Many electronic circuits, including voltage dividers and filters, produce signals that sag when current is drawn. An Emitter-Follower uses a transistor to reduce this sag by a factor of 100.

The following information is adapted from the lab materials of an electronics course at Pomona College.

## Step 1: Motivation: Understanding Sag

Most signals and voltage sources are imperfect. Generally, when we try to draw current from a voltage source, the voltage decreases. This decrease in voltage is called sag. Sag can cause significant problems in multi-stage circuits where later stages depend on receiving a stable voltage.

While there are many reasons why signals and voltage sources sag, we can usually explain sag with a simple model involving a perfect voltage source and a resistor. This model is shown in the circuit diagram above. We call the resistor in the model the Thevenin Resistance (Rth). The bigger the Thevenin Resistance,the more a voltage source sags as current is drawn.

The Emitter-Follower circuit will reduce the Thevenin Resistance of a voltage supply or signal by a factor of 100.

## Step 2: Materials

To make an Emitter-Follower, first find the following materials:

1. Power Supply with outputs of +15V , -15V and ground
2. NPN transistor (2N3904)
3. 3.3kΩ resistor
4. Sagging voltage source/signal (or the prior stage of your circuit)
6. Wires

## Step 3: Constructing the Circuit

Construct the Emitter-Follower circuit as shown in the diagram above.

1. Orient the transistor correctly and connect the collector to +15V. Be sure to check the documentation for your transistor as the order of pins can vary.
2. Use the 3.3 kΩ resistor to connect the emitter to -15V.
3. Attach your sagging signal to the base of the transistor.
4. Attach the load or the next stage your circuit to the emitter of the transistor.

## Step 4: How an Emitter-Follower Works: Understanding Transistors

When appropriate voltages are applied to the base and collector of an NPN transistor, the transistor adjusts its internal current flow until it meets the following conditions:

1. The voltage at the emitter is 0.6V less than the voltage at the base.
2. Only 1% of the current that goes out of the emitter comes from the base. The other 99% comes from the collector.

The first rule explains why the output signal of an Emitter-Follower follows the input. When we input a voltage signal to the base of the transistor, the transistor allows current to flow until the emitter voltage, Vout, is exactly 0.6V less than input voltage. The transistor adjusts so quickly that the output signal maintains the same shape as the input.

The second rule explains how an Emitter-Follower reduces sag. Since only 1% of the emitter current comes from the base, the Emitter-Follower can supply a large amount of current to a load while drawing little current from a sagging voltage source/signal. For the same current draw, the voltage source will sag 1/100 of the amount that it would otherwise. This means that the Emitter-Follower decreases the Thevenin Resistance of the voltage source by a factor of 100.

## Step 5: Potential Problems: Clipping

There are some situations where the transistor in an Emitter-Follower is unable to adjust to meet the two conditions listed in step 4. In these situations we observe clipping - shown in the graph above.

In the Emitter-Follower circuit above, the transistor can only adjust Vout by supplying current to the resistors on the right side of the circuit. When the transistor supplies current, it increases the voltage at Vout until it is 0.6V less that the base voltage.

However, since current can only flow out of a transistor's emitter, the transistor has no way to reduce Vout beyond cutting off current completely. When current from the transistor cuts off, it leaves a voltage divider consisting of the two 3.3kΩ resistors between ground and -15V. This voltage divider sets Vout to -7.5V. Since the transistor cannot reduce Vout to below this baseline, we observe the signal being clipped at -7.5V in the circuit above.

We can also observe clipping on the top of a signal. Since the transistor works by allowing current to flow from collector to emitter, it cannot output a voltage at its emitter that is larger than the voltage at its collector. Regardless of the input signal, the output cannot be more than +15V. This can result in clipping.

great read and very well explained. I build my own guitar equipment and amateur radio equipment and have seen far too many overwritten and over complicated articles.
<p>Nice explanation, thanks.</p>
<p>Thank you so much for making this slightly complicated schematic simple and easy to understand!</p>