7. Voltage and Current Division

7/31/2019 7. Voltage and Current Division

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Lesson 7(from Nilsson and Riedel, Ch. 3)

Voltage division

Adding a load Example 3.2 (resistor tolerances)

Current division

Example 3.3 Assessment Problems 3.2, 3.3


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The Voltage-Divider Circuit

At times (especially in electronic circuits) developing more than onevoltage level from a single voltage supply is necessary. One way to dothis is to use a voltage-divider circuit, which consists of two resistors inseries:

Note that the essential function of the two resistors is to divide thevoltage from the voltage source (vs) into two voltages: the voltage v1across resistor R1 and voltage v2 across resistor R2 (where, of course,vs = v1 + v2 by KVL). By putting in different resistors for R1 and R2, wecan adjust the values of v1 and v2.


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The Voltage-Divider Circuit, cont.

We want to figure out the values of v1 and v2 (assuming that vs, R1, and

R2 are given):

1. Apply KVL and OL:

2. Use OL for R1 and then substitute the result for i from #1:

21

2121 or,RR

viiRiRvvv ss

+

=+=+=

21

11

21

11RR

RvR

RR

viRv s

s

+

=

+

==


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The Voltage-Divider Circuit, cont.

Similarly, use Ohms law for v2

:

21

22

21

22RR

RvR

RR

viRv s

s

+

=

+

==

21

11

RR

Rvvs

+

=

We see that v1 and v2 are fractions of vs, i.e., they will always be lessthan vs. Note that if R1 > R2, then v1 > v2, and vice versa. In otherwords, the bigger R1 is compared to R2, the more voltage it will haveacross it (and vice versa).

Also note that if we want a specific value for v2, there are an infinitenumber of possibilities for R1 and R2. For example, if vs = 15 V and wewant v2 to be 5 V, then (R2)/(R1+R2) = 1/3, which is satisfied wheneverR2 = 0.5*R1. (Although other factors, such as power-loss

considerations, may further restrict the best choices for R1 and R2.)

We have found:


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Adding a Load

Consider taking the voltage divider circuit and adding a resistor RL inparallel with it, as below. The resistor acts as a load on the voltagedivider circuit. A load on any circuit consists of one or more circuitelements that draw power from the circuit.

Given the situation above, wed like to know what theoutput voltage vo will be.


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Adding a Load, cont.

[ ] sL

L

L

eqs

eq

eq

o

vRRRR

R

RR

RRRv

RR

Rv

221

2

2

2

1

)/(1

where,

++

=

+

=

+

=

To find the output voltage vo, first note that R2 and RL are parallel, so wecan combine them into a single, equivalent resistor Req.

Then we are left with the basic voltage divider circuit, with R1 as the topresistor and Req as the bottom resistor. So we have:

Note that as long as RL R2, the voltage ratio vo/vs is essentially

undisturbed by the addition of the load.


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Example 3.2

Real resistors usually arent manufactured to exact values, but havecertain tolerances. For the voltage divider circuit below, assume thatthe resistors have a tolerance of 10% in their values. Find the

maximum and minimum values of vo (the voltage across R2), given thepossible variations in the resistors. (Answer on next slide.)

so vRR

Rv

21

2

+

=


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Example 3.2, cont.

so vRR

Rv

21

2

+

=

From the voltage divider equation (above), we see that vo will be

largest when R2 has its largest value (10% high in this case, or 110k) and R1 its smallest value (10% low in this case, or 22.5 k).

And vo will be smallest when R2 is smallest (10% low, or 90 k) andR1 is largest (10% high, or 27.5 k).

Running the numbers gives the minimum vo = 76.60 V and themaximum vo = 83.02 V.


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The Current-Divider CircuitA current-divider circuit uses two resistors in parallel to divide thecurrent flow into two branches (i.e., in the diagram below, the current isfrom the current source on the left divides into current i1 through R1 andcurrent i2 through R2):

We want to find v, i1, and i2. First we apply OL to R1 and R2:

2211 and RivRiv ==

Then we note that because R1 and R2 are parallel, we can replace

them with a single Req. The voltage across Req will still be v, and thecurrent through it has to be is, so we apply OL to Req to get:

21

21where,

RR

RRRRiv eqeqs

+

==


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The Current-Divider Circuit, cont.

So far we have found three equations for v:

,,2211

RivRiv ==21

21where,RR

RRRRiv

eqeqs +

==

Combining the first and third equations, and the second and thirdequations, respectively, we get:

ss iRR

Rii

RR

Ri

21

1

2

21

2

1 ,+

=

+

=


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The Current-Divider Circuit, cont.

We have found:

ss iRR

Rii

RR

Ri

21

1

2

21

2

1 ,+

=

+

=

Note the physical intuition behind these equations: We expect thatwhen R1 is larger than R2, more current will flow through R2 than R1(because R2 is the path of lesser resistance). On the other hand, when

R2 is larger than R1, we expect more current will flow through R1. Andthats what the equations tell us.


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Example 3.3

Find the power dissipatedin the 6 resistor.

Strategy:

1. To find the power well need to know the current through the 6 resistor.

2. If we can find the current through the 1.6 resistor, then we canuse current division to find how much of that current flows throughthe 6 resistor.

3. To find the current through the 1.6 resistor, well try simplifying

the right side of the circuit using series-parallel simplifications.


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Example 3.3, cont.

A8)10(

416

16

21

1=

+

=

+

= so i

RR

Ri

We start by simplifying the parallel 4 and 6 resistors into anequivalent 2.4 resistor (from the formula for two parallel resistors).

Then we see that we have a 1.6 resistor and a 2.4 resistor inseries, which is equivalent to a 4 resistor. We thus have simplifiedthe circuit to the following:

Using the current-division formula (the 10 A is divided between the 16 and 4 resistors) gives us the labeled io:


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Example 3.3, cont.

So far we have simplified the original circuit on the left to the circuit onthe right (replacing the 1.6 , 4 , and 6 resistors with an equivalent4 resistor) and have found that io = 8 A.

Next we note that the 10 A in the left (original) circuit is divided between

the 16 resistor and the 1.6 resistor. The equivalent circuit on theright has to act the same way. In other words, the 10 A current willdivide into the 16 and 4 resistors, with the 16 resistor currentbeing the same as in the original circuit. Therefore the current through

the 4

resistor in the right circuit must be the same as the currentthrough the 1.6 resistor in the original circuit (i.e., 8 A).


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Example 3.3, cont.

We have found that the current through the 4 resistor in the rightcircuit must be the same as the current through the 1.6 resistor in theoriginal circuit. And we have calculated it to be 8 A.

In the original circuit, this 8 A flowing through the 1.6 resistor gets

divided between the 4 and 6 resistors. The current through the 6 resistor is then as follows (by current division) and the correspondingpower is:

W44.61)6()2.3(andA,2.3)8(46

4 26

===

+

=

pi


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Summary of Voltage-Divider and

Current-Divider Equations

Voltage divider:

sv

RR

Rv

21

22

+

=

svRR

R

v21

1

1+

=

si

RR

Ri

21

2

1 +

=

si

RR

Ri

21

1

2+

=

Current divider:

(Note the similarities and differences.)


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Consider the circuit above for a practice exercise. We will outline themethods that we would use to answer various questions about it. (Its

recommended that you try to answer them yourself before moving to thenext slide, where the answer is. Then, once you check the method ofsolution, try to work through it to get the numeric value. Numeric answersare given in the textbook.)

(a) Find the no-load value of vo (the value when RL is not there, i.e., whenits an open circuit).

Assessment 3.2


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(a) Find the no-load value of vo. Solution: Simply use voltage division.

Assessment 3.2, cont.


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(a) Find the no-load value of vo. Solution: Use voltage division.

(b) Find vo

when RL

is 150 k.



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(b) Find vo

when RL

is 150 k. Solution: First calculate the equivalentresistance of the parallel 75 k resistor and RL. Then use voltagedivision to find the voltage across the equivalent resistance (which willbe v0).



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(b) Find vowhen R

Lis 150 k. Solution: First calculate the equivalent

resistance of the 75 k resistor and RL. Then use voltage division.

(c) How much power is dissipated in the 25 k resistor if the load terminalsare accidentally short-circuited (i.e., RL becomes 0).



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(b) Find vo

when RL

is 150 k. Solution: First calculate the equivalentresistance of the 75 k resistor and RL. Then use voltage division.

(c) How much power is dissipated in the 25 k resistor if the load terminalsare accidentally short-circuited? Solution: Short-circuiting the loadterminals effectively removes the 75 k resistor, i.e., the voltage acrossthe 25 k resistor is the full 200 V from the source. (That is, all thecurrent will bypass the 75 k resistor and take the short-circuit path. Iftheres no current in the 75 k resistor, then by OL the voltage across itis 0.) So we can use OL for the 25 k resistor to find the current through

it and then p = i2R for the power.



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(d) What is the max power dissipated in the 75 k resistor? (Tip: Thinkabout the conditions under which max power will occur.)



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(d) What is the max power dissipated in the 75 k resistor? Solution: Maxpower occurs with max current (i2R) through the resistor, or equivalently

with max voltage (v2/R) across it. Both occur when RL is infinite (i.e.,open circuit or no load). So use p = v2/R where the value of v is the vovalue calculated in part (a).



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Heres another good practice problem. We will outline the solution as withthe previous problem.

(a) Find the value of R that will cause 4 A of current to flow through the 80 resistor above.

Assessment 3.3


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(a) Find the value of R that will cause 4 A of current to flow through the 80 resistor above. Solution: Use the current division formula (with the 40

and 80

resistors in series equivalent to a 120

resistor).



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(a) Find the value of R that will cause 4 A of current to flow through the 80 resistor above. Solution: Use current division (with the 40 and 80

resistors equiv. to 120

).(b) How much power will the resistor from part (a) be dissipating?



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).(b) How much power will the resistor from part (a) be dissipating? Solution:

First calculate the current through R using current division. Then use p =i2R.



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).(b) How much power will the resistor from part (a) be dissipating? Solution:

First calculate the current through R using current division. Then use p =i2R.

(c) How much power will the current source generate, given the value of Rfrom part (a)?



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).(b) How much power will the resistor from part (a) need to dissipate?

Solution: First calculate the current through R using current division.Then use p = i2R.

(c) How much power will the current source generate, given the value of Rfrom part (a)? Solution: Use KVL around the outer loop to solve for theunknown voltage across the current source, and then use p = -vi to findthe power (the minus sign assumes that we assign + polarity to the top

of the current source and polarity to the bottom, i.e., the PSC).



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End of Lesson 7

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7. Voltage and Current Division