Before we start let’s think a little bit about what we’re doing:
Our goal is going to be to find the pole(s) of the circuit.
Okay? What is a pole and why do I care where it is?
A pole is a frequency at which the gain of the device rolls off. (remember that when it rolls off , it will be at the -3dB frequency with a slope of -20dB/decade)

We care because if the gain of a device rolls off at a certain frequency, then we won’t be able to amplify a signal above that frequency very well because the gain will be decreasing by 20dB/decade.

The procedure is nearly identical whether we are using a BJT of a MOSFET, but we will work each of them side by side just in case there might be any confusion, and we’ll follow these steps as we go through.  (we will also use some values that came from the output file when running a simulation of this circuit in Cadence (or PSPICE) )

1. Take a look at one of the circuits and see what you notice, how about the MOSFET.  This step is just to help us with our knowledge understanding of the circuit.
- At a glance it just looks just like another MOSFET right? Sure is, but let’s take a look at a few things just for kicks. Notice that it is using a bypass capacitor at the source so we don’t have to worry about (at when working with high frequency).  Since the capacitor bypasses to ground, you should notice that this is a common-source amplifier.  You could notice the Values for and and start to think about what the Gate voltage is and how that may affect the circuit.
2. We are talking about frequency response so that means we are probably going to want to draw the small signal equivalent circuit.
Remember that the capacitors and will act like short circuits at high frequencies so we will ignore them, but we will have to account for some of the capacitance internal to the device.

Both devices have internal capacitances that are very similar.  As you can see from the small signal models for a MOSFET (above) and BJT (below), the only significant difference is that the BJT has an additional resistance Rpi between the Base and Emitter.

Most of the analysis we will do is based on the small signal model. Note that small signal models are not typically used in PSPICE so this picture may look a bit odd, especially the controlled source but for our purpose it is good to have a visual reference. To start we will point out what everything is. Cgs is an internal capacitance betwe

en the gate and source. The

values for Cgs was similar to one the a PSPICE simulation may give.  CM1 and CM2 are Miller capacitances which we will find values for later.  ro is a Norton equivalent resistance that makes the model more ideal.  And just pretend that the G2 looks more like a voltage controlled current source and that their gains are gm*Vgs and gm*Vpi. For the BJT CM1 and CM2 are both Miller capacitances, Cpi is similar to Cgs and Rpi the additional component used for BJTs but not MOSFETs. The other part should look familiar from the other figures.

ON TO THE ANALYSIS!!!

We will find the device gain, overall gain, equivalent input and output capacitances, and the input and output poles. The process for both is essentially the same.

Device Gain: This is the gain from the control source to the output so we are looking for Vout/Vgs (or Vout/Vpi for a BJT). We will ignore CM2 for this process. Notice the resistances ro, RD, and RL are in parallel. Vout should be given by that equivalent resistance times the current though it which is gm*Vgs from the control source. So the equation for device gain is,

   (MOSFET)

  (BJT)

Overall Gain: This will be the gain from the source (Vs) to the output (Vout). We already know what Vout/Vgs is so if we find Vgs/Vs, we can multiply them to get Vout/Vs = (Vout/Vgs) * (Vgs/Vs).  Vgs/Vs is a simple voltage divider. Hopefully you can see this from the small signal model (remember that we are ignoring the capacitors for now but they will play a part later).  The equations we will get for Vgs/Vs and the overall gain are.

  (MOSFET)

Overall Gain:   (MOSFET)

  (BJT)

Overall Gain:   (BJT)

Now we will find the input and output poles.  For this we will need to look at the capacitances and use a formula to find the Miller capacitances, CM1 and CM2.  Any explanation for the miller capacitance will have to wait for another post or check out your Electronics Book, Wikipedia, Google, etc. but we will need to use a couple of special equations.  Overall we will need to find the input resistance and input capacitance for the input pole and the output resistance and output capacitance for the output pole.

Each pole will be at a frequency w=1/RC where the R and C are the equivalent R and C at that point, so to find the input pole, we will need to find the input resistance and the input capacitance.  These are found by looking into the input (the left side of the small signal model).  The voltage source will  act like a short so we see Rs in parallel with R1//R2 for the MOSFET (the BJT will have Rpi in parallel also).  The input capacitance will be Cgs in parallel with CM1 (the BJT will be the same).  The output resistance and capacitance are found the same way only looking in from the output (the right side of the small signal model).

      (MOSFET or BJT)

So the input pole will be: (MOSFET)

  =  950                                     =

  =                                               =

=                                                                          =

(BJT)

and the output pole will be: (MOSFET)

(BJT)

=                                  =

=                                                 =

=                                        =

To Do:

finish input & ouput R, input C, Pole (& calculate answers)

Amplifiers

Definition (for Bioinstrumentation): Circuit that makes a small signal, usually voltage but occasionally current or power, big enough to do something useful–including excite an output mechanism.

Common uses:

• Biological measurements of small signals
• Audio engineering: a large current is needed to drive speakers
• Wireless communications: far from originating antenna, signal is very weak and must be
amplified to be useful.

Secondary applications:

Many amplifiers are also filters, preferentially amplifying some frequencies over others

• Don’t want to amplify noise along with signal
• Only interested in low- or high-frequency portion of signal
• Active filter provides amplification as added bonus

General Amplifier Characteristics

Common Mode Rejection Ratio (CMRR)–ratio (usually in dB) of the amplifier’s common-mode gain to its differential-mode gain. Common-mode signals are input signals common to both + and – inputs and are usually unwanted noise–60-Hz, thermal, etc; differential signals are applied to only one input.

Gain–voltage out over voltage in, or current out over current in. May be given in dB.  Bioamp requirement: often adjustable; should be 1000 or greater, should be calibrated.

Input Impedance–what the input source sees as its load working into the amplifier: if the entire amplifier circuit were modelled as a resistor, what would be the value of the resistor? Typical bioamp: Rin = 10MΩ–signal source need not provide much current.

Output Impedance–same as input impedance but from the output end: model the entire amplifier as a source, and this is its internal impedance. Bioamp requirement: Ro << Rload.

Frequency Response–over what range of frequencies is the gain constant? Graphically illustrated with a Bode plot of gain vs. frequency.

DC offset–usually an amplifier has an operator-adjustable DC offset knob, to null out any offset associated with non-ideal amplifier or sensor behavior. A DC offset signal results in an incorrect reading unless removed or filtered out (high-pass filter).

Operational Amplifier:

basis for most instrumentation-related amplifiers, cheap, readily available, easy to work with. “Operational” = good for mathematical operations (+, -, log, …)

Meanings and advantages:

• equal input voltages –> within the limits of external power supplies, an op amp outputs whatever current is needed to drive the two input voltages equal. Result is that the output voltage follows the input, scaled by a large gain.
• infinite input resistance means the op amp never loads down the source, even if the source cannot supply much power.
• zero output resistance means the op amp is an ideal voltage source, with output voltage independent of whatever load impedance it must work into.
• infinite open-loop gain means the amplification properties of a circuit containing an op amp are independent of the op amp internal properties.

Carr and Brown go through several common op-amp configurations and show how to derive their voltage gains. Suffice it for now to know that if you want to build an amplifier, an op amp is a good place to start.

Common op-amp circuit configurations:

• Inverting and non-inverting amplifiers
• Summing and difference amplifiers
• Integrating and differentiating amplifiers
• Log and anti-log amplifiers
• Instrumentation amplifier
• Low-pass filter
• High-pass filter
• Band-pass and notch filters
• Buffer (voltage follower, or unity-gain buffer)

Op Amp Equivalent Circuit

This schematic illustrates the important properties of the op amp, and of any amplifier. It can also make it easier to understand circuit operation.

Note the open-circuit inputs– Rin = infinity. The
output voltage supply is a dependent voltage
source. Also, since the gain A is infinite, v2 – v1
must be zero to get a finite output.

Difference (Differential) Amplifier

Example: derive the gain relationship for the basic differential amplifier shown, assuming U1 is ideal and Vin = V2 – V1.

To get equal gain of both V1 and V2, set R2/R1 = R4/R3. Then Vo = R2/R1(V2-V1).

To get a high gain, R2 >> R1, but to get high input impedance R1 (and/or R3) should be large, making R2 and R4 even larger…Result: high gain and high input impedance are difficult to achieve together.

Instrumentation Amplifier

A difference amp with input buffer/gain stages to increase input impedance and gain. To analyze, realize that the same current must flow in R5, R6 and R5 (since no current flows into the op amps). Set R1=R3, R2 = R4; then Vo = G1* (v3(U1) -v2(U1)), where G1 = R2/R1 = gain of second (differential) stage.

Gain of input stage is 1 + 2*R5/R6 = G2. Overall gain is G1*G2. Making R6 a potentiometer allows compensation for inequalities in the two R5s, as well as for variable gain of the entire circuit.

Overall Gain: A practical difference amp can have a gain of 100, so it is not hard to get an overall gain of 10,000 from an instrumentation amp.
Input Impedance: equal to that of the op amps U1 and U2–very large. Use FET-based amps for extremely high input impedance
Output Impedance: close to that of the op amp U1–very small: the amp will provide whatever current is needed to maintain the output voltage regardless of load impedance.

Equal resistors: in practice one cannot buy matched discrete resistors; however it is fairly easy to manufacture them within an integrated circuit. Monolithic diff-amps are available.

Non-idealities of amplifiers

Gain:  TANSTAAFL–you cannot have gain without a power supply to provide it. Real gain is limited by the external power  supplies (+/- 12 or 15 V, for op amp circuits) Exceeding the limits of the power supply results in Saturation, or “hitting the rail”.

Output impedance: a zero output impedance means the circuit will provide whatever current is needed to maintain the requested output voltage. Practically, however, an op amp can only provide some 20mA, meaning RO is negligible only for RL>>15V/20mA = 750 Ω.

Frequency dependence: to avoid oscillation or saturation, circuitry must often be added that limits the bandwidth of an amplifier.
• To keep DC offset signals (from polarizing electrodes, for example) out of the amplifier, a high-pass filter is used to cut off DC (and lower-frequency ac) signals.
• If the load to be driven contains substantial capacitance, the current output limit again becomes a problem, limiting gain at high frequencies, where capacitors look like shorts.

Input bias current: real op amps do have non-zero input currents, which produce voltage drops at the input–another source of DC offset. This source can be minimized by using FET op amps.

Impedance Bridge

Often the measurand is the relation between voltage and current (one applied, the other a response) rather than a biologically generated source. An example in Carr and Brown uses a wire heated by an applied current as an airflow sensor:  air flow from a breathing patient cools the wire, changing its resistance. Similarly, a voltage applied to a membrane induces a current flow; the ratio of voltage to current is a resistance. Such relations are best measured using a Bridge, and if the bridge is made solely of resistors it is called a Wheatstone Bridge.

Usually drawn as a diamond, this configuration of resistors is “balanced” when V+ – V- = 0. If Rtest then varies a little, a differential amplifier across V+ and V- will register a potential difference proportional to
the change in Rtest.

The impedances can have capacitance and/or inductance associated with them, in which case the bridge can measure both energy storage and  resistive loss in an element.

A return path to ground for (DC) bias currents is automatically provided by this circuit to prevent saturation.

Well there you have it, a few common amplifier configurations and some useful terms pertaining to them.  Remember important concepts such as amplifier saturation, Input Impedance, Output Impedance, and Gain.  A solid understanding of these concepts is sure to impress somebody!  Amplifiers  part II will continue to elaborate on more fun amplifier concepts.

References: References: Carr and Brown ch. 7; Webster chs. 3, 6; Neamen, Electronic Circuit Analysis and
Design (McGraw Hill, 2001) ch. 9

Since these integrals can be tedious and certain functions tend to reoccur, a table of Laplace Transforms has been linked:

Laplace Transforms Table

This table can be a little complex to use at first so an example is provided below to get you started. In this problem we implement the Laplace Transform and Inverse Laplace Transform to solve for .

The first step is to take the Laplace Transform of both sides of the equation. Use element 1 of our table for the right side and element 18 for the left side. Note that the initial conditions are necessary to take the Laplace Transform of the left side.

Inputting our initial conditions:

Assuming you are an engineering student and can do a little alegebra, our next step is to find the terms that have in common and factor it out. Our goal is to find because, after all, :

After solving for and factoring the denominator:

Now we arrive at the trickier part of this procedure. We must take the Inverse Laplace of to find . If our function does not match anything in the table, such as this case, factoring is a good place to start. This problem can easily be factored using the function on your TI-89. Just go to and enter your function in parenthesis. Using this function:

Noting and , we can take the Laplace Transform of each term independently and also manipulate the constant terms if necessary or just pull them out. Using the 2nd property in our Laplace Transform table:

To check your work you can plug into the original, differential equation and at we find that .

The best method for quickly assessing the performance of this circuit at a variety of loads is to find the circuit’s Thevenin equivalent.

Once we have the equivalent circuit, the problem can be solved using three simple voltage dividers.

Finding Thevenin Equivalents in Practice

By using the Thevenin equivalent model, the problem was solved without the need to perform a full circuit analysis each time the load changed.  However, you must learn how to acquire Thevenin equivalents before you can take advantage of them.  In practice, finding the Thevenin equivalent of a circuit is simple.

• Find V_th by measuring the open circuit voltage with a multimeter.
• Find R_th by connecting a current meter to the two terminals and dividing V_th by the measured current (called a short-circuit current).  Note that this isn’t the safest method and should never be used in practice! A safer and more proper method for finding R_th is outlined in Lab 3.  That is, find a load resistance that produces a noticeable drop in the load voltage.  Then find R_th by applying the voltage divider formula.

Thevenin Equivalent Resistance

In theory, finding the Thevenin equivalent is more difficult because we can’t rely on lab equipment to do the work for us.  On paper, the Thevenin equivalent resistance (R_th) is easier to find.

1. If there is a load, remove it – Remember that Thevenin’s theorem applies to two terminal circuits.  This implies that the circuit is unloaded (see Figure 4).  After all, V_th is equal to the open circuit voltage, as you saw above.

2. Short out any voltage sources and open up any current sources – All sources must be deactivated to find the Thevenin equivalent resistance.

3. Find the equivalent resistance looking into the two nodes – In Figure 4, the two nodes would be a and b.  The easiest way to find the equivalent resistance is to start at node a and end at node b.  If there are multiple paths from a to b, then there are parallel resistors somewhere in the circuit.

Example 1: Find the Thevenin equivalent resistance with respect to nodes a and b for the circuit in Figure 5.

Solution: Follow the steps listed above.  There is no load to remove because nodes a and b are already open.  Next, deactivate the sources by shorting the voltage source and opening the current source.  Shorting the voltage source effectively shorts out the 40 Ω and 15 Ω resistors, meaning they have no bearing on the value of R_th.  The last circuit in Figure 6 shows the two paths current can flow from a to b.  R_th is given by the equation below.

Thevenin Equivalent Voltage

Any of the circuit analysis techniques learned so far can be used to find the Thevenin equivalent voltage (V_th).  Voltage dividers, branch currents, node voltage, superposition, and source transformations (the last two will be covered soon) are all legitimate methods for finding V_th.  Of the two Thevenin equivalent values, V_th tends to be the harder one to find because there is no one best way to work all problems.  Node voltage may work well on one circuit, but may be weighed down by multiple variables on another.  Voltage division isn’t always easy, either.

Until you can recognize which method to use, perhaps the best approach is to try one method and see if it works.  If the math becomes too overwhelming, move on to a different method.  Also, realize that the equivalent voltage is measured across two nodes, as shown in Figure 7.  Be certain V_b is connected to ground before declaring that V_a is equal to V_th. In this case V_th is equal to V_a – V_b. Finally, if there is a load, remember that you must remove it before working the problem! Once again, a Thevenin equivalent is derived from an unloaded circuit.  A loaded circuit alters the equivalent values.

Example 2: Find the Thevenin equivalent voltage with respect to nodes a and b for the circuit in Figure 5.

Solution: The circuit has no ground marked.  This must be placed first.  In general, it’s most convenient to place the ground at the negative terminal of a voltage source.  Placing the ground at the negative terminal of the 17.4 V makes the voltage at the node connected to the source’s positive terminal 17.4 V.

Note: This node would not be at 17.4 V had the ground been connected elsewhere.

The next step is to find the two node voltages, V_a and V_b.  Because there is a resistor between node b and ground, V_b must be found in addition to V_a.  The current source combined with the arrangement of resistors make voltage division all but impossible.  Instead, try the node voltage method for node a.

Notice that the last term used 14 instead of 10.  This is allowable because the 4 Ω and 10 Ω resistors are in series.  Find V_b by using either node voltage or voltage division.  In this case, it’s faster to use voltage division.  If this isn’t obvious, redraw the branch containing node b as shown in Figure 9.

Now solve for V_th using the equation from Figure 7 and draw the Thevenin equivalent.

See, Thevenin’s Equivalent is a lot of fun!  Get good at this and you have mastered quite a few important engineering concepts!  Thanks to Ryan Eatinger (reatinge@ksu.edu) for contribution of this lesson.

The circuit in Figure 1 shows three resistors meeting at a node.  We’ll use this node to show how node voltage works in the most general sense (i.e. without using any numbers).  As you’ll see, the node voltage method is basically Ohm’s law applied to Kirchoff’s current law.

Note: All of the voltages in Figure 1 (and other figures throughout this article) are node voltages, or voltages at a node, which are taken with respect to ground.

1. Pick a node – This can be any node.  In the problems you’ll see, you’re told which node to use.  In reality, node voltage can be applied to any node because KCL can be applied to any node.  Let’s pick the V_x node for this example.

2. Apply KCL to the node – There are three paths or branches connected to the V_x node and each path has its own current.  These currents are labeled as I₁, I₂, and I₃.  Rather than trying to guess which way these currents flow, assume that all currents leave the node.  If we assume that all currents leave the node, then we’re also assuming no currents enter the node, simplifying the KCL equation as shown below.

Note that it doesn’t matter if our assumption is wrong.  If one of the currents is actually flowing into the node, it will show up later as a negative in the KCL equation (and we’re not afraid of negative numbers).

3. Write equations for each current – Unless one of the branches is a current source, you’ll have to place each current in terms of voltages and resistances according to Ohm’s law.

4. Derive the node voltage equation – Plug the equations you found in step 3 back into the KCL equation in step 2 to derive the node voltage equation for V_x.

5. Solve for any variables – In most problems you’ll encounter, all of the variables will be known except the voltage at the node.  In this instance, solve for V_x and you’re finished.  A node voltage equation can be written for any node in a circuit.  For multiple variables, you’ll need multiple equations (more on this later).

Example 1: Find the voltage V₁ for the circuit in Figure 2.

Solution: The first step has already been taken care of in the problem statement.  The next step is to write the KCL equation for the V₁ node.  Mark the currents as shown in Figure 3, assuming that they all flow away from the node.  The KCL equation becomes:

Next, write equations for each current.  The two voltage sources are connected to ground, making it easy to find the voltages at the nodes on the upper left and right corners of the circuit.  Notice that V_a’s positive terminal is connected to ground, making the voltage at the negative terminal -8 V.

Using these equations, derive the node voltage equation for V₁.  Solving for V₁ gives the voltage with respect to ground at that node.

Working with current sources

Example 2: Find the voltage V₁ for the circuit in Figure 4.

Solution: The approach doesn’t change when there are current sources in the circuit.  Current sources can actually simplify the math.  Once again, the node has been chosen by the problem statement.  Now write the KCL equation and solve the problem the same way as before.  Note that the current source points into the node and therefore needs to be added as a negative current.

Working with Multiple Variables

Example 3: Find the voltages V₁ and V₂ for the circuit in Figure 6.

Solution: Approach every node voltage problem in the same manner: choose a node and derive its node voltage equation.  If the equation has more than one variable, then move to the next node and write another node voltage equation.  Continue this process until the number of equations equals the number of variables.  In this example, there will be two variables (V₁ and V₂).  This means two node voltage equations are required to find the two voltages.

This time there are two nodes to choose from.  Start with V₁ first.  As always, assume the currents flow away from the node.

Deriving equations for I₁ and I₂ is the same as before, but I₃ is slightly different.  The voltage V₂ is unknown, but it can still be placed into the equation.  It will be treated as a variable just like V₁.

The node voltage equation for V₁ becomes

Now follow the same procedure to find the node voltage equation for V₂.

Finally, solve the system of equations to find V₁ and V₂.

That should just about cover quite a few node-voltage examples and concepts.  If you have any questions feel free to ask them in the questions section of the website, or simply leave a comment below!  Many thanks to Ryan Eatinger (reatinge@ksu.edu) for contribution of this post.

The general equation for a current divider is given below and illustrated in Figure 1.  I_o is the measured current, I_s is the source current (or the current entering the node, R_o is the resistance of the path I_o flows through, and R_T is the equivalent resistance of the circuit.

General Current Divider Equation: 

Notice that this equation is nearly identical to the voltage divider equation, except that R_T and R_o are inverted.  R_T is in the numerator because current dividers work exclusively with parallel resistors.  Therefore, R_T is always smaller than any of the individual resistors because of the nature of the parallel resistance formula.  The result is that I_o is always less than or equal to I_s.

The current divider equation is derived in the same fashion as the voltage divider equation.  Consider Figure 2.  All three components are connected in parallel, meaning all three have the same voltage across them.

Use Ohm’s law to place the equation above in terms of resistance and current.

Ohm’s Law: V = IR

Now use these equations to find I₁ and I₂.

Example: Find the current I_o for the circuit in Figure 3.

Solution: This problem requires two current dividers to solve.  The current first splits between R₁ and the rest of the circuit and then again between R₃ and R₄.  Because I_o is part of the second split, the current through R₂ must be found first.  Remember, though, that the current through R₂ also passes through R₃ and R₄.  The current divider equation cannot be applied without taking the resistance of the entire path into consideration.

Combine R₂, R₃, and R₄ before applying the first current divider equation.

The equation above shows that the 36 Ω resistor takes 1 A of the 4 A available to the circuit.  Therefore, the second current divider must use 3 A as the source current because that is the amount of current flowing into the node.  Also, do not try to incorporate R₂ into the second equation.  The current splits between R₃ and R₄; R₂ has nothing to do with where the current flows once it reaches the node.

Thanks to Ryan Eatinger (reatinge@ksu.edu) for the contribution of this post.

Ohm’s law defines the relationship between voltage, current, and resistance by the equation below.  The second equation better represents voltage as the difference between two electric potentials.

Note that V₁ and V₂ are voltages measured with respect to ground and V is the voltage potential measured between them.

The equation derived from Ohm’s law is incredibly useful for an electrical engineer.  Ohm’s law allows many circuits to be fully analyzed with the aid of just a few measurements.  Many circuit analysis techniques you will learn involve some combination of KCL, KVL, and/or Ohm’s law.

Ohm’s law states that there must be a voltage drop (or voltage difference) across a resistor in order for any current to flow.  If V₁ and V₂ are the same, no current flows through the resistor.  In terms of currents, a current produces a voltage drop across a resistor; if there is no current, there is no voltage drop across the resistor.

A circuit component that follows Ohm’s law has a constant resistance.  Increasing the current through the component will produce a proportional increase in the voltage drop across it.  The plot of the current through the component versus the voltage drop across it will be linear, with the slope of the line determining the resistance of the component.

Figure 2 shows I-V plot of a 1 Ω and a 2 Ω resistor, both of which follow Ohm’s law.  Figure 3 shows the I-V curve of a component not following Ohm’s law.  You will learn about two components that don’t follow Ohm’s law later in the course (diodes and MOSFETs).

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

This is my favorite approach to a problem like this one:

1) Identify meshes in a planar circuit.

2) Identify currents unknown.

3) Write KVL for each mesh.

4) Simplify and Solve.

Easier said than done I suppose, so let’s do it:

It is important to understand how many equations you are going to need.  Here is how you do that:

Using Node-Voltage:   #Equations = Ne – 1

Using Mesh-Current: #Equations = Be – (Ne – 1)

Where Ne = Number of Node,  Be = Number of circuit components.

Here we have:

Ne = 4

Be = 6

Be – (Ne – 1) = 3 equations.

Now you are ready to start writing your mesh equations.  You should only do one loop at a time.  This is simply KVL at work here, adding up the voltages around the loop and making sure they sum up to zero or to the source voltage.

Mesh 1: 5 ( i1 – i3) + 26 (i1 – i2) = 80

Mesh 2: 26 ( i2 – i1) + 90 (i2 – i3) + 8 (i2) = 0

Mesh 3: 30 (i3) + 90 (i3 – i2) + 5 (i3 – i1) = 0

We now have 3 equations and 3 unknowns, this is what we want and we should go ahead and move the “80″ in the Mesh 1 equation to the left side so that all equations sum to zero.

Confused by this? “( i2 – i1)”

That is normal, and probably the most confusing concept.  All 3 of the currents in all 3 loops are heading in the same direction, but in relation to each circuit component (such as 90Ω resistor), they are opposite.  If you are in loop number 1, this current (i1) is your reference current for this given loop, and the current through a resistor in that loop is equal to that reference current minus the opposing current on that given component.  If the currents were heading in the same direction, they would simply add instead of subtract.  This way we have summed up the total current through each component and can therefore calculate the voltage.

You can either use a TI-89 to solve for these unknowns…

OR

=

Solving the 3×3 system

I1 = 5 A

I2 = 2.5 A

I3 = 2 A