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Educational Supplement

Discrete Operational Ampliﬁer

Principles of Operation

What’s inside a ‘741 integrated circuit,

and how does it work?

§1.0 Introduction: Op-amps and the ‘741

An operational amplifier, or "op-amp" as they are more As an analog circuit element, an op-amp performs these
commonly known, is one of the most useful and common operations on the electrical signal itself – the voltage, or
types of building blocks for making analog electronic circuits. electrical potential, at its terminals – and not on a digital
representation of that signal.
Op-amps are two-input devices – almost always integrated
circuits – that perform mathematical operations upon their A real op-amp has additional pins for power supply
inputs. Placed in properly designed circuits, they are commonly connections, and sometimes for additional external adjustment
used to amplify, add, subtract, differentiate, integrate, filter, as well.
compare, regulate, mix, and control electronic signals. If you
look inside just about anything electronic, you’ll find op­amps The most popular op-amp of all time – the quintessential op-
hard at work. amp – is the μA741, designed by Dave Fullagar while working
for Fairchild in 1967. While it was not the first integrated circuit
op­amp (the μA702 holds that title), it became a smash hit
   thanks to its built-in frequency compensation which made it
   the easiest and simplest to use. Previous op­amps needed
  
carefully designed resistor and capacitor circuits to control the
bandwidth and stability of the amplifier. Imagine having to do all
that work even before connecting the inputs and the outputs!
The schematic symbol for an op-amp has two inputs: the “non-
inverting” input (+) and the “inverting” input (–), as well as one While newer op-amp designs may outperform it in just about
output. The op­amp takes the signal from the + input, subtracts every possible respect (speed, noise, voltage range, and so on),
the signal from the – input, and then multiplies the result by a the original ‘741 remains widely beloved and in active
large factor called the gain. production today — over 45 years later.

XL741: Principles of Operation (Rev 1.0.4, August 2015) 1


 

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Figure 1: XL741 Block Diagram

§1.1 The XL741

The XL741 Discrete Operational Amplifier kit is a functional The differential amplifier:
transistor-scale replica of the μA741 integrated circuit. It The input differential amplifier subtracts the – input voltage
implements the "equivalent circuit" from the original Fairchild from the + input voltage. Thus, it finds the difference between
μA741 datasheet, using discrete components such as resistors the two signals. Another part of the circuit controls the tiny
and individual transistors. It is, so to speak, a “dis-integrated currents (input bias currents) that need to flow into the inputs
circuit,” containing essentially the same components that you so that the input transistors can function.
might find on the die of a ‘741 IC.
The bias generator:
As with an integrated circuit ‘741, you can build working The bias generator is the “power plant” of the circuit. Using the
amplifiers, oscillators, and other analog circuits with the XL741, VCC+ and VCC– supply voltages, it generates an internal
and hook up with solder connections, test probes, or alligator reference current which provides power to many of the other
clips to monitor what happens at the pins. However, unlike the transistors.†
chip version, it’s also easy to insert your own probes inside the
circuit, to monitor what happens at any point inside what The gain stage:
would otherwise just be a black box. The gain stage performs the multiplication function. At its
heart, it is a simple transistor amplifier which multiplies the
This ability to peek inside the circuit makes the XL741 a unique voltage difference between the + and – terminals by a factor
educational tool. In what follows, we’ll work through the circuit of about 200,000. This helps the op­amp react to very tiny
diagram, discuss the theory of operation of the ‘741 op-amp, changes in the input voltages.
and present some opportunities for experiments and further
exploration. Most op­amp circuits operate by creating a balance such that
the op­amp tries to make both input voltages exactly equal.
§2.0 The High-Level Overview The output signal of the op-amp usually feeds back into a node
in the external circuit that surrounds the ‘741. Typically, this
Let’s begin by looking at the ‘741 block diagram, Figure 1. external circuit connection involves the – input to create
Here, we simplify the many individual components into a much negative feedback.
smaller set of functional blocks:
The output stage:
• A differential amplifier, Finally, the output stage takes the low­current signal from the
• a bias generator, gain stage and puts some muscle behind it, allowing the op­
• a gain stage, and amp to provide more current to the outside world.
• an output stage.

† The names “VCC+” and “VCC–”mean the positive and negative power supply inputs to the ‘741, respectively. In this case, DC voltages up
to 18 V (for VCC+) and down to -18 V (for VCC–).

XL741: Principles of Operation (Rev 1.0.4, August 2015) 2

     
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Figure 2: Detailed schematic diagram

§2.1 The Overall Schematic

The detailed schematic for the XL741 circuit board is shown in

Figure 2. As we walk through the details of the different blocks
that make up the circuit, it will help to keep this overall
diagram nearby.
“Collector current” Ic
You may notice – from this schematic or the parts included
with the kit – that the electronic components in the circuit “Base current” IB
consist primarily of resistors and transistors. We’ll briefly
review these two components before we dive into the circuit
diagram. NPN Transistor

§2.2 Component review

Aside: An NPN transistor also has a second internal diode, which
Review: Resistors conducts (only) from the base to the collector. In most use cases,
The defining property of resistors is that they are circuit a circuit that uses this kind of transistor is designed to always keep
elements that follow Ohm’s law, which is V = I × R, where V is (or “bias”) the collector at a higher voltage than the base, so that
the voltage across the resistor, I is the current through the the base-collector diode does not conduct current. So long as this
resistor, and R is the value of the resistor (measured in ohms). bias is maintained, we can usually ignore the presence of the
For example, if a 51 Ω resistor has current of 1 mA through it, second diode.
then the voltage difference between its two ends is 51 m V.

While there are many types of transistors, the ones in this An important detail is that there is a small voltage loss across a
circuit are bipolar transistors, in which a small current controls diode, usually about 0.7 V. (The exact loss depends on the type
a larger one, acting as amplifiers or switches. There are two of device and the operating conditions). This means that (a)
flavors of these, NPN and PNP; we’ll pick NPN to look at first: when current is flowing from the base to the emitter, the
The three terminals of an NPN bipolar transistor are named voltage at the emitter pin is usually about 0.7 V below that at
“base,” “collector,” and “emitter.” The base and emitter pins the base and (b) current does not begin to flow from the base
comprise a diode, indicated by the little arrow on its schematic to the emitter until the voltage of the base is about 0.7 V
symbol. A diode is a unidirectional circuit element, so under above that of the emitter. This typical voltage difference of 0.7
normal circumstances, current can only be made to flow from V is usually referred to as a “diode drop.”
the base to the emitter, and not from the emitter to the base.

XL741: Principles of Operation (Rev 1.0.4, August 2015) 3

Finally, there is the matter of switching and amplification. When
current flows from the base to emitter, that current (“base “Base current” IB
current,” symbol IB) is said to “switch on” the transistor: it
allows current to flow from the collector to the emitter. The “Collector current” Ic
maximum current that can flow from the collector to the
emitter is given (to good approximation) by IC = h × IB, where
IC is the “collector current” (the current flowing from the
collector to the emitter) and h is a gain factor that depends on
the particular transistor. If h has a value of, say, 30, then the PNP Transistor
transistor acts as an amplifier where a small change in IB causes
a change 30 times larger in IC. Please, NPN: Not Pointing iN.) In a PNP, the emitter and base
form a diode where current can only flow one way, but this
The other flavor of transistor, PNP, works almost exactly the time only from emitter to base (again, in the direction of the
same way. Its schematic symbol is very similar except that the arrow). And, when current flows through that diode, it allows
little arrow is pointing in. (Mnemonics! PNP: Pointing iN a current IC = h × IB to flow from the emitter to collector.

§3.0 The Differential Amplifier

Now that we've reviewed the basics, lets dive right into the
741 schematic. The non-inverting input and the inverting input
are connected to the bases of NPN transistors Q1 and Q2, 

!! 
respectively. The collectors of these two transistors are
connected to the positive power supply VCC+ through PNP   
 
transistor Q8. Because its collector and base are connected # 
together, Q8 acts like a simple diode here (we say that it is Bias current
“wired as a diode”). The collectors of Q1 and Q2 are thus   monitoring
#    #  signal
normally kept at a voltage of one diode drop (about 0.7 V) 
  
below VCC+.   

 
Q1 and Q2 act as “voltage followers” that buffer the input
signal voltage. They amplify the input current, increasing the  
input impedance. That is to say, they provide amplification such      
that the + and – inputs work well, even while drawing only a  
control signal from
 
very small amount of input current. # 
 
#  bias generator
 

Q3 and Q4 are common­base amplifiers that act as level 


shifters. The signal that controls them comes from the bias # 
generator block; we will come back to this (and its function) 
 
Output of
 
when we discuss that block. #  #  differential
  
 
If the – terminal’s voltage is higher than the + terminal’s 
voltage, then transistors Q2,Q4 conduct more current down  

and out through Q4’s collector and transistors Q1,Q3 conduct

  
less. This causes the differential amplifier’s output voltage to
increase. We will work through how this works in greater detail " " "
in the next section (§3.1). 
!! 
Aside 1: Output polarity
Ultimately, when the – terminal input is at a higher voltage than the
Figure 3: Differential Amplifier
+ terminal, the ‘741’s output voltage should decrease, so the
paragraph above might seem backwards. However, everything
works out: the gain stage will later invert the signal back to the
proper polarity.

Aside 2:Transistor gain

The PNP transistors in the original μA741 had very poor current
gain (less than a factor of 10!), but that was all they had in those
days. The combination of a NPN transistor followed by a PNP
transistor (as in Q1,Q3 or Q2,Q4) creates a sort of “super PNP”
transistor that more or less made up for that problem.

XL741: Principles of Operation (Rev 1.0.4, August 2015) 4

Aside 3: D1 and D2
We have glossed over something else here. Between Q1 and Q3, A viable solution – added in version 2.0 of the XL741 kit – is to use
and between Q2 and Q4 are diodes D1 and D2, respectively. two Schottky diodes which have a breakdown voltage above 30V.
These two diodes are not present in the original μA741 circuit. Added in series with the transistors, they protect the 2N3906s
with a very small impact on the offset voltage.
In the original μA741, Q3 and Q4 are lateral PNP devices with a
base-emitter breakdown voltage (VEBO) greater than 30V. If you so choose, you could build up the XL741 without D1 and
However, the 2N3906s used in the XL741 have a much lower D2, by substituting a wire for each of them. Doing so is somewhat
VEBO: an absolute maximum of 5V. This would limit the differential more authentic in terms of the device layout, but not in terms of
mode range (the maximum voltage difference between the + and performance. With the added diodes in series, the input transistors
– inputs) to about ±11 V. The “obvious” fix of using a different Q1 through Q4 are protected when excessive differential mode
transistor with a higher VEBO, turns out to not be practical, as few voltage is applied to the inputs. But without D1 and D2, it is critical
or none are currently in production. to keep the + and – inputs within ± 11 V of one another.

(4a) (4b)


+1 V OFF ON +2 V +2 V ON OFF +1 V

0.6 V + 0.6 V 0.6 V + 0.6 V

output voltage output voltage
0.8 V 0.8 V


Figure 4: Simplified differential amplifier

§3.1 A model differential amplifier

In order to understand how the differential amplifier section of As current flows through the branch on the right, the point
the ‘741 works, it is helpful to simplify the circuit to its core above the “load” resistor in that branch develops a voltage. We
elements (Figure 4). We can use this model to visualize the know this because Ohm’s law tells us that the voltage across a
path that the majority of the current takes, depending on the resistor depends on the current through it. We read out this
input voltages. We will show that the input with the higher voltage as the output of the amplifier.
voltage dominates the behavior of the circuit.
Let us now consider the other case, in Figure 4b. The voltage at
First consider the case of Figure 4a. The voltage at the + input the + input is 2 V, and the voltage at the – input is 1 V. The
is 1 V, and the voltage at the – input is 2 V. On the right side of behavior in this case is just the opposite: the two left-hand
the circuit, the 2V input voltage goes through the two base- transistors are on, the two right-hand transistors are off, and
emitter junctions in the two transistors (that is, two effective current flows down the left branch of the circuit. The output
diodes, with two “diode drops” of voltage loss) losing 0.6 V + voltage is lower, because less current (zero, for these input
0.6 V = 1.2 V along the way. This gives a voltage of 0.8 V (i.e., 2 values) is flowing through the right-hand load resistor.
V - 1.2 V) on the central circuit node (wire) that connects the
bases of the two PNP transistors. A valid question at this point is “If this is a differential amplifier,
where is a difference actually being amplified?” The answer is
Now, note that the voltage between the + input and this that the difference between the input voltages controls an
central node is only 0.2 V (1 V - 0.8 V). That means that there is amplified current that predominantly flows through one of the
not enough voltage across the base-emitter junctions of the two resistors near the bottom of the diagram, developing an
two left-hand transistors to turn them on. Thus, they stay off also-amplified voltage. Thus, the output voltage is an amplified
and no current flows through the the left-hand branch of the version of the difference between the two inputs.
The ‘741 does not use load resistors like the ones shown in
Figure 4. Instead, Q5, Q6, and Q7 form a current-mirror
circuit, which we will discuss in the next section. The current
mirror acts as an active load that allows a higher gain than is
possible with simple load resistors.

XL741: Principles of Operation (Rev 1.0.4, August 2015) 5

(5a) (5b) VCC+

IIn IOut IIn IOut

2 × IB
IC Tiny!

QFive QSix QFive QSix



Figure 5: Simple and buffered current mirrors

§3.2 Current Mirrors Since even a small current difference puts the differential
amplifier off balance and creates offset errors, the ‘741 has an
Transistors Q5, Q6, and Q7 form an active load current mirror enhanced current mirror, of the type shown in Figure 5b.
circuit that increases the gain of our differential amplifier.
In this case, a third “buffer” transistor (QSeven) provides most of
A current mirror is called that because it “copies” a current the base current directly from the power supply, sipping only a
through one circuit element to a current in another element. tiny fraction of current from the input current IIn to drive its
To illustrate how that works, Figure 5 shows two current base. This allows the output current to be much closer in value
mirror circuits similar to those in the ‘741. We have named the to the input current than it would without the third transistor.
transistors here QFive, QSix, and QSeven to note their
correspondence to those in the ‘741.

Figure 5a shows a basic current mirror. Notice how QFive and §3.3 The Offset Null Pins
QSix are connected. Transistor QFive is wired up as a diode, with
its collector connected to its base, allowing current to flow Inputs 1 and 5 of a ‘741 are “offset nulling” pins. These inputs
through its emitter. Part of the input current IIn provides the allow the bias currents in each side of the differential amplifier
base current IB that turns on the two transistors. Since their to be adjusted so that any remaining offset voltage can be
bases and emitters have the same voltages, symmetry dictates zeroed out.
that the same amount of current flows from the base to the
emitter in both transistors. Furthermore, since both transistors When the + and – inputs of a ‘741 are equal (connected
have the same gain, they will also have the same collector together) the output voltage at pin 6 should ideally be zero. In
current. (Recall that IC = h × IB from §2.2.) In this sense, the the real world, thanks to imperfect component matching, the
output current IOut through the collector of QSix “mirrors” the output is never exactly zero. Typically there is an input offset
collector current through QFive. voltage, such that a slightly different voltage (perhaps as much
as 10 mV) needs to be applied to the two inputs in order to
Aside: Matched transistors make the output zero. To compensate for this with the nulling
We have just relied on two properties of the transistors being inputs, hook a 10 kΩ potentiometer between the two pins (pin
equal: the current-voltage curve of the base-emitter junction and
the transistor gain. This generally requires excellent matching
1 and pin 5), with the potentiometer’s wiper connected to
between the two transistors. Ideally, the two transistors should be VCC– . Then, adjust the pot until the output is zeroed.†
fabricated as part of the same IC die for the best possible
performance. There are other places on the ‘741 where matched It is worth noting that most circuits that incorporate a ‘741 are
transistor pairs are needed as well – notably differential input not particularly sensitive to input offsets. Accordingly, most ‘741
transistors Q1 and Q2 should be matched as should Q3 and Q4. circuits leave these two terminals permanently unconnected.

While the collector currents are mirrored, there is some

subtlety since the base currents for both transistors are
supplied by the input current: IIn = IC + 2 × IB . Given that the
output current IOut is equal to IC, we have IOut = IIn – 2 × IB .
Thus, the output current is a bit less than the input current.

† When the two inputs are connected together and the output is disconnected, the output “prefers” to be fully positive or negative, as
it amplifies even tiny differences between the inputs. However, one can always trim the pot until the offset is almost exactly zero.

XL741: Principles of Operation (Rev 1.0.4, August 2015) 6

§3.4 The Common Mode Control Loop Voltage

Much as the differential input voltage refers to the difference VCC+

between the input voltages, the common mode voltage is the VIN+
average voltage of that on the + and – input terminals. Please Differential voltage:
see Figure 6 for an illustration. If both inputs change in value VIN+ – VIN–
(say, both go up in voltage the same amount) but in such a way VIN–
that the difference between the two inputs stay the same, then
we say that their common-mode voltage has changed but their Common-mode voltage:
differential voltage has not. VIN+ + VIN–
A good op-amp should only be sensitive to the differential 0V
voltage and its output should ignore any common-mode
voltage, whether steady or changing. The ‘741 has a rather Figure 6: Differential vs common-mode voltages
clever circuit, the “common mode control loop” to help
manage this.

Transistors Q8 and Q9 form a current mirror similar to the

one in §3.2. Unlike the previous example this current mirror
uses PNP instead of NPN transistors. However, it works in
essentially the same way, and Q9 “mirrors” Q8. You could also
say that the Q9’s collector current represents a measurement
of the sum total bias current (amplified by the transistor gain)
that flows into the differential amplifier. IBias
Separately, in the Bias Generator block, transistor Q10
generates a fixed reference voltage through its collector. (We IBias
will discuss this further in §4.) Thus the measured bias current (Driven)
from the collector of Q9 flows into the circuit node (wire)
that connects Q9 and Q10, while a fixed reference voltage
flows out of it. The difference between these two current
values must be taken up by current that flows into or out of
the node connecting the bases of Q3 and Q4. Control signal,
affects IBias
(Why? Because the total amount of current that flows into and
out of a given circuit node must sum to zero. If this is not
intuitive to you, you may wish to read up on Kirchhoff's circuit

If the bias current is larger than the reference current, then the
extra current must flow towards the differential amplifier,
increasing the voltage at the bases of Q3 and Q4. This
decreases the voltage across the base emitter junctions of the
transistors Q1, Q2, Q3, and Q4, reducing the input current IRef
and therefore the measured bias current. (Driven)

If the bias current is smaller than the reference current, then

current must flow out of the differential amplifier into Q10.
Thus the voltage at the bases of Q3 and Q4 decreases,
increasing the voltage across the base emitter junctions of Figure 7: The common-mode control loop
transistors Q1, Q2, Q3, and Q4, and increasing the input
current. See Figure 7.

The end result is that this circuit sets the base voltage of Q3
and Q4 to follow the optimal common mode input voltage of
the op­amp, and at the same time, it controls the maximum
input bias current.

XL741: Principles of Operation (Rev 1.0.4, August 2015) 7

§3.5 Questions and Experiments I 4. How can you disable the common mode control loop?
Don’t forget about the bias current mirror Q8,Q9; the
1. Short the base and emitter terminals of Q7. How does this changes might be more extensive than you think.
change the performance?
5. How precisely do the transistors need to be matched?
2. On a real (integrated circuit) ‘741, try measuring the values What happens if the transistor matching is poor?
of R1 and R2. You can do this with an ohmmeter connected
to VCC– and one or the other offset null pins (while 6. The XL 741 uses discrete, separately-manufactured
disconnected from power). How close are they to the transistors. Why does this still work? How does the input
nominal specification of 1 kΩ? offset voltage (for example) compare to that on an
integrated circuit ‘741?
3. Measure the voltage on the two offset pins of a real
(integrated circuit) ‘741 and compare them with the same 7. How does an active load compare to a traditional passive
measurements on the XL741. Measure the resistance load– i.e., a resistor? Could an op-amp be made with a
between that terminal and the negative rail (with the op­ resistive load instead?
amp powered off), and calculate the bias current in the
differential amplifier. Try adjusting the common mode 8. What purpose does R3 serve? What would be different if it
voltage to see if it changes the bias current. The easiest way were not there?
to control the common mode voltage is by connecting the
amplifier as a voltage follower and manipulating the voltage
supplied to the noninverting input.

§4.0 The Bias Generator

     
The core of the bias generator circuit is formed by diode-
connected transistors Q11 and Q12. These create a reference   

current set by the supply voltages and by R5. Q12 and Q13      
"  "  " 
act as a current mirror that provides bias current to the gain   


 

 
Transistors Q10, Q11, and resistor R4 form what is called a 

Widlar current source. The output current of the Widlar 

source is the reference current divided by a fixed ratio 

determined by R4. A small fraction of the reference current   " 
flows into the collector of Q10, and this reference current is !
used by the common mode control loop.

Unlike most modern analog ICs which have a reference

current that depends only on temperature (PTAT, or 
Proportional To Absolute Temperature), the ‘741 reference  
" 
current changes when the supply voltage changes. This is 

probably why the datasheet specifies performance for supply

voltages of +/­15V only. Modern op­amps use internal 

compensated current sources which make them operate 

 
" 
consistently over a wide supply voltage range. "  "   
 
" 
§4.1 Questions and Experiments II   
 
1. Try changing the value of R5. How does this impact the
performance? How about the supply current?

Figure 8: Bias Generator

XL741: Principles of Operation (Rev 1.0.4, August 2015) 8

§5.0 The Gain Stage §5.3 Inside the gain stage

The gain stage performs the multiplication function on the the The Darlington pair Q16,Q17 form the heart of the gain
signal coming from the differential amplifier. Before we get into stage. It amplifies the signal coming from the differential
the heart of it, let’s look at a couple of the building blocks that amplifier. Since it is arranged as a common-emitter amplifier, it
we’ll use. also inverts the input signal. Remember how the differential
amplifier output seemed backwards? The gain stage takes this
§5.1 Background: Darlington transistors inverted output and changes the polarity back to the intended

Q13, which mirrors Q12, is the load of the amplifier. It is an

active load to maximize the gain of the circuit. You can think of
an active load as a current source (it is a current mirror, after
Figure 9:
all) in parallel with a large, fixed resistor. This effective resistor
The Darlington plays the role of the load resistor in the common-emitter
amplifier shown in §5.2. However, as an active load, it has a
very high impedance (meaning that the voltage across it varies
Two of the transistors in the gain stage, Q16 and Q17 are strongly as a function of current) while maintaining a relatively
wired together emitter-to-base in what is known as the constant amount of bias current through it.
Darlington configuration.
In the Darlington configuration, the emitter of one transistor is Operating point
hooked to the base of another. This effectively makes one new
“super transistor” out of the two, since it is the amplified
current out of the first collector that serves as the base
current for the second transistor. If the gain of a single Resistive load
transistor were 30, then the effective gain of the Darlington
pair would be 30 × 30 = 900.
This huge improvement in gain is not without cost. Since there Current Operating point
are two transistors, there are also two diode drops to be
overcome; the Darlington pair will not begin to turn on until its
base is at least 1.4 V above the emitter.
Active load
§5.2 Background:The common-emitter amplifier
VCC+ Figure 10:
Common emitter Figure 11: Resistor versus active load

Output Q18, working with R7 and R8, is called a VBE generator. It

generates a fixed voltage drop of roughly 1.2 V regardless of
Input what the amplifier is doing. The gain stage therefore has two
outputs: one is the signal directly coming from the collector of
Q17, and the other is the same voltage shifted up by 1.2 V. If
the VBE generator were not there, the output would exhibit
The transistor circuit snippet above is called a common emitter crossover distortion.
amplifier. “Common” in this context means that the emitter is Output
shared reference point between the input and output of the
circuit. In this particular instance, common is also ground, and Input
this configuration is sometimes called a “grounded emitter” Crossover
amplifier. distortion

Thanks to the gain of the transistor, a small current into the

input causes a larger current to flow from VCC+ to ground,
going through the “load” resistor along the way. Since the
voltage drop across the resistor increases as the current
through it increases, the voltage on the output decreases as the
input current increases. That is to say, this is an inverting
Figure 12: Crossover Distortion

XL741: Principles of Operation (Rev 1.0.4, August 2015) 9

This distortion happens when the output voltage crosses 0 V This method is known as dominant pole compensation. It is very
(either rising or falling) and both transistors are shut off or much the same thing as adding a large RC low-pass filter within
close to being shut off. If the VBE generated too much voltage, the circuit, which costs gain at higher input frequencies, but makes
then both transistors could be on at the same time, which the op­amp’s characteristics predictable and easy to work with.
would waste power.
§5.4 Questions and Experiments III
C1 is the compensation capacitor (and the only capacitor in
the ‘741), also known as a Miller capacitor. This capacitor 1. What happens to the supply current if you increase the
stabilizes the behavior of the op-amp at the cost of some gain. voltage generated by Q18 by increasing R8? What happens
The introduction of this capacitor into the op-amp itself was if you decrease the voltage generated by Q18, or short out
the key innovation that led to the success of the ‘741. At the Q18’s collector and emitter? Try configuring the op­amp as
time of its introduction, all other op­amp ICs required an a voltage follower driving a 1 kΩ load, with a sine wave
external compensation network and thus were harder to use. input. Compare the input to the output on an oscilloscope.

Aside: Frequency compensation 2. What happens if you remove C1? What happens if you
increase C1?
An analysis of the ac performance of the ‘741 is outside of the
bounds of this discussion, but we can mention the broad strokes as 3. Can you measure and reproduce the curve in figure 13?
they relate to the capacitor. (We will use some new terms that you This is easiest with a variable-frequency signal generator and
might want to research separately!) an oscilloscope. At what frequency is the “knee” of the plot?
What happens at the crossover frequency? Compare the
Open-loop Low Frequency results for the XL741 to that with an integrated circuit ‘741.
Gain Dominant Pole
(log scale) 4. Short Q16’s base and emitter. What happens? Try building
Slope: 20 dB
some high gain amplifier circuits and measure the actual
per decade gain.

Frequency 5. What happens if you replace Q13 with a resistor? Try

(log scale) removing Q13 and connecting a 39 kΩ resistor from the
collector connection to the emitter connection. You can test
it with a voltage follower configuration and a pulse
Crossover frequency: the GBP generator.
(gain-bandwith product) of the op-amp

Figure 13: Frequency Response

    
The addition of the C1 creates a low­frequency “pole” in the
frequency response that dominates all the other effects caused by 
parasitics. That is to say (in essence, and glossing over some details)  
that we add one large capacitor, which overwhelms various stray #  
capacitances from all of the components within the op-amp, and  
# 
leaves the output performance (gain as a function of frequency) 
 

instead constrained by an intentionally chosen component. # 

 

 


  
 # 
§6.0 The Output Stage "

!  

The output stage uses two transistors configured as emitter 

followers. Q14 can source current to the output, and Q20 can 
sink current from the output. This allows the op­amp to drive 
current in either direction depending on the needs of the # 
 
circuit connected to the output terminal.  
# 
 

Q15, in combination with R9, acts as a current limit when the 

# 
op­amp is sourcing (outputting positive) current. When high  

# 
current flows through R9, it increases the base-emitter voltage 

across Q15 enough that it can begin to conduct. As Q15 turns  
on more, it “steals” current from the base of Q14, reducing the ! 
amount of emitter (output) current.

Figure 14: Output stage

XL741: Principles of Operation (Rev 1.0.4, August 2015) 10
When the op­amp sinks (outputs negative) current, a fraction 3. Is it possible to change the current to limit at ± 50 mA?
of this incoming current flows through R11. As the voltage on How would you go about doing this?
R11 increases, the current limit transistor Q22 begins to turn
on and steal current from the Darlington pair Q16,Q17, which 4. Suppose that you wanted to build a high-power ‘741,
increases the output voltage, reducing the sink current. The capable of sourcing or sinking 1 A of current by changing
XL741 can sink much more current than a the original ‘741 the components in the output stage. How would you go
because the 2N3906 PNP transistor has more gain than the about doing this? What components, besides those that limit
lateral PNP used in a ‘741 integrated circuit. the current would need to be changed, and why?

§6.1 Questions and Experiments IV 5. What is the total range of the XL741’s output? How does it
depend on the total power supply voltage? What sets this
1. Do Q14 and Q20 conduct current at the same time? In range?
other words, does the VBE generator create enough of a
voltage drop that both transistors are on at the same time? 6. Some op-amps have a “rail to rail” output feature, where the
Why might this be done? output voltage can get to within millivolts of the positive or
negative power supply pins. How do these circuits work?
2. What is the current limit of the XL741? Is it the same for Could you design an output stage for the XL741 that has a
sourcing and sinking current? How “sharp” is this limit? larger output range?
Compare its behavior to that of a integrated circuit ‘741.
7. What sets the minimum power supply voltage required for a
‘741 to work properly?


1. Main XL741 Documentation page:

2. XL741 kit product page:

3. Original Fairchild μA741 datasheet:

4. Tube Time, the web site of Eric Schlaepfer, designer of the XL741 circuit board:

5. Texas Instruments App Note 31 has a great number of basic op-amp example circuits:

6. Wikipedia is an excellent resource for learning basic electronics. Some potential topics of interest include:

7. Analog Devices Op Amp History (Short url: )
A detailed history of operational amplifiers, beginning with vacuum tubes.

8. Huijsing, Johan H. Operational Amplifiers, Theory and Design. Boston: Klewer, 2001. Print. pp. 292­294.
A brief but more technical description of the μA741.

9. Grey, Paul R., Hurst, Paul J., Lewis, Stephen H., and Meyer, Robert G. Analysis and Design of Analog Integrated Circuits (4th Edition). New York: Wiley,
2001. Print. pp. 454­472.
A very good technical description of the μA741 with a lot of detail and mathematical analysis. Be sure to get the 4th edition; the section on the
μA741 was removed from subsequent editions.

This document is a continuing work in progress. Your corrections and suggestions are welcome.
Please write to us at or use the web form at

XL741: Principles of Operation (Rev 1.0.4, August 2015) 11