Opamps I   

 

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Operational Amplifiers (opamps) are one of the most essential and ubiquitous elements of analog synthesizers.  This concise document will attempt to communicate some of the more important characteristics and applications of these devices.

The following picture shows the basic schematic representation of an opamp with the more important pins called-out.

The power supply pins are frequently not shown on schematics employing opamps.  This is usually done for clarity of demonstrating the more significant role in which the opamp is used.  Each power supply pin is typically bypassed with a 0.1 microfarad capacitor to ground.  Those bypass capacitors need to be physically close to the power pins of the opamp package to be effective.

An opamp has a very high voltage gain.  It is so high in fact, that a fraction of a millivolt between the input terminals will cause the output to swing over its entire range.  Numerically, opamps have voltage gains from about 100,000 to several million!  Such gain diminishes as the frequency applied goes up: higher frequency = lower gain.  Feedback is used on an opamp to tame such high gains and make the opamp perform with a high fidelity.  (More on that, later.)

The two pins "+" and "-" are the inputs of the opamp.  The "+" pin is called the non-inverting input.  Signals sent to this pin of an opamp will be in-phase with the output signal.  That is, the "+" input is non-inverting with respect to the opamp output terminal.  The "-" pin is the inverting input.  Signals sent to this pin of an opamp will appear 180 degrees out of phase with respect to the opamp output terminal.

Said another way, the "+" (non-inverting) input tries to push the output to the same polarity as is being seen on the "+" input.  The "-" (inverting) input tries to push the output to the OPPOSITE polarity as is being seen on the "-" input.

With two inputs, each capable of pushing the output towards opposite polarities, what actually will happen?  The answer can only be found when a feedback circuit is used with the opamp.  The feedback circuit is what tames the opamp.  An opamp is really a difference amplifier.  That is, it amplifies the difference between the "+" and "-" inputs.  Hang in there and read on...

 

 

There are three "rules" for opamps.  They are a bit of a generalization but if heeded, will make learning about opamps a bit easier:

Rule #1:

The output of an opamp attempts to do whatever is necessary to make the voltage difference between the inputs equal to zero.  This does not mean that the opamp output can somehow change the voltage on the inputs!  But, the output "watches" the inputs and swings the output terminal such that the feedback circuit brings the input differential voltage to zero (or as close as it can).

Rule #2:

Opamp inputs draw no current.  That is essentially true.  Real opamps draw a very small amount of input current.  For some opamps the input current is measured in pico-amperes, while less specialized opamps have input currents in the range of 1.0 or less nano-amperes.  In many (but not all!) cases, the input current can be ignored.

Rule #3

Opamp outputs are rather strong in their ability to achieve and maintain a given voltage.  That is, opamps have a fairly low output impedance.  They can be "forceful" in making certain their voltage is dominant.  A low output impedance is the characteristic needed when the output of one opamp is intended to drive the inputs of many other circuits in parallel.

 

 

Remembering our first two rules, the following basic configuration isn't too difficult to understand.  Because of rule #2, the "+" input of the opamp is effectively tied to ground.  (With essentially no input current, there can be no voltage drop across R3.)  Since the "+" input is at ground, rule #1 says that the "-" input must also be at ground:

(Now for the fun part!)  Since the "-" input is at ground, the output voltage appears across R2 and the input voltage appears across R1.  Therefore, using rule #2, we have:

Vout / R2 = -Vin / R1

Or put another way...

Voltage Gain = Vout/Vin = -R2/R1

Can it really be that simple?  Yes, it is!  (By the way, the "-" in the above equation is a consequence of the fact that the signal is applied to the inverting input of the opamp.  Remember, the inverting input drives the output towards the opposite polarity.  So the "-" is necessary to indicate that with a positive input voltage, the output voltage will be negative.)  The combination of R1 and R2 comprises an elemental feedback arrangement.  As we hinted before, it is the feedback arrangement that tames the incredible gain of an opamp.

The above example almost obscures with simplicity, what is actually happening.  Let's look at that a little closer and plug-in some real-world component values.  Then you'll have an excellent understanding of what really goes on.

To understand how the feedback arrangement actually functions, imagine that the input is held at +1.0 volt.  Now imagine that for some reason the output is uncooperative and is sitting at 0.0 volts.  What will happen?  R1 and R2 comprise a voltage divider that will hold the "-" input at +0.91 volts.  Since the "+" input is at 0.0 volts (rule #2) the opamp sees a huge difference between the "+" input (0.0 volts) and the "-" input (+0.91 volts).  Since the "-" input drives the output towards the opposite polarity of the "-" input, the output is forced negative.  How much negative?  Far enough to reach equilibrium (when the difference between the "-" and the "+" pins = 0).  That will only happen when the output goes to -10.0 volts.  That is because the resistance of R2 is 10 times the resistance of R1 and it will take 10 times as much voltage to cause the same current flow through the 100K (R2) resistor, as flows through the 10K resistor (R1).

Our "Real-World Basic Inverting Amplifier" is in fact an inverting gain-of-ten amplifier that really works!  If you need more or less gain, change the ratio of R2 to R1 per the equation:

Voltage Gain =  -R2/R1

What happens when R2 = R1?  That is unity gain.  You get out a negative voltage of equal amplitude to the positive input voltage: an analog inverter!  Okay, so what happens when the value of R2 is less than the value of R1?  You guessed it... you get an inverting amplifier with a "gain" less than -1.

We really haven't addressed why R3 is there and why the value of R3 should be the product of R2 and R1, divided by the sum of R2 and R1.  R3 is included to help minimize output drift due to temperature effects.  The details aren't all that interesting but can be a bit wordy.  So I'm going to cop-out and simply say: include R3 in your circuits and make its real value, the nearest 5% standard resistance as calculated.

 

 

Gee, rule #3 didn't figure in very prominently did it?  True.  For simple voltage amplification applications the low output impedance doesn't mean much.  But if you want to drive something more current hungry, like an LED, the low output impedance comes in real handy.  Suppose you wanted to light up an LED by "eavesdropping" on a synthesizer control voltage.  [Hanging an LED on the control voltage line by itself would screw-up things royally.]  Remember rule #2?  That says an opamp input draws essentially no current (very little in practice).  Since the opamp input won't hardly load the circuit we are trying to sense, and the opamp output is rather "sturdy" (low impedance), wouldn't an opamp be just about perfect to monitor a control voltage without harming its accuracy?  Indeed it would.

Before we jump onto that application we need to explore another arrangement of opamp circuits.  All we have seen so far is an inverting amplifier.  It is very useful in many cases but there are some drawbacks.  For one, the input impedance is equal to the value of R1.  Since R1 will typically have a value in the range of 1K to maybe 5 meg-ohms, that seems kind of "crappy" when compared to the very little input bias current the amplifier needs.  This situation gets worse as the gain is pushed higher (and R1 gets smaller).  Are there different solutions that better utilize the great input characteristics of an opamp, without the drawback of rather low input impedances?  There are a number of such solutions and one of them looks like this:

 

In this circuit, the analysis is straightforward.  VX must be = Vin.  But the voltage at VX comes from the voltage divider comprised of R2 and R1.  Therefore:

VX = Vout * R1 / (R1 + R2)

Setting Vx = Vin results in the basic gain equation:

Voltage Gain = Vout / Vin = 1 + R2 / R1

Unlike the inverting amplifier, the non-inverting amplifier cannot have a gain less than unity.  That usually not much of a problem as one can use a non-inverting amplifier as the front-end, followed by an inverting amplifier to provide a precision gain of less than unity.  Or one could also use a resistive divider on the output to reduce the signal level if succeeding stages are relatively high impedance.

 

Herewith is the LED driver circuit.  A bipolar LED is used to indicate the polarity of the input voltage and the relative LED brightness provides a gross approximation of the input voltage amplitude.  If used to monitor critical control voltages, this circuit will not load-down the control voltage itself (a good thing).

 

This circuit works but could use some refinements.  A friend of mine asked for an LED circuit that could monitor the control voltage output of his MIDI-CV unit.  He wanted to be able to "see" if something was happening.  Using not much more information than presented herein, the following circuit addresses that bit of functionality.

The MIDI-CV in question is an Encore Expressionist®.  It can output control voltages in the range of -3V to +10V.  The bank of 8 LEDs to be used were the same ones found in some  Synthesis Technology MOTM modules.  Those LEDs have a maximum continuous current rating of 20 milliamperes and a typical operating voltage of about 1.85V.  Here is a circuit that barely lights up the LED when the control voltage is -3V and lights up the LED with about 17 milliamperes of current when the control voltage is +10V.  The opamp selected must be a Burr-Brown OPA2134.  Few opamps have the current drive capability of the OPA2134.  And yes, the OPA2134 is probably my favorite, general purpose opamp!

 

Did this help anybody?  What else should be added?  It is a lot of work to create such pages but I'm willing to put forth the effort if you are willing to read and comment on what you've seen so far.

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