Va) V90 = (90/150)30 = 18 V
(Vb) V130 = (130/290)30 = 13.45 V
(Vb) V160 = (160/290)30 = 16.55 V
We can either use down resistors:
Vab = Vb - Va = 16.55 – 18 = -1.45 V
...or down resistors:
Vab = -Vb + Va = -13.45 + 12 = -1.45 V
Vab = ~-1.448 V = -1.45 V
3# In this circuit, find Vab if:
Vin = 12 V
R1 = 32 Ω
R3 = 56 Ω
R2 = 24 Ω
R4 = 50 Ω
V32 = (32/88)12 = 4.36 V
V56 = (56/88)12 = 7.64 V
V24 = (24/74)12 = 3.89 V
V50 = (50/74)12 = 8.11 V
Vab = Vb - Va = 8.11 – 7.64 = 0.472 V = 470 mV
For the circuit shown, find Vab if:
Vin = 18 V
R1 = 30 Ω
R3 = 60 Ω Each branch has 18 V
R2 = 100 Ω
R4 = 500 Ω POT
Bottom = 500 x 0.3 = 150 Ω
Top = 500 x (1-0.3) = 350 Ω
@ 30% = 350 Ω
Note: "@ 30%" notation means the potentiometer is at 30% from one end--the top is 350 Ω and bottom is 150 Ω.
As only the top is part of the circuit, the value we use:
VR1 = [30/(30+60)]18 = 6 V
VR3 = [60/(30+60)]18 = 12 V
VR2 = [100/(100+350)]18 = 4 V
VR4 = [350/(100+350)]18 = 14 V
Vab = Vb – Va = 14 – 12 = -4 + 6 = 2 V
Loading effects
Loading effects
Loading effects are the influence that internal resistance of an ammeter or voltmeter or other measuring device has on a circuit.
Importance If loading effects are significant enough, results from the device may not be accurate enough as a value.
This circuit's expected voltages are:
VR1 = 8 V
VR2 = 16 V
The meter has a 10 MΩ standard internal resistance.
If it should say 8 V, but why it says 5.7 V? The meter reads the 6 MΩ resistor as if it's in parallel with the meter's 10 MΩ internal.
This is only an issue with large resistive values.
2# An ammeter's resistance is low.
On this circuit, the meter says a 5 Ω resistance.
Expected:
IT = 36 mA
IR1 = 6 mA
IR2 = 30 mA
The loading effects are more pronounced
if used to measure current in a branch with resistance value close to the meter's internal resistance value.
Quizzes
Quizzes
[Q2] Exercises - Canva
[Q3] Exercises - Canva
[Q4]
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