---
layout: post
title: Transistor Design For Newbies
author: Dylan Müller
author_url: https://linkedin.com/in/lunarjournal
---
> Transistors are important electronic devices that find use in a wide range of
> applications. Learn how to design circuits with them.
1. [Principle of operation](#principle-of-operation)
3. [Transistor as a switch](#transistor-as-a-switch)
4. [Transistor as an amplifier](#transistor-as-an-amplifier)
5. [LTSpice](#ltspice)
# Principle of Operation
There are various analogies that you will most likely come across when first
learning about transistors, a useful analogy is that of a mechanically
controlled water valve.
![\[insert image\]](https://lunarjournal.github.io/images/3/valve2.png)
![enter image description here](https://lunarjournal.github.io/images/3/07.png)
Here it is important to reference the water analogy of current and voltage. In
the water analogy we picture a column of water moving through a pipe.
We define current as the movement of water (charge) through the pipe (wire), or
in mathematical terms the rate of flow of water (charge) past a given point with
respect to time:
![enter image description here](http://lunarjournal.github.io/images/3/01.png)
Voltage is analogous to the pressure differential between two points. For
example, suppose we suspend water in a pipe and then apply a high pressure at
the top and a lower pressure at the bottom. We have just set up a 'water
potential difference' between two points and this tends to move water (charge)
from the higher pressure region (voltage) to the lower pressure region. The
higher the water potential, the faster the column of water (charge) moves
through the pipe when it has the chance.
In reality, voltage arises due to the presence of electric fields. For a given
electric field between two points, a positive test charge may be placed at any
distance along the electric field lines, that is, its 'field potential' varies
and a positive charge placed closer to the positive end of the electric field
feels more repulsion (and therefore has a higher potential to do work) than at
the negative end of the field.
Potential difference (voltage) is just a differential measure of this electric
'field potential' or put differently, the capacity of charge to do work in the
presence of an electric field:
![enter image description here](http://lunarjournal.github.io/images/3/02.png)
With this in mind the idea of a water valve then makes sense. The valve consists
of three ports, one attached to one end of the pipe, the other port to the end
section of the pipe and then the valve itself, sitting in the middle and
regulating the flow of water between both ends.
By rotating the valve we adjust the water flow rate (current) through the pipe.
This is the basic principle of operation of a transistor. However rather than
applying a mechanical torque, we apply a potential difference at the base to
regulate current flow.
You may think of the degree to which the mechanical valve is open or closed as
proportional to the voltage applied at the base of the transistor. This means
that we can control a potentially larger current through the transistor using a
smaller current through the base (through the application of a base voltage) ,
this is one of the useful properties of transistors.
Bipolar Junction Transistors (BJTs) usually consists of three semiconductor
layers which can be of two types: n or p. The individual silicon layers are
crystalline structures that have what are known as dopants added to them. These
are individual elements (phosphorus, boron) added to neutral silicon (and
replace the corresponding silicon atoms) in order to change the electrical
properties of the layer.
![enter image description here](http://lunarjournal.github.io/images/3/10.png)
For example, boron [B] dopant has a valency (number of outer electrons) of 3,
while silicon has a valency of 4. This means that when boron and silicon bond
covalently (sharing of each others electrons) there is a mismatch (3 < 4)
between their valence electrons, leaving a 'hole', which needs to be filled with
an electron in order to match silicon's valency. This results in a crystal
structure with a net positive charge, the p type layer.
In contrast phosphorus [P] dopant has a valency of 5, again there is a mismatch
(5 > 4) with silicon's valency (4), allowing for the extra electron of
phosphorus to move freely through the crystal structure and giving the overall
crystal layer a negative polarity, the n type layer.
![enter image description here](http://lunarjournal.github.io/images/3/04.png)
If we were to place an n region and p region together we would form an
electronic device known as a diode. A diode is a 2 terminal device (with the n
side connected to the negative terminal (cathode) and p side connected to the
positive terminal (anode) ) that only allows current flow in one direction. It
is also worth nothing that by placing an n and p region next to one another
there is a localised effect at their layer boundary that results in a small
number of electrons (from the n type region) migrating to the p type region in
what is known as the depletion region.
![enter image description here](http://lunarjournal.github.io/images/3/11.jpg)
The migration of electrons from the n type region to the p type region at the np
boundary sets up what is known as a barrier potential, a secondary electric
field at the np layer boundary in opposition to the primary E-field (between p
and n).
This is the amount of voltage (pressure) required to force n layer electrons
through the np barrier (the secondary E-field) where they can flow into the
positive terminal (anode) of the diode.
It is equivalent to having a water valve initially shut tight and requiring a
torque in order to get water flowing. A typical value for the barrier potential
of garden variety diodes is between 0.3v-0.7v.
![enter image description here](http://lunarjournal.github.io/images/3/03.gif)
A bipolar junction transistor (BJT) may be viewed as a combination of two diodes
(shown below for an NPN transistor):
![enter image description here](http://lunarjournal.github.io/images/3/05.gif)
An NPN BJT transistor has two current paths, one from the collector to emitter
and the other from the base to emitter. The current flow from collector to
emitter represents the water flow in the pipe containing the valve, while the
current flow from base to emitter represents the degree to which the valve is
open or closed.
You might be wondering why conventional (positive) current flows backwards
through the base-collector diode (from collector to emitter) for an NPN
transistor. As it turns out, current can actually flow in multiple directions
through a diode. However it takes much more voltage to 'push' charge through a
diode in the direction it's meant to block than in the direction it is meant to
flow.
The ratio of base-emitter current to collector-emitter current is known as (β)
and is an important consideration in the design of circuits using transistors:
![enter image description here](http://lunarjournal.github.io/images/3/06.png)
Both transistor current paths have an associated voltage drop/potential
difference across them.
For the current flow from base to emitter, there is the base-emitter voltage
drop VBE and from collector to emitter there is the collector-emitter
voltage drop VCE :
![enter image description here](http://lunarjournal.github.io/images/3/07.gif)
The values of VCE, VBE and VCB have predictable
values for the three modes of operation of a transistor, these are:
* **Cut-off** (The transistor acts as an open curcuit; valve closed).
VBE << 0.7V
* **Saturation** (The transistor acts as a short circuit; valve completely open)
* VBE >= 0.7V
* **Active** (The transistor acts as an amplifier; valve varies between closed
and completely open)
# Transistor as a switch
When using a transistor as a switch we place the transistor into one of two
states: cut-off or saturation.
The following switching circuit is usually employed (with an NPN BJT) (shown
together with an LED):
![enter image description here](http://lunarjournal.github.io/images/3/12.jpg)
The circuit is seen consisting of a base current limiting resistor RB
as well as a collector-emitter current limiting resistor RLIM.
RB serves to set up the correct base current, while RLIM
serves to limit the maximum current through the LED (shown in red) when the
transistor is switched fully on (driven into saturation).
To calculate the values for resistors RB and RLIM we use
the equation relating base current to collector current defined earlier:
![enter image description here](http://lunarjournal.github.io/images/3/06.png)
The first question becomes what collector current IC we desire. This
value depends on the device/load you are trying to switch on/off. It is worth
noting that when a transistor is switched fully on (is in saturation mode) the
equivalent circuit (simplified) is as follows (shown without the LED, you can
assume the LED follows resistor RC):
![enter image description here](http://lunarjournal.github.io/images/3/08.jpg)
Thus at the collector a direct connection to ground is made. However this
connection is not perfect and there is an associated voltage drop from collector
to emitter of typically around 0.2v (VCE) rather than 0v. Determining
the relevant value for IC is then just a matter how much current your
load (LED in our case) requires.
For example, a typical green led requires around 15mA of current to light up
brightly so we set IC = 15mA. A green LED also typically has a 2v
drop across it. To calculate RLIM we use ohms law:
![enter image description here](http://lunarjournal.github.io/images/3/14.gif)
Given the LED and collector to emitter voltage drops of 2v and 0.2v
respectively, we can further reduce the above expression above to:
![enter image description here](http://lunarjournal.github.io/images/3/15.gif)
Choosing VCC is just a matter of what you have at hand. For example,
a 5v or 9v supply would be adequate to drive the transistor into saturation as
long as VCC >> 0.7v (due to the base emitter voltage drop) and Vcc >>
2v (for the led).
Assume VCC = 5v, then RLIM = 186.7 Ω
In calculating the required base current, we use the transistor's β value. This
can be found on the transistors datasheet and typically varies from anywhere
between 20 to 200. The rule of thumb is to use the minimum value of β for a
specific transistor type. For the standard garden variety 2N2222 transistor, the
minimum value of β is around 75. Therefore to calculate IB, we have:
![enter image description here](http://lunarjournal.github.io/images/3/16.gif)
You might have noticed an additional factor called SF for (safety factor). This
is a factor typically around 5-10 that we multiply our calculated IB
with in order to ensure we drive the transistor into saturation. This gives a
value of around 1mA for IB.
Given IB, calculating RB becomes trivial as we know the
voltage across RB as: VCC - VBE (think of
VBE as a 0.7v diode) and so we apply ohms law once again:
![enter image description here](http://lunarjournal.github.io/images/3/17.gif)
Now you can connect a switch between the base resistor and Vcc or connect the
base resistor directly to the output of a 5V-TTL micro-controller in order to
turn the LED on and off! The benefit of using a transistor to do that is that we
require a relatively small current (<1mA) in order to switch a much larger
current through the LED (15mA)!
In conclusion:
1. Determine required collector current IC
2. Calculate RLIM (ohms law)
3. Calculate IB using lowest value for β
4. Multiply IB by safety factor 5-10
5. Calculate RB (ohms law)
The simple LED transistor circuit was modelled in LTSpice, with the LED
represented as a series voltage source (representing the 2v voltage drop).:
![enter image description here](http://lunarjournal.github.io/images/3/18.png)
A simulation of the DC operating point of the circuit yielded:
![enter image description here](http://lunarjournal.github.io/images/3/19.png)
Here we can see the ~1mA base current (Ib) driving ~15mA collector
(IC) current. All current values are shown in S.I units of amperes
(A).
# Transistor as an amplifier
Here we operate the transistor in its active mode to achieve linear
amplification. Linear amplification means that our output should be a
proportional scaling of our input. For example if we feed in a sine wave we
should ideally get a scaled sine wave out, i.e with no distortion/clipping.
There are various circuit configurations used to achieve amplification using
transistors, a useful 'template' is known as common emitter configuration (shown
below with an NPN transistor):
![enter image description here](http://lunarjournal.github.io/images/3/37.png)
Here we model a 20mVp (20mV amplitude) sinusoidal signal source with a
resistance of 50 Ω, but your input can be practically anything.
It should be noted that there are two electrical 'components' of the above
circuit, these are AC (the fluctuating component) and DC (the static component).
When analysing a circuit from a DC perspective there are a few rules to follow:
* Capacitors become open circuits
* Inductors become closed circuits
This means that at the base of Q1, C3 becomes an open connection, i.e the base
of the transistor cannot see signal source V2 or the 50Ω. resistor .
Additionally, capacitor C1 becomes an open circuit and therefore has no effect
(it's as if all the capacitors weren't there in the first place).
Capacitor C3 is known as a DC blocking capacitor and is used to remove the DC
component of the input signal at the feed point (base of Q1). All signals have a
dc component:
![enter image description here](http://lunarjournal.github.io/images/3/21.png)
Effectively C3 serves to isolate the fluctuating(AC) component from the net
signal, that is, we need a signal that moves along the line y= 0.
Capacitor C2 is also a DC blocking capacitor and also serves to remove any DC
offset at the output of the amplifier.
The role of capacitor C1 is a bit more involved and requires and understanding
of AC circuit analysis, specifically the AC signal gain/amplification
Av which, for common emitter configuration, is given by:
![enter image description here](http://lunarjournal.github.io/images/3/22.png)
Here zout represents the output impedance of the common-emitter
amplifier which is given by the parallel combination of Rc and your
load resistance, RL (connected to C2).
![enter image description here](http://lunarjournal.github.io/images/3/23.png)
From an AC perspective:
* Capacitors become short circuits
* Inductors become open circuits
* Voltage sources become grounds
The term r'e is known as the transistor's AC base-emitter junction resistance
and is given by:
![enter image description here](http://lunarjournal.github.io/images/3/24.png)
The introduction of capacitor C1 nulls out the term Re from the
expression for Av. This is typically done to achieve higher values
of Av than would otherwise be possible if resistor Re was
still present. For lower, more controlled values of Av, resistor
Re should not be bypassed by capacitor C1.
The first step in the design of the amplfier is choosing Rc such that
zout isn't affected by changes in RL. For example, for a
large value of RL choose Rc << RL.
For the purposes of our example we assume RL = 100KΩ We then choose
Rc = 5KΩ
Next we determine the maximum AC gain possible given a fixed zout:
![enter image description here](http://lunarjournal.github.io/images/3/25.png)
It is usually good practice to give 20% of (Vcc/2) to Re and 70% to Rc. Higher
ratios of Vcc(Re) to Vcc(Rc) might lead to higher ac gain (Av) but
could sacrifice operational stability as a result.
Given Vcc = 5V, we get Av = 70. This is the highest
expected voltage gain for this amplifier.
We know that:
![enter image description here](http://lunarjournal.github.io/images/3/26.png)
Thus, given Av = 70, zout = 5KΩ we have IE =
0.35mA. We are now able to calculate Re:
![enter image description here](http://lunarjournal.github.io/images/3/27.png)
For Vcc = 5V, IE = 0.35mA we get Re =~2.1KΩ
A useful parameter for common emitter configuration is the AC input impedance
(looking in from C3) and is given by:
![enter image description here](http://lunarjournal.github.io/images/3/28.png)
Here Rbase represents the AC input impedance of transistor Q1
(looking into the base):
![enter image description here](http://lunarjournal.github.io/images/3/29.png)
We know how to calculate r'e from earlier and we use the minimum value of β (75
for 2N2222) to calculate Rbase:
![enter image description here](http://lunarjournal.github.io/images/3/30.png)
Thus Rbase = 5.4 KΩ
Returning to our DC analysis, we calculate the expected voltage at the
transistor base:
![enter image description here](http://lunarjournal.github.io/images/3/31.png)
We know that VRe is 30% of Vcc/2, which gives VB = 1.45V.
Now given IE = 0.35mA we can again use our minimum value for β to
calculate our required base current:
![enter image description here](http://lunarjournal.github.io/images/3/32.png)
Thus IB = 4.57uA
At this point we need to ensure that small changes in the value of base current
(which occur due to variations in β ) do not significantly effect the DC
operating point of the amplifier circuit.
In order to ensure a stable operating point we 'stiffen' the voltage divider by
ensuring the only a small fraction of the total resistor divider current flows
into the base of transistor Q1.
A good rule of thumb is to allow for 1% of the total divider current to pass
into the base of the transistor.
![enter image description here](http://lunarjournal.github.io/images/3/33.png)
We can therefore assume that IR1 ~= IR2 and solving the
above expression yields IR2 = 0.456mA. Since we know the voltage
across R2 (given by VB) we can calculate the resistance
value:
![enter image description here](http://lunarjournal.github.io/images/3/34.png)
This gives R2 ~= 3.2KΩ. Finally we calculate the value of
R1:
![enter image description here](http://lunarjournal.github.io/images/3/35.png)
R1 ~= 7.8 KΩ
The values of capacitors C3, C2 and C1 are chosen such that the capacitive
reactance (resistance at AC) at the desired signal frequency is minimal.
Capacitive reactance is given by:
![enter image description here](http://lunarjournal.github.io/images/3/39.png)
Now that we have all the required component values, we build the circuit in
LTSpice:
![enter image description here](http://lunarjournal.github.io/images/3/38.png)
A simulation of the DC operating point was performed:
![enter image description here](http://lunarjournal.github.io/images/3/40.png)
Here we can see our expected Vbase of around 1.45V and an emitter
current of around 0.38mA (instead of 0.35mA), not too bad! Let's measure the
voltage gain( with the signal source set to a peak amplitude of 1mV and a 100K
load attached):
![enter image description here](http://lunarjournal.github.io/images/3/43.png)
Our output across our load is seen reaching an amplitude of 60mV and so we have
a voltage gain of ~60.
# LTSpice
You can download LTSpice from
[https://www.analog.com/en/design-center/design-tools-and-calculators/ltspice-simulator.html](https://www.analog.com/en/design-center/design-tools-and-calculators/ltspice-simulator.html)