npn or pnp?

When looking at a BJT, the easiest way to decide whether it is npn or pnp is to look at the emitter, which is always modeled as the arrow. If you remember that the arrow tail is always at a ‘p’ node and the tip is at an ‘n’ node, you can easily decide whether the BJT is npn or pnp. Remember that the collector and emitter are always either both ‘n’ or both ‘p’.

Voltage Polarities

It is important to know which direction the voltage’s will appear positive when we begin using nodal analysis to solve BJT circuits. Typically, there will be a voltage drop of .7 V over the nodes that will be used in these calculations. Whether or is positive is decided by the type of BJT. The voltage polarities are flipped between pnp and npn BJTs. Obviously, the only difference in the symbols between the two types of BJTS is the arrow, which is the emitter. If we remember the tip of the arrow is the lower voltage, we are able to deduce that for an npn BJT and for pnp.

To be in the active mode, a BJT’s collector-emitter voltage must be above approximately .3 V. As above, this voltage polarity is reversed between npn and pnp BJTs. To determine, whether or should be positive, we can use our deduction of the base-emitter voltage polarity. The voltages, in active mode, drop from collector to base to emitter in npn BJTs and from emitter to base to collector in pnp BJTs. So, if we have figured out that we are using an npn BJT because was a positive .7V, we know that the base voltage is higher than the emitter voltage. From here we know the collector must be higher than the base, and therefore, higher than the emitter. We have just figured out that must be greater than the .3V to be working in active mode. Using the same logic, must be greater than .3V for a pnp BJT to remain in active mode.

Current Directions

Current directions are very simple to figure out. Just use the arrow. The collector and emitter currents always go in the direction of the arrow in active mode. The base current is a little more tricky to figure out, but is also fairly obvious when using the arrow as a reference. As you can see in the above npn circuit, where the arrow is ‘pointing’ away from the base, the base current flows towards the BJT, in the direction the arrow is pointing. Oppositely in the pnp circuit, the base current flows away from the BJT, in the direction the arrow is pointing. There is a table of basic equations listed in my post titled “BJT Transistor Nodal Analysis” which would allow us to calculate each current using a different current, but using Kirchhoff’s Current Law, knowing two currents, we could calculate the third. For a npn BJT, and for a pnp BJT . Note that both of these equations evaluate to .

It is also important to know that can be modeled as .

These equations are not very informative by themselves so a few examples are demonstrated below. In both examples we will assume is very large. What this means for our calculations is . Since we also assume that .

Finding missing voltages in a BJT circuit

Example 1. Solve for V3:

There are several ways to find . The more “difficult” way is to first find the emitter current, , and then use Ohm’s Law. Since we know , we can find the collector current, , and then solve for .

The easier way to find is to recall that behaves like a diode. For this pnp BJT: .

We know that so

may not always be a very large number. Had that been the case here, we would have started by finding the collector current (since it’s voltage drop and resistance are given) and since anymore, we would use the formulas above to the find the base and collector current.

Finding BJT Bias Voltages and Currents

Example 2 Solve for V2 and I1:

Here we will want to start by finding . also equals which approximately equals and this collector current will allow us to find .

Notice that was given as .  If this had not been given, we would have been able to find it because and .

Similar to the previous example, if was not a very large number. We would first find the emitter current and then use the equations in the table to find the other branch currents.

Note that both of these examples used pnp BJTs. The difference in an npn BJT is the base-emitter voltage is reveresed. You would use .

In General:

Most of these problems are very simple to solve. Typically is given and you will need to use Ohm’s Law to identify one of the currents. After one of the currents is found you will be able to solve for the other currents using the basic equations listed above. If one of the currents is not immediately obvious, the base-emitter voltage is likely needed. Most problems have you deduce the emitter voltage from the base, but it is easily possible to find the base voltage from the emitter voltage and then use that to find the base current.

To begin I will introduce our variables:

concentration of free electrons (donors)
concentration of holes (acceptors)
= number of free electrons and holes in a unit volume

In thermal equilibrium(or no doping):

and, therefore

However, doping is common in most examples. To increase the concentration of free electrons, an element with 5 valence electrons is used (i.e. Phosphorous). The resultant material is said to be n-type. To increase the number of holes, an element with 3 valence electrons is used (i.e. Boron). The resultant material is said to be p-type.

This introduces subscript n’s and p’s along with our concentration of free electron and hole variables.

In n-type silicon:

concentration of free electrons (in n-type silicon)
concentration of holes (in n-type silicon)

In p-type silicon:

concentration of free electrons (in p-type silicon)
concentration of holes (in p-type silicon)

Note: The subscript indicates whether the material is n-type or p-type.

Calculations

Typically you first want to identify whether the material you are working with is p-type or n-type. This introduces two new variables. which refers to the concentration of donor atoms and which refers to the concentration of acceptor atoms.

In n-type silicon:

Here you will use the variables , , , and .

In p-type silicon:

Here you will use the variables , , , and .

In most cases and or will be given and you will be able to find or . Then you will find or from the equations above.

Cartesian:

Cylindrical:

Spherical:

In most cases you will only need to work from Cartesian to Cylindrical or Spherical OR back, so I will only supply those equations. If you need to work between Cylindrical and Spherical, it would be one more simple step working from one of those, to Cartesian, then on to the other.

Cartesian Cylindrical:

Cartesian Cylindrical:

Cartesian Spherical:

Cartesian Spherical:

Plug the values from any given points into the correct equation to convert to a different type of coordinate system.

You would begin by finding the vector between these two points. The direction of this vector may be important so look for key words such as . Once we have established the direction we’re going in, to in this case, we subtract the beginning point from the end point. . This will give us the vector we are looking for. The next step would be to convert this vector into a unit vector, by dividing it by it’s magnitude.

These are the two formulas we are looking at:

Note:
Here I use as my vector from to and denotes the unit vector from to .

Implementing these formulas:

Hmm.. The End

RULE 1:

This is a scalar product that is equal to the two magnitudes multiplied together and multiplied by the cosine of the angle between them.  If  the two vectors are perpendicular to each other then the angle between them is 90 degrees, which will make the dot product equal zero.  This is an equivalent equation.

RULE 2:

When you finish your dot product, you should have a number, not a directional vector.  So if you get something like this you did something wrong: .  If you ended up with (any #) then you don’t have to completely rule out your answer.

A few vectors to practice with:

Okay, now we have found our dot product by applying RULE 2 above.  We can use this value along with our individual vector magnitudes to apply RULE 1 and obtain the angle .

We can calculate the projection of the vector A onto the vector B by this relationship:

Note that this is a scalar quantity, and that we can also define the projection of B onto vector A in a similar fashion:

Perform a dot product in MATLAB like so:

>>A = [ 1 2 3];

>>B = [2 3 4];

>>dot(A,B)

Enjoy

Vector

In MATLAB: >>A = [1 2 4];

Vector

In MATLAB: >>B = [2 3 4];

When you add vectors together, you add each individual directional component ().  Subtraction works the exact same way.  Lets go ahead and do a few, C will represent the vector that results from the addition and subtraction.

These vectors are all in 3-dimensional space with a X, Y and Z component.  The number in front of each directional component is the weight or magnitude of that particular directional component of the vector.  There it is, short and sweet

Capacitor stores energy in an electric field between two parallel metal plates which are separated by an insulator (dielectric) such as air, glass, paper, mica, teflon, tantalum oxide or electrolyte.

Parallel plate capacitor: in Farads (F) where F / cm.

Dielectric permittivity ε_o is a measure of the easiness with which lines of electrical force are established within the material. In a sense, it is the electrical equivalent of magnetic permeability. Relative dielectric permittivity (dielectric constant) is equal to ε_r = ε/ε_o. Capacitance is proportional to the dielectric constant, ε_r of the insulator material.

Relative dielectric constants for selected materials are: air 1.0, diamond 5.5, mica 7.0, polyester 3.4, quartz 4.3, and water 78.5. To achieve small-volume capacitors, a very thin insulator having a high dielectric constant is desirable. Real capacitors have a maximum voltage rating at which the dielectric material breaks down.

Charge stored is proportional to an applied voltage: Q V or or where the constant of proportionality is called the capacitance C.   Change in charge: which leads to or in more general terms:

Characteristics of capacitors (opposite, dual, to an inductor):

• Voltage cannot change instantly (instantaneously) in a capacitor; it tries to keep voltage constant in a circuit
• Impedes flow of low-frequency signal (blocks DC) but easily conducts high frequency ac current. Capacitor provides both ac coupling and dc isolation.

Stores energy in an electric field:

Time Constant: expressed in seconds, where the resistance R is the Thevenin’s equivalent resistance of the circuit as seen from the terminals of the capacitor.

Many capacitors are constructed by rolling alternating layers of aluminum and dielectric. Some are made in a form of ceramic disks. Voltage can be applied in both directions (may be reversed) to capacitors constructed from mica, Teflon, mylar, polyethylene or ceramic disks. In electrolytic capacitors, one plate is metallic aluminum or tantalum and the dielectric is an oxide layer formed by chemical reaction with a solution called electrolyte. The result is a large-value capacitor with a small volume. Electrolytic capacitors are “polarized” and only one polarity of voltage is allowed. Application of opposite polarity voltage causes the electrolyte to chemically dissolve the dielectric and this leads to capacitor failure.

Real capacitors have some parasitic elements in the form of lead inductance and series resistance as well as a parallel conductance due to leakage through the dielectric.

This should give you a simple understanding of a capacitor, non-graphically.  The next lesson will discuss reactance as it applies to Inductors and Capacitors.  Wahoo!

An inductor is made up of many turns of wire wrapped around a cylindrical, toroidal, or H-shaped core. The core may be empty (air) or it may contain a magnetic material (iron or ferrite) with high magnetic permeability to enhance the field.

One kind of an inductor is a solenoid which is a long, thin coil of wire. Its length is assumed to be much greater that the diameter of the coil. The magnetic flux density B within the coil is practically constant and is given by

where μo is the permeability of a free space, N the number of turns, I is the current and L the length of the coil. B-field has units of Tesla.

Magnetic flux measured in units of Weber (1Wb= Maxwells) where

is the cross-sectional area of the coil. Magnetic permeability of free space equals

whereas the magnetic permeability of the iron (ferrite) core μm=μr∙μo may be found from tables and is usually much larger than μo. The coil is wound up around a magnetic material such as iron, iron oxide, or ferrite to increase the magnetic flux for a given current. Iron cores are usually composed of thin sheets called laminations to reduce losses due to Eddy currents. Relative permeability μr is: iron 200, nickel 100, permalloy 8,000, mumetal 20,000.

Faraday’s Law applied to a coil leads to a derivation of electrical quantity called an inductance.  As you know by now from the Faraday’s law, the induced voltage in a coil is proportional to the rate of change in magnetic flux:

If the current changes with time the magnetic flux will also change in time with the same frequency:

Combining the above two equations one gets the following:

where the constant in front of a derivative $latex \frac{dI}{dt} is called an inductance

Electric current cannot change instantly (instantaneously) in an inductor. The current resists rapid changes; think of an inductor as having a large inertia like a heavy rotating wheel. It tries to keep the current constant in a circuit
Unit of inductance are Henrys (H). One Henry of inductance generates one volt of electricity when the change in current equals 1A/s.

Energy stored in a magnetic field equals to and is expressed in Joules.
Time constant: expressed in seconds, where the resistance R is the Thevenin’s equivalent resistance of the circuit as seen from the terminals of the inductor.

Real capacitors have some parasitic elements in the form of lead inductance and series resistance as well as a parallel conductance due to leakage through the dielectric.

The general equation for sinusoidal voltage signal is given by

Where A is the amplitude, is the angular frequency, t is time, is phase shift, and is a constant voltage.

Likewise, current follows the same general equation.

The equation for the sine wave in Figure 1 is .  The following sections will explain how to derive the equation for this sine wave as well as any other sine function that you may encounter.

Amplitude

The amplitude (A) is found by taking the difference between the highest point and the lowest point of a waveform and dividing by two.  In Figure 2, the highest point (VH) is 6 V while the lowest point (VL) is -2 V.  The amplitude, therefore, is 4 V.

Period, Frequency, and Angular Frequency

The period (T) of a waveform is defined as the time it takes for that waveform to complete one cycle.  Figure 2 shows two periods, or cycles, of a sine wave.  For this sine wave, the period is equal to 1 ms.  Identifying the period length allows you to derive the frequency (f) and angular frequency (ω) of the signal.

Frequency is defined as the number of cycles a waveform completes in one second.  One cycle per second is known as a Hertz (Hz).  The waveform above has a frequency of 1000 cycles per second, or 1 kHz.

The units for angular frequency are radians per second (rad/s).  Angular frequency is similar to the frequency measured in Hertz, except now we are measuring radians rather than cycles.  Because one cycle is equal toradians (360°), the angular frequency is derived by multiplying the frequency by.  A common error in homework problems occurs when students neglect to convert one measure of frequency to another.  Pay attention to which form you should be using and always remember the equation .

Offset Voltage

When a DC voltage is applied to an AC voltage signal, the AC signal is said to have a DC offset voltage.  A signal’s offset voltage can also be seen as the signal’s average voltage.  You can use integration to find the average value of a time-varying function, but for sine functions, it is easier to use observation.

Consider the sine wave in Figure 3.  The equation for this sine wave is . The waveform is oscillating around zero and has no offset.  In contrast, the sine wave in Figure 4 has an offset of 2 V.  This time, the waveform is oscillating around 2 and its equation is .

When determining the offset of a sine function, remember that an offset is a constant value.  Adding a constant to a sine function only changes the level about which the sine wave oscillates.

Phase Shift

A sine function has a phase shift when it does not begin its cycle at t = 0. A phase shift is illustrated in the figure to the right. To calculate phase shift, use the equations,

where t1 is the waveform’s original starting point, t2 is where the wave’s starting point has been shifted to, and T is the period.  

Example: Find the phase shift of the sine function in Figure 5.

Solution: The sine wave has been shifted from its original starting position (t1) of 0 ms to its new starting point (t2) at -0.4 ms.  Therefore,
t1 = 0 ms, t2 = -0.4 ms, and T = 1 ms.  Solving for degrees, the phase shift is equal to 144°.

Defining phase shift with respect to t = 0 works fine in mathematics, but real life won’t tell you when time is equal to zero.  In labs this semester you will be observing signals on an oscilloscope.  For a single signal, you can assume that there is no phase shift.  Phase shift becomes important when more than one signal is involved.  There are times when knowing the phase shift between two signals is crucial (the design of an audio system is one instance).

In instances with multiple signals, the phase shift of one signal is measured relative to another signal. One wave is your reference point from which you measure all other waves.  Figure 6 shows V₁ as the reference and V₂ as the wave that has been shifted.  Figure 7 shows the opposite.  Once you decide on a reference, the same formula given above applies.

The sign of the phase shift will vary depending on which signal you use as a reference.  The phase shift from V1 to V2 in Figure 6 is positive because V2 is shifted to an earlier time.  This is a confusing concept for many people.  Though it appears that V1 is ahead of V2, this is not the case.  V1 starts its cycle a full 3 ms before V2 does.  V1, therefore, is leading V2.

Another way of explaining this is by using the equation.  In Figure 6, t1 – t2 must be positive because t1 is larger.  In Figure 7, t1 – t2 must be negative.

Written by Ryan Eatinger (reatinge@ksu.edu).  Thank you!