Tag: analog electronics

Back To Basics Episode 10: The Art Of Biasing, Or How To Use A Transistor

Ever wondered how a tiny piece of silicon actually turns a quiet audio signal into a loud one? In this episode of our Back to Basics series, we’re moving past the theory of how transistors work and looking at how to actually use them in real-world circuits.
We’ll demystify biasing—the art of setting our transistor’s “resting state” (the Q-point)—using simple water valve analogies, walk through the “back-of-the-napkin” math to design a working 9V Common Emitter amplifier, and head over to the test bench to watch it all happen live on the oscilloscope.
No scary textbook equations required—just simple, practical rules of thumb you can use on your own breadboard today!

In our last episode, we looked at how a transistor acts like a tiny electronic valve. We used it as a switch—turning an LED fully ON or fully OFF. That’s easy, and it’s the foundation of digital electronics.

But what if you want to amplify an analog signal, like music or your voice?
If you just plug an audio signal straight into a transistor, you’re going to get a horrible, distorted mess. Half of your sound wave will simply vanish.


To fix this, we need to master the art of biasing—which is simply setting our valve to be exactly half-open when there is no input signal. Today, we’re going to demystify biasing, look at the incredibly simple math behind it, build a working 9-volt amplifier, and see how it all behaves on the oscilloscope.

Why Biasing?


Think of our transistor like a water tap.
If the valve is completely closed, no water flows. In the transistor world, we call this Cutoff.

If the valve is spun wide open, water flows at maximum capacity. We call this Saturation.


An audio signal is an alternating current. It swings positive, and then it swings negative.


If our valve is completely closed at rest, it can only open when the audio signal goes positive. The moment the signal goes negative? Nothing happens. The valve is already closed! The bottom half of your music gets completely chopped off.


So, how do we fix this?
We ‘bias’ the transistor. We use a couple of resistors to set our electronic valve to be exactly half-open when there is no input signal. This ‘happy medium’ is called the Quiescent Point, or Q-point. Because the valve is half-open at rest, the audio signal can ride up and down freely without hitting the floor or the ceiling, as long as it is not too large.

The Classic BJT Circuit


Let’s build the most classic transistor amplifier in history: the Common Emitter amplifier.


We’ll use a standard 9V battery, a 2N3904 NPN transistor, and four cheap resistors.
The math to design this is incredibly simple if we use these five steps.


Step One: Determine the specifications

First, the power supply. We can use a 9V battery for our simple circuit. Therefore Vcc = 9V.

Based on the transistor datasheet, We have an average hFE of 200, and a VBE of 0.7V.

Now, if this small amplifier will be used just to amplify a small signal voltage, we can set IC to 10mA and we can approximate IE to the same value, since the base current is going to be 200 times smaller and, therefore, negligible.


Step Two: Find the emitter resistor Re


We can simply calculate the Emitter voltage as 1/10 of the power supply voltage. This will prevent variations of the hFE to affect our design.

We can round up this value to 1V.

Now, we can apply Ohms law to calculate the emitter resistor:


Step Three: find the collector resistor Rc


For that, we need first to determine the collector voltage VC. To be able to deal with symmetrical signals, we can set Vc to half the value of the power supply, which is 4.5V.


Now, we can apply again Ohm’s law to find Rc:


Which we can round to the closest available standard value of 470 ohm.


Step four: Find the value of resistor R1


For that, we need to have a couple more information.


First, the value of the voltage at the base of the transistor. That will be


Second, in order to have a collector current of 10mA, we will need a base current of


In order to have a stable amplifier, we will need a current at least 10 times that flowing through resistor R1, so we can approximate the current through R2 to the same value. If we call such current I12, its value will be:


And now we can use one more time Ohm law and calculate R1:


which we can round to the nearest standard value:


Step five: the calculation of R2


For that, we use one last time Ohm’s law and we obtain


which we can round to the closest standard value of 3.3k.

Making the transistor an amplifier


Now, our transistor is biased and sitting happy at its Q-point, but we still need to get our audio signal in and out. To do that, we need to add two 10uF coupling capacitors.


These capacitors act like high-pass filter guards. They let our AC audio signal go right through, but they completely block the DC voltages. This ensures our bias levels don’t leak out, and whatever we plug into the input doesn’t mess up our careful calculations.

But now the question is: how much louder will our signal get?


For a simple Common Emitter circuit like this, the voltage gain Av is roughly the ratio of the Collector resistor to the Emitter resistor:


Since we used a 470 ohm collector resistor and a 100 ohm emitter resistor, our gain is roughly 4.7. If we inject a tiny 200mV signal, we will get a clean 0.94V signal out!

If you would like to see this amplifier in action, and review one more time what we just said, you can watch this video:

The MOSFET Alternative


Now, you might be wondering: what if I’m using a MOSFET instead of a BJT?
The basic idea of biasing is exactly the same, but MOSFETs are voltage-controlled, instead of current-controlled. The Gate pin of a MOSFET draws absolutely zero current. That means our voltage divider math on the input is actually perfect in the real world—no approximations, no loading effect! Just a feeble current through R1 and R2 will suffice.


However, instead of needing just 0.7V to turn on, a MOSFET typically requires a much higher Gate-to-Source threshold voltage—often between 2V and 4V—before it even begins to open. To bias a MOSFET in a Common Source amplifier, we simply adjust our input voltage divider to sit comfortably above that threshold voltage. Same concept, different valve!


Here is the design procedure for such an amplifier. It is very similar to the one for the BJT.


Step One: Determine the specifications

We will use a power supply voltage of 9V, as in the case on the BJT. We will also use the same Q-point current of 10mA, which is the drain current Id.
We will set the drain voltage to half the supply voltage, to guarantee the symmetry of the output voltage swing: Vd = 4.5V. We will also set the source voltage to one tenth of the supply voltage, to help prevent thermal runaway and secure the Q-point stability.


Step Two: Find the drain resistor Rd


This can be easily done using Ohm’s law, after which we then round to the closest standard value:


Step Three: find the source resistor Rs

Similarly, we use again Ohm’s law to calculate and then round the value of the source resistor:


Step four: Calculate the Needed Gate Voltage 𝑉g


This will be just the sum of the voltage on Rs and the Vgs of the MOSFET, which we need to take from its data sheet:


Step five: Calculate the Gate Divider Resistors R1 and R2

Because the gate draws no current, we do not need to follow the 10 times rule we used for the BJT. We just need a small current flowing through the two resistors R1 and R2. However, do not make it too small to prevent issues with current leaks and noise. A good rule of thumb is to set the total divider resistance to something in between 1 and 10 Mohm.


For example, we can choose 1M for R2.

From there, we calculate R1 based on the voltage ratio:

And that’s all for today!