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In this module, we will talk about analog-to-digital and digital-to-analog

conversion, try to understand how these things are done in practice.

We have seen that sampling discretizes time and

quantization discretizes amplitude.

How is it done in practice?

The answer is cheaply and in a very compact way.

This photograph shows you the few discreet electronic

components that you have to put together to build an analog-to-digital converter.

And if you did so, you would probably end up one of the largest digital-to-analog

converters in existence, because in most cases like in your cellphone

the whole circuit would be integrated and be neutralized.

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At the heart of both digital-to-analogue and analogue-to-digital converters

is a circuit called the operational amplifier, or op-amp for short.

The op-amp has two inputs.

The inverting input and the non-inverting input and one output.

And the input, output, relationship is that the voltage at the output

is equal to the difference between the voltages at the inverting and

non-inverting input times the gain factor.

We idealize the op-amp by saying that it has Infinite input gain,

and zero input current.

And although these are, of course, idealizations, practical implementations

of correctional amplifiers are remarkably close to this ideal.

And not very complex either, because this, for instance, is the schematics for

an operational amplifier, and you can see that it requires only six transistors, and

one resistor.

The operational amplifier can be used in two configuration,

the open loop and closed loop.

In the open loop configuration, the op-amp works as a comparator.

You put a reference voltage at the inverting input and

a test voltage at the non-inverting input.

As soon as the difference between these two voltages is non-zero the output

due to the infinite gain will shoot up or down to the maximum or

minimum allowed output voltage which coincides with the voltage

of the power supply that powers the op-amp.

In the closed loop configuration we have a feedback path

that links the output to the inverting input.

So what happens when we apply a voltage to the non-inverting input?

Well, if the voltage at the inverting input is less than x,

then the difference between inverting and non-inverting would drive the output up.

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Conversely, if the voltage here was larger than x,

the difference would be negative and this would drive the output down.

So the only equilibrium point for

the system is when the voltage at the inverting input is exactly equal to x,

which of course means that the output voltage is equal to x as well.

So the input-output relationship of the closed loop is output equal to the input.

This configuration is known as a buffer stage,

because although the output has the same voltage as the input, the second property

of the ideal op-amp prevents any current from flowing into the amplifier.

And therefore there is a separation between this side and

that side of the amplifier.

This is very useful for

taking measurements without perturbing the measured quantity.

We can complexify the closed loop a little bit by adding a couple of resistors.

In this case we have what is called an inverting amplifier.

This is only slightly more complex than the simple closed loop we saw before.

Again, the system will stabilize when the difference between inverting and

non-inverting inputs is zero.

So we can say that voltage here, let's call it V0 is equal to 0

because the non-inverting input is connected to ground.

That means that the current flow-in in the first resistor will be according to Ohm's

law, i0 = x/R1.

This current will not be able to flow into the operational amplifier

because of the second law, so it'll have to flow into this resistor here.

And therefore the output voltage will be just

the voltage's drop over R2 caused by i0.

And so y = -R2/i0, which gives us the final

input-output relationship for the inverting amplifier.

The output is a fraction of the input with a change of sign.

We're now ready to look at an A/D converter.

The device starts with the sample and hold circuit that performs the sampling.

So we have our analogue input here.

Here we have an op-amp in buffer configuration and

here we have a mosfet which is really like a switch.

And here we have a train of pulses at the sampling frequency.

When a pulse reaches the gate at the mosfet,

the mosfet closes very briefly and allows the first buffer

to charge this capacitor to the instantaneous level of the input signal.

The capacitor is connected to another buffer stage that will put out the value

measured by the capacitor for the duration of the interval between pulses.

We put a buffer stage here so that we can put out a constant voltage

without discharging the capacitor between sample and instance.

An analog-to-digital converter needs not only to sample the input signal, but

also to quantize it.

So the sample on hold provides a stable voltage level between sampling times and

now we need a circuit that converts that to binary format.

Here we have a simple diagram of a 2-Bit quantizer.

What this quantizer does is take a maximum positive voltage V0 and

a minimum negative voltage -V0, this could be the A and

B extrema that we used in our quantization example.

And then it uses a series of four resistors of equal value

to produce intermediate boundary levels, the iKs

in our quantization example, that will be used to define the quantization regions.

Here you have +0.5 volts, you have 0 volts, you have -0.5 volts.

So we're dividing the interval from V0 to -V0 into four equally spaced intervals.

These reference voltages here are used with the bank of competitors,

so operational amplifiers in open loop to determine which

quantization interval the current sample value belongs to.

So let's work out an example with an input voltage of 0.2 V0.

So what we have here is that the first comparator will have

0.2 volts at the non-inverting input with the reference of 0.05 volts.

So the reference is higher, so here the out put will be a negative voltage

This competitor will compare 0.2 volts to 0 volts.

And so, this competitor will output a positive voltage, and similarly,

this one will compare 0.2 to -0.5, so again, the output will be positive.

Next we have a logic network that will encode

this comparison levels into a binary value.

These are exclusive or gates and the truth table for these gates is the following.

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So this gate will output a positive level and

this gate will output a negative level.

Now we have a network of dyads that acts as a parallel to serial converter.

So we have a negative voltage here that will be blocked by these two dyads, so

nothing happens here.

A positive voltage here that will go through, and

will indicate the most significant bit out of the quantizer.

Since this is positive, the most significant bit will be 1.

And here, the negative voltage will not go through the diode, so the least

significant bit will be connected to the ground and it will be equal to 0.

So the voltage of 0.2V0 will be encoded by the binary notation 1,0.

Finally, let's look at the D/A converter.

Suppose we have a sample in binary notations,

so x(n) = a set of bits from 0,

b1 to b R-1, so we use R bits per sample.

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We use this binary representation to drive a bank of voltage

generators at voltage V0.

So in the previous example we had a 2-Bit quantizer that gave

us a representation for our current value of 1,0.

So in that case, if we were to apply that value to this bank of generators,

this voltage generator would be on and this voltage radio will be off.

This generator are connected to resistor structure that is

known under the name of R-2R ladder.

Now the details of this structure are a little bit tedious to work out.

But if you're patient and you apply Thevenin's Theorem

to this ladder you will see that the system seen by this

point in the circuit is equivalent to a resistor of value

R connected to voltage source of value V equal to the sum for

K that goes from 0 to R-1 of V0 divided by 2 to the K times bK.

So in other words, each bit in the binary representation of

the sample contributes a voltage source of value v0/2K,

where K is the position of the bit in the binary representation.

So if this is equivalent to a voltage

source of value K V0/2K b0,

plus a resistor of value R.

This is simply an inverting amplifier and the value out here will be -V0 times