Voltage/current dividers - Gary Tuttle's ISU web...
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EE 201 voltage/current dividers – 1 Voltage/current dividers Trivial to understand, but still very useful. Use in combination with equivalent resistances to quickly find a particular voltage or current in a circuit. + – V S R 1 R 2 R 3 – v R2 + Want to know v R2 . Easily solved with KCL, KVL, & equivalent resistances. i S L 6 = 9 6 5 HT = 9 6 5 + 5 + 5 Y 5 = L 6 5 = 5 5 + 5 + 5 9 6 That’s it. The total voltage across a series string is divided among the resistors according to a simple ratio. Voltage divider
Transcript of Voltage/current dividers - Gary Tuttle's ISU web...
EE 201 voltage/current dividers – 1
Voltage/current dividersTrivial to understand, but still very useful.
Use in combination with equivalent resistances to quickly find a particular voltage or current in a circuit.
+–VS
R1
R2R3 –
vR2+ Want to know vR2.
Easily solved with KCL, KVL, & equivalent resistances.
iS
L6 =965HT
=96
5� + 5� + 5�
Y5� = L65�
=5�
5� + 5� + 5�96 That’s it.
The total voltage across a series string is divided among the resistors according to a simple ratio.
Voltage divider
EE 201 voltage/current dividers – 2
+–VS
R1 R2R3 –
vR2+
–vR1+
– vR3 +
Easy to find the other voltages, too.
Y5� =5�
5� + 5� + 5�96
Y5� =5�
5� + 5� + 5�96
Put in some values.
5 kΩ15 kΩ15 V
+–
–vR2+
–vR1+
– vR3 +
10 kΩ
Y5� =��Nż
�Nż + ��Nż + ��Nż (��9) = �9
Y5� =��Nż
�Nż + ��Nż + ��Nż (��9) = �.�9
Y5� =�Nż
�Nż + ��Nż + ��Nż (��9) = �.�9
EE 201 voltage/current dividers – 3
Current dividerSame idea, but with current.
ISR1 R2 R3iR2 Want to know iR2.
Easily solved with KCL, KVL, & equivalent resistances.
vR
+
–
Y5 = ,65HT
=,6
�5� + �
5� + �5�
=�5�
�5� + �
5� + �5�,6
L5� =Y55�
The total is divided according a simple ratio determined by the resistors.
EE 201 voltage/current dividers – 4
ISR1 R2 R3iR2iR1 iR3
The other currents are just as easy.
L5� =�5�
�5� + �
5� + �5�
,6
L5� =�5�
�5� + �
5� + �5�
,6
Plug in some numbers.
3 k! 15 k!9 mA 5 k!
L5� =�
��Nż�
�Nż + ��Nż + �
��Nż(�P$) = �P$
L5� =�
�Nż�
�Nż + ��Nż + �
��Nż(�P$) = �P$
L5� =�
�Nż�
�Nż + ��Nż + �
��Nż(�P$) = �P$
iR2iR1 iR3
EE 201 voltage/current dividers – 5
Example 1
VS +–
R1
R2 R3 R4
R5
vR4
+
–12 V
24 !
48 !36 !
36 !
12 !
In the circuit, find vR4.
vR4 = vR3Combine parallel combinations.
VS +–
R12
R34
R512 V
16 ! 18 !
12 !
vR4
+
–
Y5� =5��
5�� + 5�� + 5�96
=��ż
��ż + ��ż + ��ż (��9)
= �.��9
Use voltage divider.
EE 201 voltage/current dividers – 6
Example 2IS
R2
R3
R5 R4
R1
R6
iR3
In the circuit, find iR3.
IS
R45
R13
R6
iR13
120 !
40 !10 !
60 ! 20 !
80 !
40 !
80 !
80 !
Find the equivalent resistance in each branch.
Use a current divider to find iR13.
L5�� =�
5���
5�� + �5�� + �
5�,6
=�
��ż�
��ż + ���ż + �
��ż(�.�$) = �.��$
0.5 A
0.5 A
Referring back to the original circuit, iR13 divides between R2 and R3.
L5� =�5�
�5� + �
5�L5��
=�
��ż�
���ż + ���ż
(�.��$) = �.����$
EE 201 voltage/current dividers – 7
Example 3
VS +–
R1
R2 RL
1 k!
3 k!
In the simple divider circuit at right, if RL is attached in parallel with R2, the voltage across R1 doubles. What is the value of RL?
Without RL,
Y5� =5�
5� + 5�96
Y�5� =
5�5� + 5HT
96
= � 5�5� + 5�
96
From the two expressions for v’R1
5�� +
5�� = 5� + 5HT
5HT =5� � 5�� =
�Nż � �Nż� = �NżWith RL,
�5HT
=�5�
+�5/
Then
5/ =�
�5HT
� �5�
= �.�Nż
EE 201 voltage/current dividers – 8
Example 4
R1 R2 R3IS
iR2iR1 iR3
For the simple current divider at right, design it (i.e. choose resistor values) so that the currents are in the ratio iR1:iR2 :iR3 = 1:2:4 and the total power dissipated in the resistors is 10 mW.
From the current divider equation, we know that the currents are proportional to the inverse of the resistance in each branch.
�5�
:�5�
:�5�
= � : � : �
Therefore,
5� : 5� : 5� = � : � : �
R2 = 2R3 and R1 = 2R2 (R1 = 4R3).
�5HT
=�5�
+�5�
+�5�
=��5�
+��5�
+�5�
=�.��5�
The equivalent resistance of the three in parallel is
5HT =5��.��
Find Req,
3 = ,�65HT 5HT =3,�6
= ��Nż
R3 = 1.75Req = 17.5 kΩ; R2 = 2R3 = 35 kΩ; R1 = 2R2 = 70 kΩ
1 mA
Finally:
