EEGR2111 L7

Circuit Analysis using equivalent resistances

Use equivalent resistances to analyze the following circuit:

Find i₁, i₂, V_R1, V_R2

To find i₁, note that the 4, 8, and 8 Ohm resistors are in parallel with R_eq1 = 2
Then this 2 ohms is in series with the 1 ohm resistor resulting in an equivalent series resistance R_eq2 = 3
Finally, Ohms law yields i1 = 6 / 3 = 2 Amp

Then, V_R1 = 2 Volt

And, V_R2 = (2 Amp) R_eq1 = 4 Volt

The two 8 ohm resistors in parallel have an equivalent resistance of 4 ohms
So the current divides equally between the pair of 8 ohm resistors and the 4 ohm resistor,
So i₂ = 1 Amp

Example L7-1:

Earlier, we saw how a dependend source modeled a BJT and could generate voltage gain

What is the equivalent circuit of the following?

Above: V1 = i / k
So by comparison with Ohms law v = i / R, the above is equivalent to a (1/k) ohm resistor

Example L7-2:

Below is a model of a 2N2222 BJT amplifier circuit with external biasing resistorsR1 and R2, and load resistor RL:

Find the current gain i_o / i_i when:
R1 = R2 = RL = 1000, R_\pi = 2000 and g_m = 0.05

First, two bias resistors and R_\pi are in parallel with equivalent resistance:
R_eq = 1 / { (1/1000) + (1/1000) + (1/2000) } = 400
so V_be = R_eq i_i
then i_o = g_m V_be = g_m R_eq i_i
So:
i_o / i_i = g_m R_eq
= (0.05) 400 = 20

Would increasing R_eq increase current gain?

What would be the maximum current gain?

What is the power gain?

Equivalent resistance

Generally: we may replace any circuit portion comprised of resistors with a single equivalent resistor:

What is R_eq?

Redraw the circuit

Note:

When dependent sources are present, be careful to account for the controlling voltage or current when replacing resistors with their equivalents.

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Chapter 4

Circuit analysis

In previous chapter we focused on circuits with one loop or two nodes (or used equivalent resistance transformations to simplify the circuit)

Now we introduce two new methods

Nodal analysis: any number of nodes

Mesh analysis: any number of (planar) meshes
- applies only to planar circuits

Nodal analysis

Given circuit with N nodes, and N KCL equations
N - 1 KCl equations are required
(recall 2-node example where we showed second KCL to be redundant)

The remaining node is called the "reference node"
(often this node is "ground" or zero volts)

PROCEDURE:

Label nodes a, b, c, ... or 1, 2, 3 ...

Generally let the ground node be node 0, the reference (but any node may be chosen)

Then, the node voltages are the voltage between the nodes and the reference node, i.e.:
V₁ = V₁₀ = V₁ - V₀ = V₁ - 0
(Note that in this notation:
V₁₂ =V₁ - V₂ = V₁₀ - V₂₀ = V₁ - 0 - { V₂ - 0 } )

Apply KCL to all N - 1 nodes (not the reference node)

Solve the resulting set of N - 1 simultaneous equations

Example L7-1:

Find the nodal equations for the following circuit

4 nodes -> so need 3 equations

First, number the nodes (reference = 0)

Then, find nodal equations at each node using KCL

Node 1:
i₁ + (V₂ - V₁) / R1 =0
3 + V₂ - V₁ = 0

Node 2:
(V₁ - V₂) / R1 + (0 - V₂) / R2 + (V₃ - V₂) / R3 =0
V₁ (1/R1) + V₂ ( - 1/R1 - 1/R2 - 1/R3) + V₃ (1/R3) =0

Node 3:
(V₂ - V₃) / R3 +(0 - V₃) / R4

So, 3 equations are:
- V₁ + V₂ + (0) V₃ = -3
V₁ + (-11/8) V₂ + (1/8) V₃ = 0
(0) V₁ + (1/8) V₂ + (-1/4) V₃ = 0

In matrix form:

      
      | -1    1      0   |   |V1|      | -3 |
      |  1  -11/8   1/8  |   |V2|  =   |  0 |
      |  0    1/8   -1/4 |   |V3|      |  0 |

Note: if we do current LEAVING node for KCL we get:

      |  1    -1     0   |   |V1|      |  3 |
      | -1   11/8   -1/8 |   |V2|  =   |  0 |
      |  0    1/8    1/4 |   |V3|      |  0 |

Solve using Cramer's rule:

      
              | -3    1      0   |   
         det  |  0  -11/8   1/8  |   
              |  0    1/8   -1/4 |         -63 / 64    63
 V1 =       -------------------------  =   -------- =  --- = 12.6
              | -1    1      0   |          -5 / 64     5
         det  |  1  -11/8   1/8  |   
              |  0    1/8   -1/4 |   

Similarly for V₂, V₃

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