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Voltage/Current Dividers Voltage/current dividers Voltage and current dividers are easy to understand and use. They are so easy that it may seem not worth the bother of learning them as a separate techniques. But the divider methods, when combined with the equivalent resistances, may be the most used technique in electronics. Knowing how to use dividers will allow us to quickly recognize what is happening in a circuit and determine important voltages and currents. An engineer could certainly analyze and design circuits without having voltage and dividers in their “tool bag”, but they would be wasting lots of time writing unnecessary KVL and KCL equations. EE 201 voltage/current dividers – 1 Voltage divider + vR1 – iS Consider a portion of circuit that R1 + + has several resistors in series, like V R v S – 2 R2 the circuit at right. Suppose we – R3 want to find the voltage across R2. – vR3 + We could start by finding the current, which would be equal to the source voltage divided by the equivalent resistance of the string. VS iS = Req For the series string, Req = R1 + R2 + R3. Then the voltage across R2 is just R2 R2 vR2 = R2 ⋅ iS = ⋅ VS = ⋅ VS Req R1 + R2 + R3 The total voltage is divided among the resistor in the string. The fraction of the voltage across R2 is given by a simple resistor ratio. EE 201 voltage/current dividers – 2 The other resistor voltages are calculated just as easily. + vR1 – R2 iS R vR2 = ⋅ VS 1 + R1 + R2 + R3 + V R v S – 2 R2 – R3 R3 vR3 = ⋅ VS R1 + R2 + R3 – vR3 + The three divided voltages sum up to VS, as KVL insists. If we insert some numbers: VS = 15 V, R1 = 4.7 kΩ, R2 = 15 kΩ, and R3 = 10 kΩ. 4.7 kΩ v = (15 V) = 2.37 V R1 4.7 kΩ + 15 kΩ + 10 kΩ 15 kΩ v = (15 V) = 7.58 V It’s that easy. R2 4.7 kΩ + 15 kΩ + 10 kΩ 10 kΩ v = (15 V) = 5.05 V R3 4.7 kΩ + 15 kΩ + 10 kΩ EE 201 voltage/current dividers – 3 Current divider Same idea, but with parallel + resistors dividing a current. IS vS R1 R2 R3 Suppose we want to know the – iR1 iR2 iR3 current through R2. We could start by finding the voltage, which would be equal to the source current multiplied by the equivalent resistance of the parallel resistors. vS = IS ⋅ Req 1 For the parallel combination, Req = 1 + 1 + 1 R1 R2 R3 Then the current through R2 is 1 vS Req R2 iR2 = = ⋅ IS = 1 1 1 ⋅ IS R2 R2 + + R1 R2 R3 As in the case of the voltage divider, the fraction of the current through one resistor is determined by a simple ratio based on resistor values. But in the current case, resistor inverses are used. EE 201 voltage/current dividers – 4 The other resistor currents are calculated just as easily. 1 R1 + iR1 = ⋅ IS 1 + 1 + 1 I v R R R R1 R2 R3 S S 1 2 3 1 – iR1 iR2 iR3 R3 iR3 = ⋅ IS 1 + 1 + 1 R1 R2 R3 The three divided current sum up to IS, as KCL insists. If we insert some numbers: IS = 15 mA, R1 = 2.2 kΩ, R2 = 3.3 kΩ, and R3 = 6.8 kΩ. 1 2.2 kΩ iR1 = 1 1 1 (15 mA) = 7.54 mA 2.2 kΩ + 3.3 kΩ + 6.8 kΩ 1 3.3 kΩ iR2 = 1 1 1 (15 mA) = 5.02 mA 2.2 kΩ + 3.3 kΩ + 6.8 kΩ 1 6.8 kΩ iR3 = (15 mA) = 2.44 mA 1 + 1 + 1 EE 201 2.2 kΩ 3.3 kΩ 6.8 kΩ voltage/current dividers – 5 In many instances, the combination of dividers with equivalent resistances provides for fast calculation of voltages and currents. R1 Example 1 1 kΩ In the circuit at right, find vR4. R2 Since R3 and R4 are in parallel, they 470 Ω + + have the same voltage and we can V R R v S – 3 4 R4 use the parallel equivalent. We can 12 V 560 Ω 820 Ω – also find the parallel equivalent of R5 R1 and R2. R R 330 Ω R = 3 4 = 333 Ω 34 R + R 3 3 R12 R1 ⋅ R2 R12 = = 320 Ω 320 Ω R1 + R2 + + Then use a voltage divider on the V R v S – 34 R4 simplified circuit. 333 Ω – R R 3 v = 34 ⋅ V = 4.065 V R4 R + R + R S 330 Ω EE 201 12 34 5 voltage/current dividers – 6 Example 2 In the circuit at right, find iR2. R R R Since R1 and R2 are in series, 1 4 5 39 Ω 82 Ω 56 Ω they have the same current, IS R3 and we can use the series 100 Ω 2.2 A R i equivalent. We can also find 2 R2 12 Ω R6 the equivalent resistance of the 120 Ω branch with R4 - R5 - R6. R12 = R1 + R2 = 51 Ω R4 ⋅ R5 R456 = + R6 = 153 Ω I R R R R4 + R5 S 12 2 456 51 Ω 100 Ω 153 Ω 2.2 A iR2 Then use a current divider on the simplified circuit. 1 R12 iR2 = ⋅ IS = 1.19 A 1 + 1 + 1 R12 R3 R345 EE 201 voltage/current dividers – 7 Example 3 2.2 kΩ 3.3 kΩ R1 + R3 + Find vR2 and vR4 + V R v R v in the circuit. S – 2 R2 4 R4 12 V 3.3 kΩ – 3.3 kΩ – We solve this using the voltage divider calculation twice in succession. First we find vR2 using a voltage divider formed by R1 and the equivalent resistance of R2 in parallel with the series combination of R3 and R4. R + R3 R1 + + 3 + + R R v R V R v v R v 234 2 R2 4 S – 234 R2 R2 4 R4 – – – – R234 Then the voltage R = R R + R v = ⋅ V = 6 V 234 2 ( 3 4) R2 S vR2 is divided R234 + R1 R R = (3.3 kΩ) (6.6 kΩ) between 3 and 4. R4 = 2.2 kΩ vR4 = ⋅ vR2 = 3 V R3 + R4 EE 201 voltage/current dividers – 8 Example 4 R2 iR2 Find iR3 in the circuit. 22 Ω I R Similar to example 3, we can S 1 0.8 A 33 Ω cascade dividers to find the R3 iR3 R4 current in two steps. 22 Ω 22 Ω IS splits between R1 and the branch with R2 - R3 - R4. To find the current through R2, we use the equivalent resistance of that branch, which forms a current divider with R1. Then iR2 is divided between R3 and R4. R2 R2 iR2 22 Ω IS R1 R234 iR2 R234 0.8 A 33 Ω 33 Ω R3 R4 R3 iR3 R4 22 Ω 22 Ω 1 1 R234 R3 R234 = R2 + R3 R4 iR2 = ⋅ IS = 0.4 A iR3 = ⋅ IR2 = 0.2 A 1 + 1 1 + 1 R1 R234 R3 R4 = 3.3 kΩ EE 201 voltage/current dividers – 9 Example 5 Find the voltage vx indicated R1 R3 1 kΩ in the circuit at right. 2.2 kΩ + + v – V x S – 110 V By KVL, vx = vR2 – vR4. R2 R4 100 Ω We see that R1 and R2 form a 3.3 kΩ voltage divider splitting VS. The same for R3 and R4. Using voltage dividers. R2 vR2 = ⋅ VS = 66 V R R R1 + R2 1 3 R4 + + v – v = ⋅ V = 10 V V x R4 S S – R3 + R4 + + Then R2 vR2 vR4 R4 vx = 66 V – 10 V = 56 V. – – EE 201 voltage/current dividers – 10 Example 6 Find the voltage vx indicated in the circuit at right. R1 R3 1 kΩ 2.2 kΩ As with the previous example, KVL + vx – IS tells us that vx = vR2 – vR4. 60 mA We see that the series combination R2 R4 100 Ω R1 + R2 forms a current divider with 3.3 kΩ the series combination R3 + R4, splitting IS between the two branches. Using current dividers. 1 R1 + R2 iR1 iR3 R1 R3 iR1 = 1 1 ⋅ IS = 10 mA R + R + R + R + v – 1 2 3 4 I x 1 S R + R + + i = 3 4 ⋅ I = 50 mA R3 1 1 S R v v R + 2 R2 R4 4 R1 + R2 R3 + R4 – – Then vx = iR1R2 – iR3R4 = 28 V. EE 201 voltage/current dividers – 11 Example 7 In the circuit at right, the switch can be opened or closed to control the R R voltage across R3. When the switch 1 2 + + V R v is closed (R1 shorted out), vR3 is S – 3 R3 twice as big as the case when the – switch is open (R1 not shorted.) How is R1 related to R2 + R3? There are several approaches to answering this question, but using voltage dividers is a convenient method. With the switched closed, R1 is shorted out and vR3 = 2vR′ 3 R3 vR3 = ⋅ VS R2 + R3 R3 2R3 ⋅ VS = ⋅ VS With the switch open, R1 R2 + R3 R1 + R2 + R3 is part of the divider: R1 + R2 + R3 = 2 (R2 + R3) R3 vR′ 3 = ⋅ VS R1 + R2 + R3 R1 = R2 + R3 EE 201 voltage/current dividers – 12 Example 8 In the circuit at right, the two switches can be opened or S1 S2 closed to control the current IS R3 iR3 through R3. Calculate the 3 kΩ 100 mA R1 R2 current through R3 for all 1.5 kΩ 3 kΩ combinations of the switches being open closed.
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