Voltage Divider

4 Key excerpts on "Voltage Divider"

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  eBook - ePub

  Electronic Servicing and Repairs

  • Trevor Linsley(Author)
  • 2014(Publication Date)
  • Routledge

    (Publisher)

  5

  ELECTRONIC CIRCUITS IN ACTION

  _____    

  Voltage Divider

  In Chapter 8 we consider the distribution of voltage across resistors connected in series. We find that the supply voltage is divided between the series resistors in proportion to the size of the resistor. If two identical resistors were connected in series across a 12 V supply as shown in Fig. 5.1(a) both common sense and a simple calculation would confirm that 6 V would be measured across the output. In the circuit shown in Fig. 5.1(b) the 1 kΩ2 and 2 kΩ2 resistors divide the input voltage into three equal parts. One part, 4 V, will appear across the 1 kΩ 2 kΩ resistor and two parts, 8 V, will appear across the 2 resistor. In Fig. 5.1(c) the situation is reversed and, therefore, the voltmeter will read 4 V. The division of the voltage is proportional to the ratio of the two resistors and, therefore, we call this simple circuit a Voltage Divider or potential divider. The values of the resistors R 1 and R 2 determine the output voltage as follows:

  For the circuit shown in Fig. 5.1(b)

  FOR the circuit shown in Fig. 5.1(c)

   

  Figure 5.1 Voltage Divider circuit.

  EXAMPLE 1

  For the circuit shown in Fig. 5.2 calculate the output voltage

   

  Figure 5.2 Voltage Divider circuit.

  EXAMPLE 2

  For the circuit shown in Fig. 5.3(a) calculate the output voltage. We must first calculate the equivalent resistance of the parallel branch

  The circuit may now be considered as shown in Fig. 5.3(b)

  Voltage Dividers are used in electronic circuits to produce a reference voltage which is suitable for operating transistors and integrated circuits. The volume control in a radio or the brightness control of a CRO requires a continuously variable Voltage Divider and this can be achieved by connecting a variable resistor or potentiometer as shown in Fig. 5.4 . With the wiper arm making a connection at the bottom of the resistor, the output would be 0 V. When connection is made at the centre, the voltage would be 6 V, and at the top of the resistor the voltage would be 12 V. The voltage is continuously variable between 0 V and 12 V simply by moving the wiper arm of a suitable variable resistor such as those shown in Fig. 2.3

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  Physics of Electronic Materials

  Principles and Applications

  • Jørgen Rammer(Author)
  • 2017(Publication Date)
  • Cambridge University Press

    (Publisher)

  For voltages of interest, the current is proportional to the small voltage, Ohm’s law, 102 Electric Circuit Theory R I I Figure 5.3 Electric circuit symbol for a resistor. V = RI (5.7) where the positive proportionality constant is called the resistance of the resistor. The resistance depends on the resistor’s geometry and the strength of scattering processes inside it. A simple model of a resistor was discussed in Section 4.8 . The electric circuit symbol for a resistor is often chosen as a rectangular box or as depicted in Figure 5.3 . The unit of electric potential and thereby that of voltage is, in SI units, the volt (V). Charge times electric potential has the dimension of energy, which is measured in joule (J), defined as newton meter (N m), the newton (N) being the unit of force equal to kg m/s 2 . A charge of 1 C at a potential of 1 V (relative to a reference point) has an electric potential energy of 1 J. The ohm ( ) is the unit of electrical resistance, equaling one volt per ampere (V/A). A resistor is typically made from powdered carbon mixed with insulating material such as ceramics, the mixture held together by resin and wrapped in a plastic covering with two metallic leads sticking out at opposite ends. Its resistance then depends on the fraction of good conducting material, the amount of carbon. 5.3 Voltage Divider In an electric circuit, definite output voltages are needed at terminals in response to input voltages, and to achieve this resistors are put to essential use. Consider a Voltage Divider as depicted by the circuit symbols in Figure 5.4 . In Figure 5.4 we have introduced a circuit symbol, at the bottom, which we display separately in Figure 5.5 and is referred to as ground (or earth in the UK), but is in practice just a large good conductor, a charge reservoir, say a metal casing, so that its electric potential stays at the same constant value, which is taken as voltage reference, i.e.

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  Introduction to High Power Pulse Technology

  • S T Pai, Qi Zhang;;;(Authors)
  • 1995(Publication Date)
  • WSPC

    (Publisher)

  Diagnostics 279 T 0 oscilloscope Ri > Ri (a) Fig.8-1 (a) Idealised resistive Voltage Divider; (b) Equivalent circuit of a real resistive Voltage Divider. 8-2 Voltage Measurement (a) Resistive Voltage Divider Owing to its structural simplicity and wide-range applicability, the resistive Voltage Divider is one of the most widely used devices for voltage measurements. The basic structure of an ideal Voltage Divider is shown in Fig.S-l(a). It consists of two resistors R and R 2 which respectively serve as the high voltage arm and low voltage one. The low voltage V 2 across the resistor R 2 is directly measured and the unknown voltage V] is determined by using the given values of R, R 2 and V 2 in conjunction with the formula V 2 = Hi +R 2 (8-1) This description is only for an ideal situation. In actual cases, circuit inductance and stray capacitance cannot be totally ignored. The equivalent circuit of a real resistive Voltage Divider should be approximately represented by that shown in Fig.8-l{b) in which the circuit inductance V and the stray capacitances C s , p g have been included. Analysis of a circuit as such is rather difficult. For simplicity, let's first assume that the stray capacitances are negligible, the circuit can then be represented by that shown in Fig.8-2(a). The relationship between the measured voltage V 2 and the unknown voltage V in this case is 280 fit(roifuc(ion to High Paver PvUe Technology V, v, Input waveform 'Output waveform (a) (b) Fig.8-2 (a) Equivalent circuit with inductance included; (b) Actual output wave-form. where L g is the equivalent total inductance of the circuit.

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  Essentials of Electrical and Computer Engineering

  • J. David Irwin, David V. Kerns, Jr.(Authors)
  • 2022(Publication Date)
  • Wiley

    (Publisher)

  Note the power balance that is exhibited, and the fact that the 12 V source is actually absorb- ing power. VOLTAGE DIVISION Voltage division is a technique we will employ on numerous occasions in our study of circuit analysis. It is a very simple technique that indicates the manner in which a voltage divides between resistors in series. Consider the circuit in Figure 2.14. Applying KVL, yields the following equation: v(t) = i(t)R 1 + i(t)R 2 and thus i(t) = v(t)∕(R 1 + R 2 ) But v R 1 = i(t)R 1 and v R 2 = i(t)R 2 So, v R 1 = [R 1 ∕(R 1 + R 2 )] v(t) v R 2 = [R 2 ∕(R 1 + R 2 )] v(t) These equations indicate that the source v(t) divides between the two resistors in direct proportion to the total resistance R 1 + R 2 . + – v(t) R 1 V R 1 R 2 i(t) + – V R 2 + – FIGURE 2.14 Circuit used to describe voltage division Example 2.7 Given the circuit in Figure 2.15, let us determine the voltages V 1 and V 2 . Applying voltage division V 1 = [2∕(2 + 4 + 6)] 24 = 4 V V 2 = [(4 + 6)∕(2 + 4 + 6)] 24 = 20 V And, of course, 4 V + 20 V is equal to the source voltage. We could also calculate the voltage across the 6 Ω resistor and find that it is 12 V. Note carefully, that voltage division is simply equivalent to calculating the current and then applying Ohm’s law – it is just done in one step. 30 CHAPTER 2 The Basic Laws of Circuit Analysis FIGURE 2.15 Circuit used in Example 2.7 + – 24 V V 1 + – V 2 + – 2 Ω 4 Ω 6 Ω Example 2.8 Let us apply voltage division to determine the voltage V x in the network in Figure 2.16. FIGURE 2.16 Circuit used in Example 2.8 + – + – V X + – 3 Ω 2 Ω 1 Ω 24 V 12 V The net sum of the voltages that causes a clockwise current is 24 − 12 = 12 V. The sum of all the resistors in the network is 1 + 2 + 3 = 6 Ω. Therefore, the voltage across the 3 Ω resistor is V x = [3∕6] 12 = 6 V KVL provides a quick check for this value. The KVL equation is −24 + 1I + 2I + 3I + 12 = 0 which yields a current of 2 A, and V x = 3I = 6 V.

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