E72 Lab #3

Diodes, Transformers and all that

In this lab you will be introduced to an extremely common non-linear device, the diode, and some of its more common uses. You will also learn a little bit about transformers, a component that is in almost every piece of electronic equipment that you own.  All of the equipment for this lab is in Hicks 310.

Diodes

The diode is a passive non-linear circuit element that, ideally, passes current in only one direction. Its symbol is shown below, as well as an image of some real diodes.

 

On the real diode the cathode is marked by a stripe (on the right side of both diodes in the image).

For an ideal diode current can flow in only one direction, anode to cathode, and there is no forward voltage drop across the diode. An accurate model of a diode obeys the relationship

where I_s and n are two parameters that define the characteristics of the diode. I_s is called the saturation current, n is the emission coefficient (1<n<2) and v and i are voltage and current as defined above. The quantity V_T=kT/q is called the thermal voltage and is equal to about 25 mV at room temperature; q is the charge on an electron, k is Boltzman’s constant, T is absolute temperature. We will call this model the "real" model of the diode. Another, simpler, model of a diode (made from Si, as are all those in the lab) states that if v=0.6 volts the diode is on and conducts current in the direction of the arrow, if v<0.6 volts it is off and does not conduct current. The current-voltage relationships for the three models are shown below.

We will normally use the 0.6 volt model which represents a good compromise between the oversimplifications inherent in the ideal model and the great difficulties inherent in using the transcendental relationship in the more accurate model.   The quantity of 0.6 volts is an approximation, and 0.5 or 0.7 volts may often be more appropriate.  Some kinds of diodes have lower voltages  (e.g., germanium=0.3 V), and some are higher (e.g., Light Emitting Diodes=1.0).

Zener Diode

There is a special kind of diode called a zener diode that acts like a regular diode when the voltage across it is positive. But, if there is a large enough negative voltage across it then it will conduct in the reverse direction. The symbol for the zener diode and its I-V characteristic are shown below.

In addition to forward conduction at 0.6 volts, the zener diode also conducts in the reverse direction when the voltage gets to V_z, called the zener or "knee" voltage. This voltage depends on the way the diode is constructed and can be manufactured from about 2 to 20 volts, depending on the intended use.

 

The Light Emitting Diode (LED)

Another type of diode that we will use is the light emitting diode, or LED.  The LED has similar characteristics to a standard Silicon diode but it has a higher voltage drop, typically between 1 and 2 volts, depending on the color (LED's at higher wavelengths (and hence energies) have higher voltage drops).  Standard colors for LED's include (infrared), red, orange, yellow, green, and blue.

Some applications of diodes are shown later in the procedure section.

Transformers

Transformers can be used to increase or decrease AC voltages. They are used in electrical equipment to convert the 120 Volts coming from the wall socket to lower and safer voltages for use in equipment. The schematic diagram and defining equation for a transformer are shown below.

     

Where V₁, V₂, I₁ and I₂ are defined as shown, and N₁ and N₂ are the number of windings on the primary and secondary coils, respectively. Power transformers are usually not described in terms of N₁ and N₂, but instead are described in terms of the voltage output, and assume a 120 volt input. Also specified is the maximum current output which is limited by the size of the wire used in the transformer.

The transformer shown above is called a single tap transformer because there is only one output. It is possible to build transformers with multiple taps, and a common way of doing that is to make a connection to the middle of coil 2; this is called a center-tapped transformer.

In this lab you will be using a 12.6 Volt center tapped transformer (which can also be hooked up as two 6.3 volt transformers with a common connection in the middle). Note that the two 6.3 V outputs (V2) will be in phase with each other so that the voltage at the top node (on the right) will always be equal in magnitude, but opposite in sign, to the voltage at the bottom node, when measured with respect to the middle node, which is usually grounded. Remember also that all of these voltages are RMS (Root-Mean-Square) values -- the peak voltages will be the square root of 2 times larger.

 

Procedure:

The transformers are in the drawers of the cabinets under the bench in the back of the room; the diodes, capacitors, op-amps and resistors are in the cabinets that are on the wall at the back of the classroom portion of room 310..  

Power supplies

1) Hook up the circuit shown below with R=470W (it doesn’t matter which transformer output you use as long as you connect ground to the center pin), and use a 1N400X diode (the X can be any number). Look at the input voltage (on the anode of the diode) and output voltage (on the cathode) and explain the output. This circuit is called a half-wave rectifier. Note especially that the rectified output remains about 0.6 to 0.7 volts below the input while the diode is on (in accordance with a diode modeled by a constant forward voltage drop).   Do a neat job with your wiring, you will be making modifications to this circuit for parts 2 through 8.

2) Repeat 1) with a 47 mF capacitor placed in parallel with the resistor. Note that you will have to use an electrolytic capacitor which has a preferred polarity. Make sure that the positive lead of the capacitor is connected to the diode, and the negative lead to ground. If you hook it up backwards, unpleasant liquids may ooze from the capacitor. Explain the output.

3) Repeat 2) with a 330 µF capacitor. What you have built is a simple power supply to convert AC power to DC power. Note that it is not ideal because there is still ripple at the output. Measure the ripple (the amplitude of the variation in the output voltage), and derive an equation to predict it. Compare your calculation with the measured value.

4) Connect the circuit shown below, sketch or print v_i, v_o and v_x (each measured relative to ground). Obviously this is a better power supply. we have added some regulation to the crude power supply from the previous step, but decreased the output voltage. Make sure you use a 1/2 Watt resistor for the 120 W resistor , else you'll  burn out the resistor with too much current.  Use a 6.2V zener diode, 1N4735.

5) Add another 470 W resistor in parallel with the first one (for a total combined resistance of 235 W) and repeat the previous part. Compare the quality of the voltage regulation with 470 W and 235 W, and explain any differences. If you haven't seen a large change in output, keep adding 470 W  resistors until you see a distinct change in the output voltage as seen on the oscilloscope.  How many 470 W resistors did you use before seeing distortion?

6) Consider replacing the single diode with a diode bridge, as shown using a  DF01 or DF04 full wave bridge rectifier module (you needn't actually construct it).  The module has four diodes and is oriented as shown (The "~" markers are where AC power is applied, the "+" and "-" markers correspond to the positive or negative rectified signal).  Why might you expect this circuit to work better than the one from part 5?  

 

7) What you have built is almost a practical DC power supply.  It has 120 VAC as input, and a regulated DC output.  However, the regulation is not very good (it varies with fairly small changes in load).  To do the last stage of regulation in a real power supply, a voltage regulator is usually used.  The diagram below shows a 5 volt regulator, the 7805.  Link to 7805 datasheet. The two capacitors on the output are used by the 7805 to decrease variations in the output.

7805 Pinout

Input = Line Voltage Output = Vreg (regulated voltage)

Build the circuit shown below.  Be very careful with your location of grounds -- it is easy to melt a new spot in the breadboards.   Measure Vout.  Add a few 470 W  resistors in parallel with the first, and you should see no change in the output.  The output will not change until the input drops below about 7 volts (this minimum difference between input voltage and output voltage for a regulator is called dropout voltage).  You don't need to add all of the resistors necessary to see this drop.  This circuit is typical of what you would find in a commercial product (thought the 7805 is relatively old technology).

The 7805 is a "series" regulator, because the regulator is in series with the load (the 470 W  resistor).  The zener regulator that you built is a "shunt" regulator, because the regulator (the diode) is in parallel with the load (this is called a shunt).  The series regulator is generally more efficient because it dissipates less power than that of a shunt regulator.  For equivalent currents through the regulators, the power will be less in the series regulator than the shunt because the voltage drop across it is usually less.

8) Build the circuit shown below using 2  470 W resistors, and a full wave bridge rectifier module.  Be very careful with your location of grounds -- it is easy to melt a new spot in the breadboards.   Note that the wiring of transformer and diode bridge have changed substantially.   Print out, or carefully sketch, Vout1 and Vout2 (measured with respect to ground), and make sure you understand  how the circuit works. 

9) Design (but don't build) a power supply that will generate both +5V and -5V at 50 mA (imagine a 100W, 1/2 Watt resistor as your load).   Use the 50 mA current rating to specify a capacitor.  You will need both a +5 V regulator, (7805, datasheet) and a -5 V regulator (7905, datasheet).  Note, the pinout for the 7905 is different than for the 7805.  You will need to consider the "dropout voltage" of the regulator. 

In order for the 7805 to work, the input voltage must be greater than the output voltage.  The regulator is guaranteed to work as long as the input is greater than the output by an amount specified by the dropout voltage.  For example, if the dropout voltage was 1.5 volts, the input would have to stay above 6.5 volts (5 V output plus 1.5 V dropout).  This can help you determine the amount of allowable ripple voltage, which can help you choose the size of the capacitor.

Precision Rectifiers - the superdiode

To turn in: include (at least) the following for the specified parts of the procedure.

1. For part (1) of the lab show v_i and v_o, and give a brief explanation of the output.

2. For parts (2 & 3) show a derivation of the ripple in terms of R and C.  The ripple is the amount that the voltage drops from one peak to the next.

3. For parts (2 & 3) show v_i and v_o, a measurement of the ripple.   Compare measured and calculated ripple.

4. For parts (4 & 5) show v_i, v_o, and v_x.

5. For parts (4 & 5) present a qualitative comparison of v_o in parts 4 and 5.  Explain the change in output that you observed.  It is important that you understand this -- we'll come back to it in another lab.  How many 470 W resistors did you use before seeing distortion?

6. Why would you expect the circuit for part 6 to work better than that for part 5?

7. What would happen for part 7 if instead of using a transformer, the input of the diode bridge was attached to a 10 V DC source with the positive terminal at the top of the bridge, the negative terminal at the bottom? What if the DC source were reversed in polarity?  Wall transformers (the little black transformers that you plug directly into a wall) can have AC output, or DC outputs -- with either polarity.  The devices they power (cell phones, battery charges, ...) often have a circuit similar to this one in them so that they will work even if the incorrect transformer is plugged in -- but don't try this at home unless you are sure that your device has such a circuit.

8. For part (8) explain show vi, Vout1 and Vout2.  Describe in detail how the full wave bridge rectifier works.

9. For part (9), give a schematic and explain how you chose your value for the capacitance.

10. For part (10) Show v_i, v_o and the output of the op-amp and give a description of how the circuit works.

11. For part (10) give an explanation of what causes the glitch in v_o.

12. For part (11) show how v_i and v_o would look if v_i(t)=sin(2*p*100*t), and describe how the circuit works. 

13. Simulate the circuits from parts 10 and 11 and compare the results to your experimental results.  You should fee free to use the simulated results when discussing circuit operations.  Use the MultiSim model of the 741 op amp instead of the 411 -- the 411 model doesn't saturate.