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.

Â This is Dr. Robinson.

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

Â In the previous lesson,

Â you were introduced to voltage transfer characteristics.

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

Â to examine circuits that perform this conversion.

Â Now AC or alternate current is used in power transmission.

Â So if you measure the voltage between the hot terminal and

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

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

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

Â example Tobias op amps.

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

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

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

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

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

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

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

Â Here we have an input sinusoidally varying voltage.

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

Â sine wave from a wall outlet.

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

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

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

Â Here's the AC voltage input.

Â And it's typically applied to a transformer.

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

Â voltage to another.

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

Â the transformer is known as a step up transformer.

Â If the output voltage is smaller than the input voltage,

Â it's known as a step down transformer.

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

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

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

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

Â where a DC component has been introduced into the waveform.

Â The input here, you can see that for

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

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

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

Â But in a rectifiered wave form,

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

Â Now this can technically be considered a DC voltage,

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

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

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

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

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

Â The low pass filter reduces the AC components and

Â smooths the output, so that, at the output,

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

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

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

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

Â Here's the AC wall plug.

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

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

Â That sine wave is applied through a fuse, and

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

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

Â the pilot light turns on to indicate electricity is applied.

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

Â its output side.

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

Â the voltage here would be 80 volts rms.

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

Â which rectifies the signal, and

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

Â The voltage here,

Â smoothed by the capacitor, produces the positive DC voltage.

Â And the voltage here smoothed by this capacitor produces the negative

Â DC voltage.

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

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

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

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

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

Â In this picture I'm showing the power supply for

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

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

Â Here's another example power supply.

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

Â slides ago.

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

Â designed by professor Marshall Leach from Georgia Tech.

Â Here is the power transformer.

Â This and this are the filter capacitors.

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

Â This square here is the bridge rectifier.

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

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

Â So, this portion here is the power supply that

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

Â right channel amplifiers for this audio amp.

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

Â I've made one change to it.

Â This is the bridge rectifier,

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

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

Â to smooth the rectified output voltage.

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

Â input transformer in our DC power supply, and

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

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

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

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

Â Here in red is our input voltage.

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

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

Â But by adding that capacitor to the circuit to act as

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

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

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

Â However, the value of the capacitor that I chose for

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

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

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

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

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

Â every half-period of the input.

Â So the time between this peak and

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

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

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

Â the resistance and the capacitance.

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

Â discharge through the resistor.

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

Â the resistor.

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

Â assuming that there's no filter capacitor.

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

Â different values of filtering capacitor.

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

Â The capacitor charges up to this value from the,

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

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

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

Â So you have this charge, discharge, charge,

Â discharge cycle that produces the output voltage.

Â And because of this small time constant,

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

Â producing this large output ripple voltage.

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

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

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

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

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

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

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

Â 168 volts.

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

Â We know that for a sine wave the peak voltage is

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

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

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

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

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

Â We then substitute the relationship between V peak and

Â V rms, we get this equation.

Â And if we substitute the voltage values that we know and

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

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

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

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

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

Â a rectifier, and a filter capacitor.

Â And we saw that the filter capacitor affects the performance of

Â the DC output voltage.

Â Large filter capacitors which result in larger time constants result in

Â a smoother output voltage, or a smaller ripple voltage.

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

Â Thank you and until next time.

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