Solving the Differential Amplifier – Part 2

Design a Differential Amplifier with the Coefficients Identification Method

In the first part of this series, MasteringElectronicsDesign.com: Solving the Differential Amplifier – Part 1, I wrote that the common usage of the differential amplifier is as a gain circuit for the differential voltage at its inputs.  When the circuit in Figure 1 has the resistor ratios equal

image001 (1)

the amplifier transfer function is

image002 (2)

and the circuit amplifies the difference between the input signals.

differential_amplifier_1

Figure 1

But the resistors’ calculation  becomes a bit of a challenge, when one might be faced with designing a differential amplifier with a certain transfer function.  The example I took in the first article was as follows:  Given an input range of, -0.5V to 5.5V, the output has to swing between, -1.25V and +2.365V.  I solved the problem by using the amplifier transfer function and a system of equations.

In this article I am going to write about designing the resistors of this differential amplifier using the method of coefficients identification.

Starting from the differential amplifier transfer function,

image004 (3)

we note that this is a linear function Vo, with two variables:  V1 and V2.  However, if we consider one of these two variables a known value, say, V2, Vo becomes a simple linear function with one variable.  Let’s note this function with Vo(V1).

The design requirements are as follows:

If Vin1 = -0.5V, then Vout1 = -1.25V and
If Vin2 = 5.5V, then Vout2 = 2.365V,

where by Vin1 and Vin2 I noted the input range limits, and by Vout1 and Vout2 I noted the output range limits.

A linear function of first degree is a straight line which is determined by two points in the (x,y) plane (see Figure 2).

Linear Function

Figure 2

If we know one point (x1,y1) and the second point (x2,y2), we can determine the slope of the line, which is the tangent of the angle α between the x2-x1 and the hypotenuse of the triangle formed by the segments x2-x1 and y2-y1.

image010 (4)

If we take an arbitrary point on this line between (x1,y1) and (x2,y2) and call it (x,y), the slope of the line has to be the same between the segment to the left and the one to the right of (x,y) point (see Figure 3).

image0111 (5)

linear_function_2

Figure 3

Solving for y in equation (5), the result the well known linear function y(x), that we know it goes through the first point (x1,y1) and the second point (x2,y2).

image013 (6)

Having said that, now we can compare the differential amplifier transfer function (3) with the linear function (6).  When these two functions are identical, Vo(V1) is y(x) and V1 is the variable x.  These are two linear functions that can be identical only if they have the coefficients identical, hence the name of the method.

image014 (7)

As y(x) is determined by its two points in plan, so is Vo(V1).  The given data points (Vin1, Vo1) and (Vin2, Vo2) determine the transfer function Vo(V1).  Therefore, (7) can be rewritten as the following system of equations.

image015 (8)

After replacing the known values Vin1, Vin2, Vout1 and Vout2 and after calculations, the system becomes

image016 (9)

which is exactly the result we had in part one of this series.

The system of equations (9) can be solved in the same manner as in the first article.   In brief, we choose the voltage reference V2, based on the available voltage references we have in the system, then we calculate the ratio .  Knowing this ratio we can calculate.  Then, knowing the resistor ratios, by choosing a pair of resistors say, R1 and R3 we can calculate R2 and R4.

Therefore, if we choose V2 = 2.5V, R3 = 10 kOhm, and R1 = 10 kOhm, the result is R4 = 3.795 kOhm, or a standard value of 3.83 kOhm, with 1% tolerance.  Also, R2 = 7.754 kOhm, or a standard value of 7.68 kOhm, with 1% tolerance.

Injecting AC into the DC Power Supply Rail

In allaboutcircuits.com forum, a question was posted: How can I combine an AC source of known frequency and amplitude with a DC power supply?

I thought this is an interesting problem, so here is the solution.

Subsequent messages clarified the requirements: 5V power supply, and the load needs 0.5A, which would make the load 10 ohms. The signal that rides on top of the DC voltage has to be 20 mVpkpk.

First, we need an inductor in series with the power supply to block the AC component. In this schematic, V1 is the 10 MHz generator and V2 is the 5V power supply.

image090426_1

The inductor’s impedance should be at least 10 times higher than the load, so that most of the generator energy goes into the load. If we note XL the inductor impedance, then XL = 10·RL,  so XL = 100 ohms, and L1 can be calculated

img2375

The result is L1 = 1.59 uH, with a standard value of 1.8 uH.

Then, we need a capacitor in series with the generator, to block the DC component from coming into the generator output. The generator, through this capacitor, “sees” the load in parallel with the inductor, considering that the power supply output impedance is low, close to zero.  Because of that, the capacitor impedance, together with the parallel combination load and inductor impedance form an attenuator. The signal amplitude on the load, can be written as follows:

img23821

where VL is the AC signal amplitude on load, VG is the generator amplitude, RLparXL is the load and inductor in parallel

img2377

and XC is the capacitor impedance. Using the above relation the capacitor impedance becomes

img2384

Taking into consideration that VL = 20 mVpkpk, L = 1.8 uH and choosing the generator amplitude VG = 100 mVpkpk, C1 becomes

img2394

with the result C1 = 433.07 pF. A standard value for C1 is 430 pF.

The generator amplitude can be slightly adjusted around 100 mVpkpk, so that the signal on load is exactly 20 mVpkpk.

And what is the current the generator sources on load? This can be easily calculated noting that the generator “sees” a capacitor in series with RL in parallel with L1. Since XC + RLparXL = 45.94 ohms, the generator output current is 2.18 mApkpk.

A 0.5 A, 1.8 uH inductor can be easily found. The generator sends into the power supply approximately 0.2 mApkpk signal. If this is not desirable, the inductor value can be increased with the disadvantage that its size increases.

Note that this method considers the load as being pure resistive. If the load has reactive components, RL has to be replaced by the combined impedance values of the resistive and reactive components.

Solving the Differential Amplifier – Part 1

Design a Differential Amplifier Based on the Input and Output Voltage Level Requirements

The differential amplifier, also known as the difference amplifier, is a universal linear processing circuit in the analog domain.  Why?  Because you can achieve any linear transfer function with it.  It can be reduced to a simple inverter, a voltage follower or a gain circuit.  It can also be transformed in a summing amplifier.

differential_amplifier_1

Figure 1

So, what is the common usage of the differential amplifier in Figure 1?  When the resistor ratios are equal

image001

the amplifier transfer function is

image002

and the circuit amplifies the difference between the input signals.

However, there are times when the electronics designer is faced with the following design requirements:  Given an input range of, say, -0.5V to 5.5V, the output has to swing between, say -1.25V and +2.365V.  It is clear that this requires an amplifier with a certain gain and an offset different than zero.  How can we design the differential amplifier to achieve such a function?

Read more…

Why this website?

After many years of engineering in electronics design, I decided it is time to start a website with my views about this art.  Analog and mixed-signal design is not black magic or cryptology.  This is art, real and powerful and relevant.

The purpose of this website is to provide elegant solutions to analog and mixed-signal design problems that have a strong basis in math.  This website is not about Electronics in general.  This website is about the art form of designing electronic circuits.   My goal is to show hobbyists, engineering students and engineers that analog design is not complicated, so long as one understands the fundamentals and the reasoning behind the many aspects of the design.  Hopefully, if they see what a wonderful art electronics design is, more will be encouraged to pursue it.

As in Mathematics, where some solutions to problems are regarded as elegant, in Electronics one can enjoy the same treat, that fuzzy warm feeling of achievement, after a circuit is designed and works as expected.

Math is a central tool, for this website, which I use to prove a circuit or path of reasoning.  I disagree with articles, commentaries and books that take pride in the fact that they show a concept with no math involved.  In those cases students are invited to learn formulas by heart without any understanding of underlying physics, or where the formulas came from.  This deprives students of a powerful reasoning tool.

In the last decades, computer aided circuit design tools created a path of ease in electronics.  More and more people today apply a SPICE based program and then, when the prototype is built, wonder why it does not work.  In my view, SPICE should be used as a proofing tool after an engineer designed every aspect of his or her circuit.

To prove this alternate method of design, I will use SPICE and math at the same time.  I hope that I will start a conversation about electronics design where people can exchange ideas or get some help.

They say that analog and mixed-signal design engineers are in demand these days.  This is due to students’ reluctance to tackle analog design because it is too complicated, or because it involves too much Math and Physics.  I would like to change that.  Analog design is not complicated.  Analog design as well as mixed-signal design relies on Physics, from the knowledge one gets in high-school about electricity, to Maxwell’s equations.

Why analog and not digital design?  Well, there’s nothing wrong with digital design.  After all, everything today is digital, isn’t it?  Not quite.  The world, as we know it, is analog and the interface between the digital processing circuits and the outside world is handled by analog processing circuits.  In a lot of applications, the fine specifications or the secret sauce are handled by the analog circuitry.  For example, it is very important on how one chooses an analog to digital converter (“ADC”), and how the signal is shaped before the ADC.  Therefore, for all these reasons, this website puts more emphasis on analog circuitry rather than digital design.

Another outcome I am hoping to achieve is that youngsters will see that math is not difficult or abstract or useless.  Math is a tool like any other for real life designs, whether they are in Electronics, Physics, or Mechanics.  During my years of teaching Electronics at a university and calculus to my daughter, I really enjoyed the “aha” look on my students’ face, when they really got it, when their circuit worked, or when they solved a math problem.  Students are always eager to learn.  They only need someone to show the path.

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