How to Choose the Right Resistor for Your Circuit Design

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Selecting the appropriate resistor is a fundamental step in circuit design. Below is a concise selection guide suitable for beginners and engineers:

1.Clarify the purpose of the resistor

How to Choose the Right Resistor for Your Circuit Design

First, it is essential to understand the role of resistors in a circuit, as different applications have varying requirements for resistance value, power rating, and precision:

1.1Partial pressure

Two series resistors distribute voltage proportionally, which is suitable for obtaining an intermediate level or reference voltage.

  • Typical scenarios: Providing reference voltage for analog signals and ADC voltage scaling

  • Use two series resistors to divide the 12V voltage into 5V and 7V.

1.2Current Limiting

By using Ohm's Law V = IR, when the voltage is fixed, the current is limited by setting a resistor.

  • Typical scenarios: LED current limiting, power input current limiting

  • Use a resistor to limit the current flowing through the LED to prevent it from being too bright or burning out.

1.3shunt resistor

How to Choose the Right Resistor for Your Circuit Design

It is often used in conjunction with an operational amplifier or ADC to measure the magnitude of the current flowing through a circuit.

  • Typical scenarios: battery management system, motor drive, power monitoring

  • Calculate the current by detecting the voltage drop across the resistor (I = V/R)

1.4 Bias

Provide a stable DC bias operating point in an amplifier or transistor circuit.

  • Typical scenarios: transistor base resistance, operational amplifier resistor network

  •  Example: Set the base current required for the transistor to enter the amplification state.

2.Calculate the required resistance value (unit: ohm)

2.1Using Ohm's Law:

V = IR → R = V/I

2.2Example:

  • To limit the LED current (working voltage is 2 v), the power supply for 5 v, target current to 20 ma:

  • The voltage across the resistor = 5V - 2V = 3V

  • R = 3V / 0.02A = 150 Ω

3.Determine the power rating

3.1The power calculation formula is as follows:

P = V × I = I² × R = V² / R

It is recommended to choose a power rating that is twice or more of the actual power consumption to enhance reliability.

3.2Example:

  • 150 Ω resistance, current 20 mA:

  • P = 0.02² × 150 = 0.06 W

  • Resistors of 0.125W or 0.25W should be selected

4.Select the appropriate tolerance (accuracy)

Tolerance Precision Description Common Applications
±5% General precision Consumer electronics
±1% Medium precision General-purpose circuits, instrumentation
±0.1% ~ ±0.5% High precision Analog circuits, measuring instruments

5.Select the packaging form

According to the installation and the current size to choose encapsulation: SMD (SMD) resistor:

suitable for automatic production, commonly used packaging, such as 0603, 0805, 1206

plug-in (DIP) resistance: for manual welding and prototype test power

metal casing resistance: used in the occasion of big electric current or high calorific value

6.Consider the environment and special parameters

Operating temperature range: such as - 55 ° C to + 125 ° C

temperature coefficient (TCR) : requires high stability when choosing low TCR resistance (e.g., plus or minus 50 PPM / ° C)

whether to non-inductive resistance: high frequency or pulse application should avoid to use winding resistance

Frequently asked questions

FAQ Q1: choose resistance will bigger more secure?

A: In current-limiting situations, if the resistor is selected too large, the current may be insufficient, affecting the functionality. Choosing too small may burn out the components. It should be selected based on the calculated value, with a slight margin.

Q2: how high power resistance does not burn?

A: First, calculate using P = I²R, and then select a model with a power rating of twice or more.

Q3: how do the LED current limiting resistor to choose?

A: It can be calculated by subtracting the LED voltage from the supply voltage and then dividing by the target current. A conventional 1/4W with an accuracy of ±5% is sufficient.

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