ECE101 – Basic Electrical and Computer Engineering

If you’ve studied the history of electricity at all, you have certainly heard of The War of the Currents.  Edison wanted DC current for power distribution, Tesla wanted AC current.  After quite the dirty little fight, Tesla came out on top with his AC current.  But why?  Despite all the arguments and under handed tactics to persuade the public, it came back to Ohm’s law to decide who ultimately was the winner.

If all wires were superconductors, there wouldn’t be an issue using DC power.  But, in reality, copper wire has a finite resistance.  And as we know from looking at ohm’s law, there has to be a voltage drop across the wire to move a set amount of current through.  As the current increases, the voltage drop increases.  In reality, a run of less than a hundred feet can have a significant voltage drop at the load. So, we take a quick glance at the power equation and can come up with a simple solution.
 

All we care about is power at the end. And as is obvious, it doesn’t matter if most of that power is because of high voltage or high current. So by increasing the voltage, we get a dual benefit. First is that we don’t have to move as much current, so there is a smaller voltage drop. Second is that the voltage drop is a smaller percentage of the overall voltage.

Yet we don’t want to have kilovolt lines coming into our house because of the obvious danger. This is where AC power is the clear winner. Whereas it is difficult and lossy to convert DC voltages, AC voltage merely requires a transformer, both easy and relatively efficient. So power companies create their power, transform it up to extremely high voltages for transmission over long distances, and then slowly step it down to the more reasonable and safe voltage level used in your home.

That’s basically it. The simple ability to transform an AC voltage put it head and shoulders above DC in the War of the Currents and won because of it.

I’ll put up a text version when time permits.

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Summing Amplifier

Note: For this and future circuits topics I’ll be using LTSpice, which is provided free from Linear Technology here.  I also recommend viewing my LTSpice Tutorial.  Also, unless otherwise stated, any future discussions involving op-amps will assume an ideal op-amp.  This will assume that the voltages at both inputs are at the same voltage and that no current enters either input.

The Basics

The summing amplifier is similar to the inverting amplifier, but has multiple inputs each with their own input resistor.  Only three inputs are shown here, but there can be two or more (having only one input would be an inverting amplifier) and my gain equation and derivation are shown with the standard “n” inputs.  The schematic and gain equation are below as well as a simple simulation.  Continue reading below for more information on the derivation of the gain equation.

Schematic:

Gain Equation:

Simulations:

Sweeping V1

Sweeping V2

Sweeping V3

Gain Derivation

The derivation of the gain equation for this amplifier is very simple using Kirchoff’s Current Law (KCL).  First, remember that we are considering the op-amp to be ideal, meaning that the inputs are at the same potential.  With the non-inverting input grounded, this creates a virtual ground on the inverting input.  Also, we assume no current enters the inputs of the op-amp.  With these two assumptions, the circuit can be simplified as shown below.

Simplified Schematic

Using Kirchoff’s Current Law in the direction indicated by the arrows and treating Vgnd as zero volts results in the following derivation.

Gain Derivation

For my simulation, I selected varying values for the input resistors, resulting in a weighted sum and swept each source from 0 to 0.5 volts.  The effect of the difference in resistor values is shown by how much the output varies depending on the source being swept.  Since V3 is weighted more than V2, which in turn is weighted more than V1, the output varies most when V3 is swept and least when V1 is swept.

Difference Amplifier

Note: For this and future circuits topics I’ll be using LTSpice, which is provided free from Linear Technology here.  I also recommend viewing my LTSpice Tutorial.  Also, unless otherwise stated, any future discussions involving op-amps will assume an ideal op-amp.  This will assume that the voltages at both inputs are at the same voltage and that no current enters either input.

The Basics

The difference amplifier, as the name suggests, amplifies the difference between two inputs with the gain being set by the two resistor pairs.  The schematic and gain equation are below as well as a simple simulation.  Continue reading below for more information on the derivation of the gain equation.



Schematic:

Gain Equation:

Simulation:

Gain Derivation

The derivation of the gain equation for this amplifier involves applying Kirchoff’s Current Law (KCL) to each node.  Since we assume the inputs to be at the same potential and that no current enters them, we will first solve the simplified circuit at the non-inverting input and then substitute that value into the simplified circuit at the inverting input.

Simplified Schematic

Gain Derivation

Using KCL in the direction indicated by the arrow on the bottom portion (the non-inverting input) V+ can be solved for according to the following derivation.

Solving the top portion of the circuit for Vout is shown in the following derivation.

Now, remembering the assumption that V+ and V- (the two inputs to the op-amp) are at the same potential, V+ can be substituted in to the last equation for V-.  This substitution and some reorganization are the final steps in the derivation.

For the simulation, I chose to make R1=R2 and Rf=Rg.  With Rf=10*R1, the gain is 10*(V2-V1).  I set V2=0.5V and varied V1 from 0 to 1V, giving a difference ranging from -0.5V to 0.5V and the output ranging from -5V to 5V.

I just wanted a quick post for people like me who think that some things can’t be as easy as they seem. In this case, using a crystal oscillator.

Looking in the data sheet for your PIC, it will show you the different ways the PIC’s clock can be driven. You’ll notice that in terms of hardware, the only thing you have to do is hook the two sides of your oscillator to the two designated oscillator pins on the PIC and then hook those same two pins to ground in series with very small capacitors. The optimal values of the capacitors is found in the data sheets of either the PIC or the crystal. That is it for hardware. For software, it is just as easy. Once again, using the data sheet as a reference, you set up the configuration word according to the crystal speed and type you have.

Once the PIC is powered, you can use an oscilloscope to check the oscillator pin. It will not be a square wave, just a very steady sine wave. However, without power, the oscillator won’t be doing anything.

And that’s it. “Wait,” you may say, “that’s really it? All that information is in the data sheets.” And that was the point. There is absolutely nothing tricky about the crystal oscillator and no reason you shouldn’t be able to use one in your projects.

Non-inverting Amplifier

Note: For this and future circuits topics I’ll be using LTSpice, which is provided free from Linear Technology here.  I also recommend viewing my LTSpice Tutorial.  Also, unless otherwise stated, any future discussions involving op-amps will assume an ideal op-amp.  This will assume that the voltages at both inputs are at the same voltage and that no current enters either input.

The Basics

The non-inverting amplifier consists only of an op-amp and two resistors.  The schematic and gain equation are below as well as a simple simulation.  Continue reading below for more information on the derivation of the gain equation.



Schematic:

Gain Equation:

Simulation:

Gain Derivation

The derivation of the gain equation for this amplifier is very simple using Kirchoff’s Current Law (KCL).  First, remember that we are considering the op-amp to be ideal, meaning that the inputs are at the same potential.  With the non-inverting input grounded, this creates a virtual ground on the inverting input.  Also, we assume no current enters the inputs of the op-amp.  With these two assumptions, the circuit can be simplified as shown below.

Simplified Schematic

Since Vin is connected to the inverting input of the amplifier, it is present also at the non-inverting input as shown in the circuit.  Using Kirchoff’s Current Law in the direction indicated by the arrow results in the following derivation.

Gain Derivation

As you can see in my simulation, I chose Rf=900k and Ri=100k giving me a gain of 10.  My simulation sweeps the input from 0 to 100mV and the output is shown to go from 0 to 1V as would be expected.

Inverting Amplifier

Note: For this and future circuits topics I’ll be using LTSpice, which is provided free from Linear Technology here.  I also recommend viewing my LTSpice Tutorial.  Also, unless otherwise stated, any future discussions involving op-amps will assume an ideal op-amp.  This will assume that the voltages at both inputs are at the same voltage and that no current enters either input.

The Basics

The inverting amplifier consists only of an op-amp and two resistors.  The schematic and gain equation are below as well as a simple simulation.  Continue reading below for more information on the derivation of the gain equation.



Schematic:

Gain Equation:

Simulation:

Gain Derivation

The derivation of the gain equation for this amplifier is very simple using Kirchoff’s Current Law (KCL).  First, remember that we are considering the op-amp to be ideal, meaning that the inputs are at the same potential.  With the non-inverting input grounded, this creates a virtual ground on the inverting input.  Also, we assume no current enters the inputs of the op-amp.  With these two assumptions, the circuit can be simplified as shown below.

Simplified Schematic

Using Kirchoff’s Current Law in the direction indicated by the arrow and treating Vgnd as zero volts results in the following derivation.

Gain Derivation

As you can see in my simulation, I chose Rf=1Meg and Ri=100k (Mega must be entered as Meg) giving me a gain of 10.  My simulation sweeps the input from 0 to 100mV and the output is shown to go from 0 to -1V as would be expected.