# Notes on Lab #1 - dc Circuits

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Physics 219 – Fall, 2002

Notes on Lab #3 – Diodes

(Note: We will skip Lab 3-1 in the Student Manual.)

Topics

3.1 Diode I-V Curve 1

3.2 Measuring the resistance of a diode 3

3.3 Half Wave Rectifier (Student Manual Lab 3-2) 5

3.4 Full Wave Bridge Rectifier (Student Manual Lab 3-3) 7

3.5 Ripple (Student Manual Lab 3-4) 12

3.6 Diode Limiter (Student Manual Lab 3-7) 14

3.7 Impedances of Test Instruments 15

The Oscilloscope Probe 16

3.1 Diode I-V Curve

In this lab we add a new circuit element, the diode. In doing so, we are entering the world of modern, semiconductor-based electronics.

[pic]

Schematic diagram symbol for a diode

Diodes are the "one-way streets" of electronic circuits. That is, they allow current to flow through them in one direction only (in the direction of the arrow).

An ideal diode would have the following I vs. V (or I-V) curve:

[pic]

I-V curve of an ideal diode

Real diodes offer a good approximation of this ideal behavior: The I-V curve of a typical silicon diode (like the 1N914 that you use in lab) is given below. Note the scale changes!

[pic]

I-V curve of an real diode (e.g. a 1N914)

Since the effective resistance is given by the ratio [pic], we see from the above curve that for a "forward-biased" diode (i.e, the voltage across the diode is positive and greater than about 0.6 V) so that it is trying to make current flow in the direction of the arrow)[pic] and the current flows without opposition. For a "reversed-biased diode (V < 0) [pic] and current will not flow no matter how much "reverse bias" (voltage) is applied.

(In our crude fluid model of electrical circuits, you can think of a diode as fat pipe with a hinged flap on one end, so that water can open the flap only when it flows so that it pushes in the right direction.)

A real diode is usually made by forming a "p-n junction" in a semiconductor such as silicon:

[pic]

p-n junction diode

Golden Rule of Diodes: When a diode with an I-V curve like the one above is “forward-biased” and is carrying appreciable current (> 10 mA say) it will have a voltage drop ~ 0.6 V across it.

3.2 Measuring the resistance of a diode

Try measuring the resistance of a 1N914 diode with a digital ohmmeter.

To understand the result you first need to know a little bit about how an ohmmeter works. The ohmmeter internally generates a current of fixed magnitude (we'll see in a bit how to generate a constant current) and sends this known current through the unknown resistance:

[pic]

The meter is capable of measuring the voltage across the unknown R and then "calculates" the resistance from R = V/I.

The magnitude of the constant current I that the meter chooses to generate depends on the scale that you have selected. The smaller resistance scale uses larger currents. (This is because the meter wants to generate a voltage drop that will be easily measurable. Thus, smaller values of R require larger values of I.)

Examination of the diode's I-V plot shows that for a forward-biased diode (the ohmmeter sends current out of the red terminal) the resistance that the meter measures will be smaller as the scale becomes more sensitive. For example, consider the I-V curve shown below. I2, which corresponds to a more sensitive scale than I1, gives a resistance measurement R2 that is smaller than the resistance measurement R1.

[pic]

If the diode is connected to the ohmmeter so that it is reverse-biased, the ohmmeter will measure a very large resistance (> 20 M ) so that you will not be able to measure it.

3.3 Half Wave Rectifier (Student Manual Lab 3-2)

A rectifier is a circuit that converts an ac signal (that is, a time varying voltage where the polarity, or sign, of the voltage is alternating in time) into a dc signal (where the polarity does not change).

One reason that this is useful is that one often requires a constant voltage supply. (For example, the integrated circuits used in most computers require a stable 5 V supply.) However, the power source that is often the most readily available, the wall socket, provides an ac voltage (60 Hz, 110-120 V rms in the U.S.). Rectification is the first step in converting this ac signal into a constant voltage.

Consider the circuit:

[pic]

[pic]

where [pic] is a sinusoidal signal with a 6 V amplitude (12 V "peak-to-peak") supplied by a transformer. As is the case so often in this course, we can view this circuit as a kind of voltage divider. For [pic] the resistance of the diode is huge (compared to 2.2 kΩ) so that [pic]. For [pic] the resistance of the diode is small compared to 2.2 kΩ so that [pic].

[pic]

3.4 Full Wave Bridge Rectifier (Student Manual Lab 3-3)

The problem with the half wave rectifier is that you are utilizing only half of the input signal. This is wasteful since there are still energy losses during the "unused" portion of the cycle (in the windings of the transformer, for example). Also, as you'll see shortly, the full wave bridge rectifier is better at reducing "ripple".

The circuit for the full wave bridge rectifier is

[pic]

To understand this circuit requires some careful thought:

1) For starters, note that neither terminal of the transformer (points A and C is the schematic) is connected directly to ground. (The transformer outputs are said to be "floating".) The transformer creates a voltage between the terminals that oscillates sinusoidally at 60 Hz. We call the voltage difference between the terminals

[pic]

[pic]

2) Now, consider those times when [pic]. For example, at time [pic], [pic]. Keeping in mind that the transformer is acting like a "current pump" that tries to pump current out of its positive terminal and suck it back into the negative terminal, and that current can only flow through a diode in the direction of the arrow, we deduce that the current flow at t = t1 will look like:

[pic]

3) Let us now try to find the absolute voltages at various points in the circuit. We start at point D, which is "tied to ground" and is therefore at 0 V. We head from point D through diode #4. Since current is flowing through this diode, the golden rule of diodes mandates that there must be a 0.6 V drop. Thus,

[pic]

4) Since we have assumed that VAC = 6 V, this means that

[pic]

5) Since VA = +5.4 V and VD = 0 V, diode #1 is reversed-biased and no current flows through it. Therefore, all the current must flow through diode #2, as shown.

6) The golden rule of diodes requires that point B must be 0.6 V below point A:

[pic]

Thus 4.8 V is the maximum voltage that develops across a load resistor.

Note that in the above circuit diodes #1 and #3 play important roles even though no current flows through them. For example, diode #1 isolates point A from point D and therefore allows these points to be at different voltages.

From the above discussion it follows that during the portion of the cycle when

[pic]

the voltage across the load resistor as a function of time looks like:

[pic]

The similar analysis follows for times such that VA - VC < 0. For example, at time t2, when VA - VC = -6 V, we have

[pic]

This time the negative terminal of the transformer sucks current from ground (VD still = 0 V) through diode #1 so that VA = -0.6 V. We calculate the voltage at point C to be

VC = VA + 6V = +5.4 V

Diode #4 is now reverse-biased and current flows through diode #3 so that VB = 4.8 V. Thus during times when VA - VC < 0 we have:

[pic]

The key point is that when VA - VC < 0 the current still flows through the load in the same direction as when VA - VC > 0.

[pic]

3.5 Ripple (Student Manual Lab 3-4)

[pic]

This effectively adds a "RC time lag" so that the output voltage [pic] cannot follow the 60 Hz fluctuations of the input voltage. The output voltage will look like:

[pic]

During those portions of the cycle when charge is being added to the capacitor (i.e., when [pic] is increasing) current is flowing through the diodes to get to the capacitor. Since a conducting diode has a very small resistance, the RC time constant during the charging portion of the cycle is very short and the output voltage can respond rapidly. During the discharging portion of the cycle charge drains off the capacitor through Rload. (It cannot flow backward through the diodes.) If Rload. is large enough so that Rload.C >> 1/120 sec then the capacitor will not discharge very much before it starts charging again, and the ripple will be small.

We define the ripple amplitude as the quantity labeled ΔV in the figure shown above. We can calculate the ripple amplitude for the above circuit as follows:

During the discharging portion of the cycle

[pic]

so that for the full wave bridge rectifier

[pic]

Since [pic] for small x we have

[pic]

Thus we see that ripple will be minimized when Rload and C are both large. Good power supplies generally contain large capacitors. Really cheap power supplies contain no capacitors; (To see what I mean, try using the “dc output” that is available on the left side of the blue boxes that you are using as variable transformers to power some modest load ,like a 1k resistor. Use an oscilloscope to see what the output of this “dc power supply”. Also try looking inside the box and see if you can tell how the “dc power supply” is constructed.

Even with large capacitor, the ripple is still often unacceptably large. Also the output voltage supplied by this scheme will vary if the input voltage fluctuates. (The amplitude of the "line voltage" supplied by a wall socket can fluctuate dramatically, particularly during "brown-out" conditions on hot summer days. Therefore a good constant voltage supply will also make use of a device called a voltage regulator to further stabilize the voltage. The dc power supplies inside your big black breadboards contain a voltage regulator.xc

3.6 Diode Limiter (Student Manual Lab 3-7)

Some instruments can be damaged or destroyed if too large a voltage is applied to the input. The following scheme is commonly used to limit input voltages to < 0.6 V.

[pic]

It works for both positive and negative input voltages. You should be able to figure out how this circuit works yourself! The Diode Clamp, which you build in Lab 3-6 is similarly straightforward.

3.7 Impedances of Test Instruments

As you know, it is desirable for any voltage measuring device to have as large an input impedance as possible, so that it will perturb the circuit you are measuring as little as possible. It is quite simple to measure the input impedance of your dc digital voltmeter. You need only build the following circuit:

[pic]

By varying R and looking at the voltage displayed by the voltmeter, you can deduce Rin. (Hint: Think voltage divider!)

For instruments used to measure high frequency signals, such as an oscilloscope, the situation is more complicated. The Thevenin equivalent circuit for the input of your oscilloscopes looks like:

[pic]

The input impedance of our particular oscilloscopes, [pic] is due to a 1 MΩ resistor in parallel with a 30 pF capacitor. The small parallel capacitance is an example of stray capacitance. We use the term “stray” to indicate that this capacitance is not put there intentionally by the oscilloscope designer. (In fact, the designer works hard to make the stray capacitance as small as possible. This is because the presence of stray capacitance ends up limiting the scope ability to observe high frequency signals. In practice the smaller the stray capacitance , the more they can charge for the oscilloscope!) Recalling that the magnitude of the capacitor's impedance is [pic], we see that at low frequencies the input impedance is close to 1 MΩ. (For example when ω = 100 rad/sec, the magnitude of the capacitor's impedance is 3.3 x 108 Ω so that the impedance of the parallel combination is dominated by the 1 MΩ resistor. However, as ω increases the capacitor causes the input impedance to decrease significantly from 1 MΩ.

Thus the input impedance of the oscilloscope is frequency dependent, getting smaller as the frequency of the input signal increases

(Note that, on account of the presence of the stray capacitance in parallel with the resistor we now refer to an input “impedance” rather than an “input resistance”.)

To see why this creates a problem, consider the following situation that arises quite often in the life of an experimentalist. Suppose one wishes to measure some time-varying signal V(t) with an oscilloscope. Suppose the box that is producing this signal has an Thevenin impedance (or “output impedance”) [pic]:

As always, if we don't want the signal to "droop" when we attach the oscilloscope in order to observe the signal. For this reason we would like to make sure that the condition that [pic].

It may be difficult to satisfy this requirement when dealing with high frequency signals. For example at a frequency of 1 MHz (ω = 2 x 106 rad/sec) [pic]. Even worse is the fact that since [pic] is frequency-dependent, the different frequency components of a signal can droop by different amounts, leading to distortion of the signal.

The Oscilloscope Probe

A clever way of minimizing this problem is to use an oscilloscope probe. The probe serves to raise the input impedance of the oscilloscope by a factor of 10 at all frequencies. It thus helps minimize the distortion caused by capacitive loading and makes it easier to satisfy the [pic] criterion. Of course, there's no free lunch; it accomplishes this at the cost of decreasing the sensitivity of the oscilloscope by a factor of 10, so you might not be able to use the probe when looking at very weak signals.

The schematic drawing below reveals that the probe is a remarkably simple device. It consists only of a 9 MΩ resistor in parallel with a variable capacitance ([pic]can be varied from 3 pF to 6 pF, usually by turning a screw located on the tip housing).

[pic]

Therefore the impedance of the probe is given by:

[pic]

while the input impedance of the oscilloscope (including the typically 30 pF of capacitance to ground present in the coaxial cable) is given by

[pic]

It is easy to see that by adjusting the variable capacitor on the probe tip so that

[pic]

it follows that

[pic]

[pic]

[pic] independent of frequency .

Thus the total input impedance of the oscilloscope and probe combination is

[pic]k

at all frequencies.

Thus when you use the probe the input impedance is ten times greater than the impedance of the oscilloscope without the probe, making it a “less intrusive” measuring device.

But remember, here’s no free lunch of course. Because of the voltage divider effect, the oscilloscope's sensitivity is reduced by a factor of 10 when you use a probe.

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