ER = Etotal � � R Rtotal � �

Follow-up question: algebraically manipulate this equation to solve for E_total in terms of all the other variables. In other words, show how you could calculate for the amount of total voltage necessary to produce a specified voltage drop (E_R) across a specified resistor (R), given the total circuit resistance (R_total).

Notes:
Though this "voltage divider formula" may be found in any number of electronics reference books, your students need to understand how to algebraically manipulate the given formulae to arrive at this one.

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As R₂ increases in resistance, V_out increases and I_R1 decreases.

Notes:
Nothing special here - just a qualitative analysis of a very simple voltage divider circuit.

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The voltage divider with the ganged rheostats will yield the greatest change in output voltage for a given change in R₂'s resistance, because only the numerator of the fraction in the voltage divider formula changes with R₂, not the denominator as well.
Follow-up question #1: what happens to the amount of current in each circuit for a given change in R₂ resistance? Explain why.
Follow-up question #2: explain how a potentiometer performs the exact function as the second circuit with the two (complementarily) ganged rheostats.

Notes:
Understanding the mathematical basis for the answer may be a significant leap for some students. If they experience trouble understanding how the voltage divider formula proves the answer, have them try a "thought experiment" with really simple numbers:

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Initial conditions:

     

R₁ = 1 W

     

R₂ = 1 W

     

V_battery = 1 volt

Now, increase R₂ from 1 W to 2 W and see which voltage divider circuit has experienced the greatest change in output voltage. Once these example quantities are placed into the respective formulae, it should become easy to see how the voltage divider formula explains the larger voltage swing of the second divider circuit.
Point out to your students that this is an example of practical problem-solving: performing a "thought experiment" with really simple quantities to numerically explore how two different systems react to change. Although there is nothing particularly difficult about this technique, many students avoid it because they think there must be some easier way (a ready-made explanation, as opposed to a thought experiment of their own) to understand the concept. Getting students over this attitude barrier is a difficult yet crucial step in them developing self-teaching ability.

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As R_load increases, R_transistor will decrease in resistance so as to maintain a constant current through the load and a constant R_total.

Notes:
This model of current mirror transistor behavior, albeit crude, serves as a good introduction to the subject of active loads in transistor amplifier circuits. This is where a transistor is configured to operate as a constant-current regulator, then placed in series with an amplifying transistor to yield much greater voltage gains than what is possible with a passive (fixed resistor) load.

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If the active load were a perfect current regulator, the voltage gain of this single-stage amplifier circuit would be infinite (�), because the Thévenin equivalent resistance for a current source is infinite ohms.

Notes:
There is more than one way to comprehend this effect, and why it works as it does. One of the more sophisticated ways is to consider what the internal resistance of a perfect current source is: infinite. Ask your students how they contemplated this effect, and what means they employed to grasp the concept.

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Of course, all the current sources are active loads, but there is one more at the lower-left corner of the schematic. I'll let you figure out where it is!

Notes:
Even if students do not yet know what an öperational amplifier" circuit is, they should still be able to identify transistor stages, configurations, and active loads. In this case, most of the active loads are obvious (as revealed by the current source symbols).
Don't be surprised if some of your students point out that the differential pair in this opamp circuit looks üpside-down" compared to what they've seen before for differential pair circuits. Let them know that this is not really an issue, and that the differential pair works the same in this configuration.

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Q₆ is an active load for Q₄

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Q₂₃ is an active load for Q₄

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Q₁₃ is an active load for Q₁₇ and Q₂₂

Notes:
Even if students do not yet know what an öperational amplifier" circuit is, they should still be able to identify transistor stages, configurations, and active loads.

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