This course introduces students to the basic components of electronics: diodes, transistors, and op amps. It covers the basic operation and some common applications.

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From the course by Georgia Institute of Technology

Introduction to Electronics

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This course introduces students to the basic components of electronics: diodes, transistors, and op amps. It covers the basic operation and some common applications.

From the lesson

Diodes Part 2

Learning Objectives: 1. Examine additional applications of the diode. 2. Make use of voltage transfer characteristics to analyze diode circuit behavior.

- Dr. Bonnie H. FerriProfessor

Electrical and Computer Engineering - Dr. Robert Allen Robinson, Jr.Academic Professional

School of Electrical and Computer Engineering

Welcome back to electronics.

This is Dr. Robinson.

In this lesson, we're going to look at AC to DC conversion.

In the previous lesson,

you were introduced to voltage transfer characteristics.

Our objectives for today's lesson are to introduce AC to DC conversion and

to examine circuits that perform this conversion.

Now AC or alternate current is used in power transmission.

So if you measure the voltage between the hot terminal and

the neutral terminal of an, of a wall outlet in your house,

you would see a 120 volt rms 60 hertz sine wave, a voltage that varies with time.

However DC or direct current is used to power electronic devices, for

example Tobias op amps.

Now we need a way to convert an alternating current signal to

a direct current signal and it turns out that diode rectifiers that we've

studied in previous lessons are a key component to doing this.

Let's look at an overview of how AC to DC conversion works.

Here, I have a block representing an AC to DC converter, and

I've also labelled this as a DC power supply, because it supplies a DC voltage,

or DC power, to whatever electronic circuit is connected to its output.

Here we have an input sinusoidally varying voltage.

And again, this is typically a 120 volt rms, 60 hertz,

sine wave from a wall outlet.

It's applied to the input of the AC to DC converter, and

the output is a pure DC voltage, a voltage that is constant with time.

Now, let's look at the components inside a DC power supply.

Here's the AC voltage input.

And it's typically applied to a transformer.

Now a transformer is a passive electrical device that converts one AC

voltage to another.

If the output voltage of the transformer is greater than the input,

the transformer is known as a step up transformer.

If the output voltage is smaller than the input voltage,

it's known as a step down transformer.

So at the output of the transformer, we still have an AC voltage.

It just may possibly be of a different level than the input voltage.

The AC voltage is applied to a rectifier, say a full-wave rectifier.

We know that at its output we have a waveform that looks like this,

where a DC component has been introduced into the waveform.

The input here, you can see that for

each, for half the period the voltage is greater than zero.

And for half the period the voltage is less than zero.

So that the average value of this sine wave, or its DC value is equal to 0.

But in a rectifiered wave form,

you can see that there is a positive DC voltage, or average value in the waveform.

Now this can technically be considered a DC voltage,

because in the rectifier all the current is flowing in the same direction.

However, you can see that there's a significant AC component.

So, this is known as a pulsed DC wave form.

But, to get out the pure DC voltage that we desire,

we apply this rectified voltage to the input of a low pass filter.

The low pass filter reduces the AC components and

smooths the output, so that, at the output,

we have our voltage that's constant with time, or a DC voltage.

I want to show you a schematic of an actual DC power supply.

Now this is a schematic for a bi-polar power supply.

It produces both a positive DC output, and a negative DC output for an AC input.

Here's the AC wall plug.

Pin three is the safety ground, and between pins one and

two, you would measure the 120 volt r m s, 60 Hertz sine wave.

That sine wave is applied through a fuse, and

through a power switch, to the input side of the transformer.

So, when the switch is closed the AC signal is applied to the transformer and

the pilot light turns on to indicate electricity is applied.

Now, the transformer transforms the voltage to a different AC voltage on

its output side.

In this case, this is actually a step-down transformer and

the voltage here would be 80 volts rms.

That rms voltage is applied to the input of the bridge rectifier,

which rectifies the signal, and

that rectified signal is applied to two filter caps to smooth out the output.

The voltage here,

smoothed by the capacitor, produces the positive DC voltage.

And the voltage here smoothed by this capacitor produces the negative

DC voltage.

Here are some example DC power supplies, and I'm sure you've seen this type of

DC power supply before, it's known as a wall wart, or a block transformer.

So within this block is a transformer, a rectifier, and a filter capacitor.

The input is applied by plugging the block directly into the AC wall outlet and

at the end of this chord is the output DC voltage.

In this picture I'm showing the power supply for

an audio preamplifier, here is the transformer, here are the filter caps, and

between these filter caps is a bridge rectifier that you can't see.

Here's another example power supply.

This is actually an implementation of that circuit schematic that you saw a few

slides ago.

This is the power supply from the world famous Leach audio amplifier,

designed by professor Marshall Leach from Georgia Tech.

Here is the power transformer.

This and this are the filter capacitors.

These are large electrolytic capacitors of about 12,000 microfarads.

This square here is the bridge rectifier.

Here are the output fuses that you saw on the schematic.

Over here you can see the power switch and here is the pilot lamp.

So, this portion here is the power supply that

supplies power to these two circuit boards that contain the left channel and

right channel amplifiers for this audio amp.

Now here's a circuit schematic that you should be familiar with by now, but

I've made one change to it.

This is the bridge rectifier,

a full wave rectifier circuit that we've looked at previously.

But I've added a capacitor here to act as a smoothing filter,

to smooth the rectified output voltage.

Now let's assume that the input voltage here is the output voltage from our

input transformer in our DC power supply, and

let's assume that it is a one-to-one transformer.

That means that the output voltage is equal to the input voltage.

So the AC voltage between these two points is a 120 volts rms, 60 hertz sine wave.

Now let's look at how, some of the voltages in that circuit vary with time.

Here in red is our input voltage.

Now without the capacitor in the circuit, we know that the output voltage would

look like this waveform in green, a full wave rectified sine wave.

But by adding that capacitor to the circuit to act as

a smoothing filter the output voltage instead becomes this blue curve here.

And you can see that this is a reasonable approximation to the DC voltage that we

want at the output, a voltage that's constant with time.

However, the value of the capacitor that I chose for

this example is such that we have a ripple voltage in the output.

The voltage at the output is varying between this level and this level.

And it's varying at a frequency of 120 hertz.

Because remember, the input sine wave is a 60 hertz sine wave.

But the output of our full wave rectifier we get an output for

every half-period of the input.

So the time between this peak and

this peak would be the period of a 120 hertz sine wave.

Now let's look how varying the capacitor value will affect the DC output voltage.

Remember in an RC circuit the time constant is given by the product of

the resistance and the capacitance.

A larger time constant means that a charge capacitor will take longer to

discharge through the resistor.

A smaller time constant means that the capacitor will discharge faster through

the resistor.

Now, on this plot, the red curve would be the output of the full wave rectifier,

assuming that there's no filter capacitor.

The blue curve, the green curve, and the purple curve are all different outputs for

different values of filtering capacitor.

Here, I'm using a small capacitor such that the time constant is small.

The capacitor charges up to this value from the,

from the output of the rectifier, and then as the output from the rectifier begins to

decrease the voltage on the capacitor discharges until the voltage output from

the rectifier increases again causing the capacitor to charge back up to this value.

So you have this charge, discharge, charge,

discharge cycle that produces the output voltage.

And because of this small time constant,

this small value of capacitor, the capacitor is able to quickly discharge,

producing this large output ripple voltage.

But as we increase the voltage, you can see that the output of

the power supply is becoming more and more like the DC voltage that we want.

The purple curve here is using the largest volume capacitor, and

you can see that it has a very small ripple current, or ripple voltage.

Now we know that ideally the DC output voltage from this power supply would be

a straight line that connects the peaks of this rectified sine wave.

And we can see that from the plot, the DC level at the output is approximately,

168 volts.

Now let's calculate what the DC output voltage for this power supply should be.

We know that for a sine wave the peak voltage is

equal to the rms voltage times the square root of 2.

And we also know from our previous analysis that in going from

the input to the output of our full-wave bridge rectifier we lose two Vf volts,

because of the two voltage drops across the forward biased diodes.

So our DC voltage should be equal to the peak voltage minus the two Vf volts.

We then substitute the relationship between V peak and

V rms, we get this equation.

And if we substitute the voltage values that we know and

assume that we have a silicon diode such that Vf is equal to 0.65 volts,

we have the VDC is equal to our 120 volt rms sine wave from our wall outlet.

Times the square root of 2 minus 2 drops of across silicone diodes of about

0.65 volt to get us 168.4 volts, which is the voltage we saw on our previous graph.

So, in summary AC to DC conversion is performed using a transformer,

a rectifier, and a filter capacitor.

And we saw that the filter capacitor affects the performance of

the DC output voltage.

Large filter capacitors which result in larger time constants result in

a smoother output voltage, or a smaller ripple voltage.

In our next lesson, we're going to look at diode waveshaping circuits.

Thank you and until next time.

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