Circuit Idea/Group 66a
Dafar Shaban, Dilyana Dilova, Irina Hadjieva, Miroslava Hristova, Victor Glavev, Alexandra Georgieva, Danail Dekov, Liliya Bancheva, Elina Lazarova, Mihaela Borisova, Tzvetan Tzvetkov, Hristiana Stancheva, Silviya Nakova, Nataliya Genova, Vasil Tzanov, Ivan Tzvetkov.
Lab 1: Investigating passive resistive circuits by Microlab system
Tuesday, March 18, 2008, 13.45 h
Using the heritage of group 65a
Welcome to the laboratory. The picture on the whiteboard represents the laboratory setup that your colleagues from group 65a have used for reproducing the genuine Ohm's experiment. You can use the results of their laboratory experiment to introduce the next great idea named voltage diagram. It will help us to visualize the voltages that your colleagues have measured along the resistive film.
Introducing the idea of pressure diagram (an opened water circuit)
By the way, we can derive this idea from the well-known hydraulic analogy (plumbing) that we can see everywhere around us. For example, imagine a large vessel filled of water that supplies a long thin pipe; let's first the pipe to be tapped. The question is: "What is the pressure inside the pipe?" And more precisely speaking, "What are the local pressures along the pipe?"
We can get to know, if we drill small holes at equal intervals along the pipe (if we want to be more precise, we might stick vertically thin glass pipes acting as local manometers). The result is expectable for us: all the water levels (accordingly, all the local pressures along the pipe) are equal. This picture shows the pressure distribution along the pipe; we can name it "pressure diagram".
The pressure diagram of a closed water circuit
Now open the pipe; the water will begin flowing. What are the local pressures along the pipe now? At the left end the pressure is maximum; at the right end it is minimum. Our intuition suggests that the local pressures will decrease gradually from left to the right.
Really, the levels of the water bars (accordingly, the local pressures along the pipe) decrease gradually from left to the right. The envelope of the pressure diagram is a triangle.
Conveying the idea of pressure diagram to voltage one
Let's now apply this powerful idea to draw the analogous electrical voltage diagrams. The idea is obvious: if we think of voltage as a kind of pressure, we may present the local voltages by local voltage bars in exactly the same way as we have presented the local pressures along a pipe by a local water bars! As above, the lengths of the voltage bars are proportional to the magnitudes of the local voltages regarding to ground (we might set the zero voltage level at the height of the resistor and then to draw the positive voltage bars above and the negative voltage bars below the resistor's level). The set of these voltage bars forms the whole voltage diagram. We can use the envelope of the voltage diagram instead the very set of voltage bars, in order to simplify the image.
Making a computer draw a "living" voltage diagram on the screen
Basic idea. But it is too hard and boring for us, human beings, to draw all the voltage diagrams when the circuit attributes (voltages and resistances) vary. Because these diagrams have to be not static, dead pictures; they have to be "living" diagrams (animations) that change accordingly to the circuit state. So, wee need a computer that "watches" closely what we do with the circuit under test and builds the according "living" voltage diagram on the screen. Then, let's build such a computer-based system and leave it to do this donkey work:). Then we will only look at the picture on the screen and think about the circuit phenomena behind it!
Components. First at all, we need some computer. It may be humble enough, if only it has some graphical possibilities (for example, the computer that is used in this laboratory to build Microlab system in 1986 is a version of the famous Apple II). Then we have only to connect to the PC buses a few analog-to-digital converters (ADC's) acting as analog inputs and a few analog-to-digital converters (ADC's) acting as analog inputs and our computer-based system Microlab is ready!
Power supply. Note that the analog-digital periphery "sucks" energy directly from the poor computer:); so, it acts not only as a computer but it supplies the periphery too. Then can we supply the very resistive object as well? Can we use the very DAC's to supply the resistor under test? This will allow the computer to control the obect! Only, the DAC's can give maximum 10 mA by their op-amp outputs. So, we can't supply the low-resistive (10 Ω) wire of group 65a!
Resistor. Do you remember the linear 4.7 kΩ varying resistor from the previous laboratory exercise? As far as I can remember, we carried out a "reverse engineering" then taking it to pieces:) What do we do now? Suggest a solution! A student: Yes, I do... Obviously the colleague suggests to supply such a resistor by the DAC's 10 mA outputs without any problem (calculate the maximum current). Thus we can use it instead the low-resistive wire. Then we supplied only the one of the resistor's terminals. Now, let's complicate the arrangement and make it more confusing:) by supplying both the ends (we suspect it might be very interesting). Well, connect AO1 (DAC1) to the right and AO2 (DAC2) to the left end of the resistor.
Voltmeters. What do you think, wheather to connect the two old-fashioned but attractive bipolar voltmeters V1 and V2 to observe the input voltages? Or to rely only on the abstract digit VOM's and on the screen digit meausements? I see, you like the good old meters maybe because they are something "live", moving, geometrical, spatial, real... People trust such genuine things... But is there any problem as these "antiques" have 20 kΩ internal resistance)?
Ground. Now, let's say also some words about ground. But what is ground? We can find a possible answer in the Wikipedia article about virtual ground. Shortly, ground is a reference point, regarding to which we measure voltages. The PC power supply that we use is the so-called "split supply". It consists of two 12 V supplies connected in series (- + >>> - +). The middle point serves as ground in this arrangement. The DAC's grounds are connected internally to this ground. We have also to connect the black test ends of our voltmeters to this ground.
Software. More than twenty years ago, I prepared a program that can visualize the voltage diagram on the screen as a "living" animation. For this purpose, the program make computer "interest" in the local voltages in three key points - the left end point, the slider intermideate point and the right end point. Let's then satisfy its curiosity:) by connecting, for a start, the DAC1's output to the ADC1's input and the DAC2's output to the ADC2's input.
Tuesday, April 01, 2008, 13.45 h
Building the circuit on the whiteboard
Making biasing batteries
Mounting the circuit on a prototyping PCB
Investigating the circuit by Microlab
Tuesday, April 15, 2008, 13.45 h
What is op-amp?
Inputs. Grounded voltage source... What is single-ended input? Floating voltage source... What is differential input? Why do we need a differential input? Input resistance... Do input currents flow? How does a non-inverting input behave? How does an inverting input behave? Can we use an op-amp with a differential input as an op-amp with single-ended (inverting or non-inverting) input? What do we do in these cases?
Output. How the op-amp output is made? What is the idea of a complementary, push-pull emitter follower? Output resistance...
Supplying. The idea of bipolar (split) supply (+12 V and -12 V here)... Why do we need a bipolar supply? Where currents flow? We can imagine the output part of the op-amp consists of two parts - a positive and a negative. Each of them works when its output voltage has the according polarity. By the way, is it possible to supply the op-amp with only one (e.g., positive) voltage and, at the same time, to obtain the opposite (negative) output voltage? And can the op-amp output voltage exceed the supply voltage? Maybe, it would be possible, if there were capacitive or inductive differentiating circuits inside the op-amp... But (for now) an op-amp contains only resistive and active components:(
Gain. Typical value is A > 200000. It seems too high. Can we manage to utilize all this gain? What is the maximum input voltage to keep the op-amp in the active region (assume 10 V maximum output voltage)? Can we use a bare op-amp as an amplifier? Or we can use it as another useful device having analog inputs and descrete output (write its name here)? We may conclude: If we use the op-amp as an amplifier, the voltage difference between its inputs is zero. But if we use it as a comparator, the input voltage might be whatever.
Transfer characteristic. Microlab can help us to obtain this curve on the screen. There is no time in this graphical presentation; there is only input and output quantities. We might discern three regions on this curve: negative saturation (large), active region (narrow) and positive saturation (large).
Generalization. What does the op-amp actually do? A possible answer is: an op-amp converts the steady power supply into varying voltage source; it acts as a voltage-controlled voltage source.
Building a general negative feedback follower
Driving: keeping up a constant speed, direction, temperature, volume... More negative feedback analogies: studying, teaching, keeping up a room temperature (body weight, mental equilibrium...) In all these cases we do all that is possible to realize our goals, to move toward the goal...
Structure (block diagram)
So, the simplest negative feedback follower consists of only three components: power (energy) source E, regulating element R and a subtractor (-).
This humble device operates according to a simple algorithm: if X > Y, then increase Y; if X < Y, decrease Y; if X = Y, do nothing. As a result, always X = Y. So, this is an "active copying" principle and this device is an "active follower". This "algorithm" is so simple that humble devices (tubes, transistors, op-amps...) having no any intellect can perform it.
Why do we make followers in this odd way? What are the advantages of this "negative feedback" approach?
Building an op-amp follower
Passive electric follower
Now, we have only to convey this powerful idea from real life to electronics. Well, let's begin.
VIN will represent X in our electric follower; accordingly, VOUT will represent Y. But let's first persuade ourselves what is the use of building such an active follower instead a "passive" one.
What is the simplest voltage follower? Of course, a piece of wire can act as a humle voltage follower. It works perfectly, if there is no load (there is no current flowing). Only, is it still a follower, if we connect a big load at the end of the wire (especially, if it is a thin and long wire)? And imagine what happens, if the load and wire resistances vary... For example, the wire length can vary...
Another nasty problem of the passive voltage follower is that the load consumes energy from the input voltage source.; so, it droops when loaded.
Op-amp negative feedback follower
No, the bare wire is not a perfect follower in all the cases; so, it is worth the trouble to build an active negative feedback follower...
First, we connect a power supply - a voltage source V representing the power source E from the block diagram. We assume a bipolar input voltage; so, we get a bipolar (split) power supply. Then, we connect an op-amp OA acting as the regulating element R. Finally, we need a voltage subtractor to make the op-amp compare its output voltage VOUT with the input voltage VIN. What is the simplest voltage subtractor? Of course, this is the bare loop (a piece of wire) according to Kirchhoff's voltage law (KVL). In other words, we have to connect the two voltage sources in series and in opposite directions (+ -, - +) traversing the loop, in order to subtract their voltages. Then, we have to cut the loop and to feed the op-amp input by the voltage difference.
Investigating the circuit by Microlab
We are human beings, not computers... It seems strange but, in order to understand what the abstract op-amp really does in this circuit, we have to think of the op-amp as a "slow-thinking" device, not as a fast, non-inertial, instant device... From this human viewpoint, the op-amp compares continuously its output voltage VOUT with the input voltage VIN and changes VOUT in the right direction so that to keep almost zero difference VOUT - VIN = 0. It does exactly the same what we, human beings, were doing in all the analogies above.
Where currents flow in this circuit? Draw them with whole loops in the two cases - at a positive and a negative input voltage. Remember: a current always arrives at the point from where it has gone. Another tip: in the two cases (at positive and negative input voltage) the load current will flow through different power supplies (accordingly, through the positive and negative voltage sources). Imagine also the complementary output emitter follower inside the op-amp: the upper n-p-n transistor will "blows" the output current when the op-amp "wants" to produce a positive voltage and the lower p-n-p transistor will "sucks" the output current when the op-amp "wants" to produce a negative voltage.
Disturbing the negative feedback follower
Let's now "provoke" our op-amp by various disturbancies to see how it will react:) Firstly, we may introduce and vary a line resistance. Then, we may vary the load resistance. Finally, we might vary both the resistances. If we are curious enough, we can vary the voltages of the power supply as well (the one, the other and both).
Output resistance. Can we say that we have realized one eternal dream of electricians - to build a varying perfect voltage source:)? Can we think of an op-amp follower as an (almost) ideal voltage source?
Conclusion: in electronics, we make ideal devices by appllying negative feedback.
Converting the NFB follower into an amplifier
Lab 4a: How to make perfect circuits by series NFB
Is the real diode a perfect component?
Diodes are two-terminal semiconductor elements, characterized by allowing the current to flow in the forward direction when the voltage is positive but not in the reverse one when the voltage is negative. So diodes represent the action of a valve and so were they called in the past. However, the diode needs a little (approximately 0.5-0.7V) voltage-push in order the current to start flowing freely. The result is small but noticeable voltage drop - VF. What is this voltage drop - useful or harmful? Sometimes it is useful; other times it is harmful...
"Useful" examples. When we make voltage stabilizers, we need this voltage drop. In these cases we do all that is possible to create and increase this useful voltage drop appearing across various diode component: diodes, LEDs, zeners, multiple diodes connected in series...
"Harmful" examples. In other cases, when we use a diode as a switching element, we do all that is possible to remove and to remove this harmful voltage drop... Then we need an ideal diode without forvard voltage drop VF...
Making the simplest series diode rectifier
A rectifier is an electical element that converts alternating current to direct current. Rectifiers form the basis for electronic power supplies and battery-charging circuits. Furthermore, they are used in signal processing to demodulate radio signals and in the precision conversion of AC to DC voltage in electronic voltmeters. However, keeping in mind that the diode allows a current to flow in one (positive) direction but not in the other (negative), we can easily use the term "rectifier" to name a diode.
So let's make the diode perform as a rectifier and thus participating in a half-wave rectifier circuit (the sircuit diagram is shown on the left). We have a sinusoidal source (for the experiment it is 1V) and a resistive load.
When the source voltage is positive, the diode is in the so-called forward-bias region. If the diode was ideal the source voltage will appear across the load. However, the diode we have is not ideal; so it performs a voltage drop equal to the push-up voltage of the diode (approximately 0.5 - 0.7V). So during the positive wave the output voltage is less than the source voltage (for our example it will be VL = VS - VF = 1 - 0.7 = 0.3V. The result is shown in the diagram on the right.
During the negative wave of the sinusoidal voltage the diode is in its reverse-bias region and no current flows through the load. Thus only the positive half-cycles result on the load, which makes the diode suitable as a rectifier.
Making an "ideal" diode without VF
Deriving the basic idea from our human routine
Implementing the powerful idea into an electrical circuit
Making an almost ideal "op-amp" series diode rectifier
What can act as a varying voltage source in our electronic circuit? What can "help" the imperfect diode by adding so much voltage to the input voltage as it loses across the diode? At last Lab 3 we were using transistors for such a purpose; let's now, for the sake of change, use an op-amp...
Now the op-amp has to "insert" the "helping" voltage VF into the circuit; so, how to connect it?
Eureka! We have invented an (almost) ideal diode without (any) forvard voltage VF! Let's investigate it by Microlab system.
Firstly, let's remind what an operational amplifier (op-amp) is: it's a device having a differential input (it may be current, voltage, mechanical motion?!?Circuit-fantasist (talk) 18:09, 7 May 2008 (UTC), etc.) and the output that varies according to the input but with a larger amplitude. Almost always they are used with negative feedback (the output signal is returned to the input in opposite direction to the source signal). However, they may be also used with a positive feedback (the output signal that is returned is summed with the input original signal) but much rarely. For ideal OA in a negative-feedback circuit, the network returns a fraction of the output to the inverting input and thus makes the differential input voltage toward zero.
So let's now go back to our experiment. What we are trying to do is to compensate the harmful fordward voltage drop on a real diode with an operational amplifier. But isn't the result of this try an ideal operational amplifier?!
We have a source voltage Vin and at the point named "1" we will have that voltage. However, if we want to compensate the voltage drop across the diode, we should use an op-amp that will have on its output a voltage equal to the source voltage plus an additional amount that is equal to this voltage drop. This will be the voltage we will have at point "2". If this increased voltage is applied before the diode, the voltage after the diode will be the same as our initial source voltage and that is what we have at point "3". Which is our purpose!