I have this homework problem for my electronics class and I'm in need of assistance.

Here is the original problem statement:

Original problem statement

This is what I have after drawing the circuit:

My drawing of the circuit

My professor always suggests that we start at the output and write it in terms of the first dependent source to the left, but when trying to write Vout there are two dependent sources.

I'm not sure how to handle that.

Things I tried:

Current Divider

I can't use a current divider to find the voltage across R4 because the series combination (R8 + R2) are not in parallel. And I can't do a KCL at Node H without adding another unknown variable to the mix.

KVL expression for Vout:

Similar as the current divider, I can't write an expression for Vout because there are no KVL loops that don't require me to define even more variables.

All I could figure out to do was start at the input and write equations for the variables that control the dependent sources, but that involved defining even more unknowns. As you can imagine, this got really ugly, really quickly.

All other amplifier circuits we have looked at in class are "linear"; in that there aren't two sets of cascaded amplifiers stacked vertically, like in this problem, and I can't find anything like this circuit in the chapter of our textbook.

If anyone could explain an intuitive thought process or a process to solve a cascaded amplifier circuit like this one, I'd be very grateful.


1 Answer 1


In order to use KVL equations, you should be wary of current sources. While it is still possible to include them in the equations, they tend to complicate things. There is one in your circuit, part of the EFGH two-port.

Ideal current source

It is possible to convert this to a Thévenin equivalent. For this you determine the open-circuit voltage, and the closed loop current to calculate the equivalent resistance:

$$V_{thev} = V_{oc} = G\cdot v_3 \cdot R_4$$

$$R_{thev} = \frac{V_{oc}}{I_{cc}} = \frac{G\cdot v_3 \cdot R_4}{G\cdot v_3} = R_4$$

The Thévenin equivalent then becomes


simulate this circuit – Schematic created using CircuitLab

Leading up to the following circuit, where you can apply KVL laws more easily:


simulate this circuit

This schematic nicely defines 3 distinct loops (I chose from left to right \$\Gamma_1\$, \$\Gamma_2\$ and \$\Gamma_3\$, all clockwise) for which we can write the KVL equations:

$$\begin{align} -V_{IN} + R_5\cdot \Gamma_1 &= 0 \\ &\Downarrow \\ \Gamma_1 &= \frac{V_{IN}}{R_5} \end{align}$$

(\$i_s = \Gamma_1\$)

$$\begin{align} Z\cdot \Gamma_1 + R_1\cdot \Gamma_2 + R_7\cdot \Gamma_2 + R_3\cdot \Gamma_2 + R_6\cdot \Gamma_2 &= 0\\ &\Downarrow \\ Z\cdot \Gamma_1 + (R_1 + R_7 + R_3 + R_6)\cdot \Gamma_2 &= 0 \end{align}$$

(\$v_1 = R_1\cdot \Gamma_2\$, \$v_3 = -R_3\cdot \Gamma_2\$)

$$\begin{align} -kR_1\cdot \Gamma_2 + R_2\cdot \Gamma_3 + R_8\cdot \Gamma_3 - GR_4(-R_3)\cdot \Gamma_2 + R_7\cdot \Gamma_3 &= 0 \\ &\Downarrow \\ (-kR_1 + GR_4R_3)\cdot \Gamma_2 + (R_2 + R_8 + R_7)\cdot \Gamma_3 &= 0 \end{align}$$

These equations can be solved relatively easily. Finally, \$V_{OUT} = R_8\cdot \Gamma_3\$.

Similar to current sources for KVL, voltage sources turn out to be a bit tricky for KCL laws too. KVL is preferred here because:

  • There are more voltage sources (3) than current sources (1), so we can save work
  • The only current source is shunted with a resistor, making conversion to the Thévenin equivalent easy
  • KVL laws do not need a ground node to be solved, KCL would require you choose grounds in the three loops

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