phase splitter lab1a
Figure 1. Schematic of a phase-splitter

1. Lab goals

Topics
  • Transistor operation modes: active, saturation, cutoff.

Each of the two lab sections are doing different activities.

Your task is to learn about the concepts using Schematic of a phase-splitter as your example and describe your findings to your classmates who did the other activity. The topics of focus will be similar between the two sections, however. This allows everyone several angles by which to learn the concepts. A secondary goal is to increase your technical communication / teaching skills through action.

2. Step 1

Begin with only the circut of Schematic of a phase-splitter contained in the Step 1 box.

Set Rc to zero for this part.

2.1. DC conditions

Find the values of R1 and R2 so the emitter voltage is near 0 V.

  • What is the condition required at the base node?

In order to make the design be easier, we want the base current to be less than 1/10 of the current through R1 and R2 (so we can ignore base current with little loss of precision).

  • What is the emitter current?

  • Estimate the base current.

10× to 20× this base current is then your target current through R1.

  • Find a R1 value that meets this condition and is in stock.[1]

  • Calculate the required R2 value and round to a value that is in stock.[1]

Build this circuit and verify that the emitter voltage is indeed near 0 V.

2.2. Amplifier

Now setup the circuit contained in the Step 2 box. Use the AD2 function generator output W1.

Use channel 1 to view the node voltage at B (between B and zero) and channel 2 to view out.

Set an initial amplitude of 200 mV peak-to-peak.

Verify that the amplifier is working as a voltage follower.

Now, increase the input signal amplitude until the output voltage starts to have some “issues”. Set the amplitude to 6 V peak-to-peak.

  • Your task is to figure out why the output looks the way it does!

  • Then you should make a short document describing what you did and your explanation of the last waveforms.

Circuit notes

The amplifier’s output voltage will track the input voltage with about a 0.7 V downward shift (VBE). However, at large amplitudes, the output will clip at about -1.5 V.

As the input (base) voltage and therefore output / emitter voltage decreases, the emitter current decreases. This current is easily found by fixing VE and solving for the currents at the emitter node via KCL. Emitter current eventually reaches zero and will not become negative because the transistor is now in cutoff mode. The equivalent circuit at node out in this condition is merely a voltage divider between -5 V and zero involving Re and Rload.

Section 2.2.3.D of [AoE] also discusses this behavior of a resistively-biased EF. In addition, [L-AoE] sections 4N.4.4 and 4N.4.5 add some nice explanation and points out the power dissipation problem that rears its head as soon as you try to get more voltage swing by lowering Re.

3. References


1. The GEM 163 storage cabinet has every E12 series value from 1 Ω to 10 MΩ