Analog ICs

Prof. K. Radhakrishna Rao

Department of Electrical Engineering
Indian Institute of Technology Madras

Lecture - 03
Translinear Networks (Currentmode Circuits)

It is a kind of multiplier, a current multiplier. There are two cells that we talked about,
and one such cell is an important cell that is depicted here.

(Refer Slide Time: 01:40)

We can see how this forms the translinear loop minus T3, T4, T1, T2 form the translinear
loop - type A and type B cell, we had drawn last time. This is the type B cell, where now
we are trying to input signals and see what happens to this cell in terms of the signals that
are inputted. The common base point is connected to an arbitrary base potential. This Vb
is an arbitrary base potential with this put to ground. Now let us see how this circuit
behaves. This is called Gilberts Gain Cell. This concept is important in VLSI design,
analog VLSI design, particularly.
(Refer Slide Time: 05:52)

The cell should be always something that can be used to build as a superstructure. It
should be such that it should be identical in its nature when we build a superstructure
with this kind of cell. Let us see whether that characteristic is present in this particular
cell. Let us now see the relationship. Let us see how the current relationship is there. Ia1 is
the current flowing in this, into, let us call the current flowing in this as I1, the current
flowing through this as I2. So, Ia1 into Ia1 into 1 I1 is equal to, again Ia2 is flowing through
this into I2.

Based on translinear principle current flowing through the counter clockwise diode will
be product of those will be equal to the current flowing through the diodes which are
clockwise. So Ia1 into I1 is equal to Ia2 into I. This is the basic current multiplication
relationship. In this context, let us see how this can be used for current amplifier. So, Ia1
by Ia2, therefore, is equal to I2 by I1. The ratio relationship is clear from this. Ia1 by Ia2, we
will call this as input current. The ratio of the input current, Ia1 by Ia2 is equal to I2 by I1 I2
by I1. I2 and I1 will form the part of output current.

So, the ratio of input current, is it exactly equal to ratio of the output current? This
relationship is valid, as I told you, over a large magnitude of variation of current. It is not
a small signal relationship. So, from maybe microamperes to hundreds of mA, this
relationship is valid. Now, suppose we write it as Ia1 minus Ia2 by Ia1 plus Ia2, what is it
going to be equal to? It is I2 minus I1 by I2 plus I1, the ratio relationship. And we have I2
plus I1 always equal to a constant current minus dc current minus I0. So, if we say that Ia1
minus Ia2 is the differential input current, Ia1 and Ia2 are two input currents.

Whenever we have two input currents, like in the case of two input voltages, we can have
a differential input voltage. Then, we have two input currents. We can talk of differential
input current and a common mode input current, any current. Therefore, Ia1 can be split
up as a current common to both. What is that? Ia1 plus Ia2 by 2 plus Ia1 minus Ia2 by 2.
This is the differential part. So, any current, Ia1, can be split up as a common current,
current common to both, plus half of differential path. Similarly, Ia2 can be written as
same common current Ia1 plus Ia2 by 2, minus Ia1 minus Ia2 by 2. This aspect of common
mode, and differential mode current, or common mode and differential mode voltage, has
already been exposed to as far as you are concerned. In differential amplifier, same thing
can be thought of in terms of current. So, what does it mean?
I have a common current here which can be made a DC, and differential current can only
be kept as signal current.

Suppose this common current, Ia1 plus Ia2 by 2 is a DC, let us call that as, let us say I0
prime, I0, say, I0, because I0 prime we have called this current, so, we will call this as I0, a
common current, the common current I1 plus I2 by 2. And I1 plus I2, we are calling it as I0
prime. So, if I say Ia1 plus Ia2 is equal to, let us call it as I0 common current, then we will,
for that, in order to keep it in line with this, it is I1 plus I2 is equal to I0 prime.

Similarly, Ia1 plus I Ia2, we will call it as I0, thereby, this is I0 by 2, we will call it. But
common current is a single value, whether you call it I0 or I0 by 2, it does not really
matter. It is a single current. So it is I0 by 2 plus the differential current I0 by 2 minus the
differential current. What happens to this relationship now? Ia1 minus Ia2 that is, the
differential current by Ia1 plus Ia2 is nothing but I0. This is equal to I2 minus I1 by I0 prime.

This is an important relationship. If you notice this relationship, we will see something
very nice emerging out of it; I2 minus I by I0 prime. Therefore, I2 minus I1 equals, what is
it? It is I0 prime by I0 which is nothing but the ratio of two currents minus the ratio of two
DC currents. What happens? It is a constant; it can be controlled by us. Ratio of two
currents into Ia1 minus Ia2 in differential input current. So, this is an important relationship
in what is called as current mode amplifier.

(Refer Slide Time: 10:27)

So far you have been exposed to voltage mode amplifiers. Why cannot we think
amplifiers as current mode amplifiers? That is, that seems to more basic, and that kind of
amplification can be done easily compared to voltage mode amplification. Look at the
way we have obtained it: differential input current times a gain factor is equal to output
current.

Now, this gain factor is not something very simple. In fact, this gain factor is absolutely
constant, because we have already known that we can obtain a current ratio by area ratio.
I can obtain a current mirror; I can generate a current reference, and I can obtain any
current ratio by area ratio. So, I can see to it that this ratio of current is going to be same
as the ratio of areas. It means it is just geometry-dependent. That is absolutely constant,
independent of temperature, supply voltage, everything. This is what you have been
trying best to achieve in the case of voltage amplifiers which you have failed to achieve
in a simple manner; which you could achieve only by using feedback, etc.

Here there is no feedback at all adopted. Without using negative feedback, I have been
able to demonstrate to you that we can get an amplifier minus current amplifier minus
whose amplification factor is absolutely constant. This Gilberts Gain Cell, even though it
was proposed way back in 1970s, its prominence was never understood at that time. But
in a VLSI context, analog VLSI context, you see that this gain cell is the most important
concept ever to be given to analog circuitry. Why?
First of all, its dynamic range is fantastic. There is no other amplifier whose dynamic
range can be as high as this. Then, it is independent of device property like beta, etc.
device parameters, and that you are achieving without negative feedback. If you do not
do it without forgetting to feedback, what will happen?
If you do not adopt negative feedback, then all the problems associated with negative
feedback, like need for frequency, compensation, all these things, do not exist. Therefore,
this concept can be adopted for designing what are called as wide-band amplifiers, or
video amplifiers. This basic concept is the concept that can be straightaway adopted for
video amplifiers, or wide-band amplifiers, because, there is no need for negative
feedback, and if there is no need for negative feedback, there is no need for negative
feedback because of the fact that amplifier gain is independent of device parameters.

And if you do not have negative feedback, then the consequent problem of frequency
compensation being needed, and oscillation to be prevented, all these things are avoided.
Now, apart from that, look at this structure. This is the input current. What will be the
output current? Differential output current.
(Refer Slide Time: 15:04)

Ib1 minus Ib2 is the differential output current, Ib1 is equal to Ia1 plus I2; Ib1 Ib2 is equal to
Ia2 plus I1. What does it mean? This is equal to Ia1 minus Ia2 into 1 plus I0 prime. This is
like your non-inverting amplifier. So, differential output current is equal to 1 plus I0
prime by I0 into Ia1 minus Ia2, and this can now become Ia1 and Ia2 for the next stage
which is exactly identical to this, only that we should be operating at I0 double prime
which is higher than I0 prime in order to get a gain factor.

Now this can become the input current to another identical stage which will come here
and the output of that will be again this as the input current into 1 plus I0 double prime by
I0 prime. So now you can put this cell over another cell and keep on building your
structure of current amplifier until you reach the power level of current that you desire.
You can go on doing it. So this is truly a self-structure needed for VLSI design because
this structure looks exactly identical, and it can be put one over the other, and you can
build a superstructure. So you go on doing this, and then where do you stop?

Suppose you will reach the current level that you desire, then you cannot convert this into
a voltage simply by putting two resistors to the final supply voltage and take the
differential output. So, converting a differential output current into voltage is simply by
putting resistors. So, that is not a problem at all.

This is what is done here, and it is truly wide-band, primarily because you are at all times
dealing with only currents, and the voltages involved are only diode voltages, and
therefore, the capacitors to be charged will be those capacitors across the diodes, and
therefore, they are to be charged by a very small magnitude of voltage, and therefore, it is
going to be very fast. So, that is the current mode of operation; is always fast because in
the case of voltage mode of operation the capacitors have to be charged up to a certain
value of voltage. In this case, since we are operating in current mode and the voltage is
always equal to the diode voltage you do not have to spend time in charging the
capacitor. That is why, please remember the current integrated circuit design is primarily
with current mode of operation. That is quite suitable for all designs of circuits of analog
VLSI systems. That is why we are calling it Gilberts Gain Cell.

Like we do namaskaram to Vinayaga before proceeding with any work, when I started
discussing about amplifier, I had to bring about this important concept of current
amplifier by Gilbert, that is, Gilbert cell, as that pride of place. We have to preserve that
pride of place for the Gilbert cell as a current amplifier. Remember, this has not yet
become popular in design of ICs. It is going to become popular in a very short time. Now
we will go back to the conventional amplifier which is always treated as a voltage
amplifier. This we had discussed earlier as an inverter in our digital circuit.

We said the bipolar inverter, or mass inverter, is the basic stage which can also be used as
an amplifier. Let us see how this inverter is suitable as an amplifier. This is a basic
transistor amplifier which is called common emitter amplifier, all of us know about it.

(Refer Slide Time: 19:46)

Vi is the input signal and V0 is equal to Vcc minus Ic(RC); and that is equal to Vcc minus
alpha times IE, and IE is, in turn, equal to IE0 exponent Vi by VT into RC. So, Vcc minus
alpha IE0, RC into exponent Vi by VT. So, this we already know. This is an exponential,
therefore, relationship between output and input of a transistor stage, basic transistor
stage, if you plot this, how will we see this? This, we had already seen in our discussion
about digital inverter.
(Refer Slide Time: 23:49)

Now, let us see how it looks like when we discuss its characteristic in the case of
amplifier. When Ic is, when Vi is very small, this is equal to Vcc, and therefore, it will
start with Vcc, then as Vi keeps on increasing, current will keep on increasing, and
therefore, this is the kind of characteristic, this is the exponential characteristic: as current
keeps on increasing, the output voltage here keeps on decreasing. Finally, the transistor
will enter the saturation reached where it is absolutely useless. That is, the moment this
voltage reaches about this, Vi plus V gamma, or Vi, Vi plus 0, that is, VCT is equal to 0 is
the point where it is entering the saturation region.

So, this collector base potential should be reaching 0, or VCE should be reaching V
gamma. So that is when it is entering the saturation. When it is getting forward biased,
both the junctions will be getting forwards bias. So, the voltage will be of the order of 0.1
to 0.2V. So this is the state of affairs as far as this amplifier is concerned. Now, we are
interested in operating this in this region. So far, in the digital circuitry, we were
interested in operating it either here or here in the steady state. It was only transiting from
here to here when it was going from high to low, or low to high, whereas, in the case of
amplifier design we are going to operate it in this region. So, somewhere here, let us say,
where the gain is going to be, the gain is nothing but the slope, V0 by Vi. So, gain is
nothing but the slope, that is, nothing but V0 by Vi. Let us see this. As far as this is
concerned, we have this as Vcc minus (IE0).

Now, RC exponent Vi by VT, in order to make this get located somewhere here, so that it
can act as a good amplifier. I have to apply what is called a bias voltage, so that if I do in
terms of voltage, it will be Viq from bias and voltage, plus the signal voltage which is V
by VT. What happens here? This is Vcc minus alpha IE0, then exponent ViQ by VT into RC
exponent Vi by VT. So what is alpha IE0 exponent ViQ by VT? What is this value? IE0
exponent ViQ by VT is equal to IEQ. IE0 exponent ViQ by VT is equal to IEQ cohescent
operating current and alpha times IEQ is nothing but ICQ. So, this entire thing is nothing
but ICQ. This is, therefore, Vcc minus ICQ into exponent Vi by VT into RC.

(Refer Slide Time: 24:53)

So we see here that this is Vcc minus ICQ RC. What is exponent ? Vi by VT1 plus Vi by
VT1 plus X plus X power 2 by 2 factorial and so on. So, in an amplifier, which is
supposed to be linear the Vi when it varies the output should also vary according to Vi.
That means all the other terms should become very small. These are the nonlinear terms.
So what is the condition for this to be a linear amplifier? We can see that Vi by VT has
to be much less than 1. This is an important relationship. In the case of a bipolar
transistor that Vi by VT should be much less than 1 irrespective of the operating points.
That means if Vi by VT is very small, then the nonlinearity resulting in the exponent is
going to be minimal. It does not mean saturation, as well as cutoff nonlinearity will not
come into picture. Those things are to be separately taken into account.

If saturation and cutoff nonlinearity does not come into the picture, then the nonlinearity
resulting due to the exponential relationship is going to depend upon only Vi by VT
magnitude. Then Vi by VT should be made very small compared to work. That was the
assumption that, if Vi is a sine wave, then immediately we know that there is
predominant second harmonic distortion in this because of this square term. This also is
to be noted clearly.

The predominant second harmonic distortion is responsible for the first instance to cause
final distortion. Therefore what is this equal to now? Vcc minus ICQ. RC is nothing but
VCQ. Vcc minus ICQ RC is nothing but VCQ minus ICQ, RC into , Vi by VT that is the
wanted term that into 1 plus rest of the term is the nonlinear term which has to be made
very small, which is nothing but Vi by VT2. So we want this factor to be very small
compared to 1, strictly speaking, is Vi by 2VT which should be made very small
compared to 1.
So what is this, ICQ RC by VT? Nothing but the gain which we have been calling as VCQ
minus gm the transconductance of the transistor into RC into Vi.

(Refer Slide Time: 28:36)

So you can see that phase shift coming about, that when Vi increases the collector voltage
is decreasing. So you see the phase shift of 180 degree, and also the gain of gm RC
coming to picture. This is clear. So this amplifier is not good enough for us.

Look at the problem associated with this amplifier. First of all, I have to bias it at a
voltage which is ViQ, then superimpose over that the required signal set of a pair, that ViQ
should be of the order of 0.5 to 0.6, and that itself should not change.

Suppose I am able to get a voltage source of that type, I can bias it, but, even if I bias it,
you know that this voltage itself is going to keep changing with respect to temperature.
What is the coefficient of this voltage, VBE? VBE changes with respect to temperature in
a manner which is minus 2 mV by degree C rise in temperature. You know all this. A
diode which is forward biased has a temperature coefficient of minus 2 mV by degree C
rise in temperature.

That is, VBE by T, that is the temperature coefficient of a diode which is going to
result in minus even if the signal is not varying, if the temperature is varying, what will
happen? This collector voltage is going to drift. That is the problem of drift. So, because
VBE is changing with respect to temperature, the output voltage is changing even though
input has remained constant. Input has not changed. So, output changes with respect to
temperature, and consequently, with respect to time. So, it is drifting with respect to time
because of its dependence on temperature. This is the major disadvantage of this
amplifier. Particularly for low-frequency signal this is going to be nuisance.
For high-frequency signal at least I can decouple this and only extract the high-frequency
path and couple it on another stage. So, it is not going to be a problem. But for low-
frequency signal, this is going to be a problem. That is the reason that this stage is not
preferred in integrated circuits. In integrated circuits, you would like to exploit the
property of the device to compensate for this. Straightaway I would say that one
technique is to see to it that this is modified somehow, and in an amplifier, when I have
given an input signal, I would like to give it straight to the input of the input terminal of
the amplifier. I do not want to couple it again through capacitor, etc. That means
preferably I would like the amplifier input voltage to be 0 when the output voltage is also
0.

If the input voltage, when it is non-zero, output voltage is also finite for some reason,
because of operation, etc. There is one more thing called the Offset. The only way I can
make the input 0 when the output is 0 and to become the operating point is just to see that
the device works with both positive as well as negative supply. Otherwise, I cannot say
with the single supply you can see that output has to be always somewhere in between,
which means it has to have an offset etc. That means, when the input is 0, output is not
going to be 0. So, in a true amplifier, I would like the output to be 0 when the input is 0.
Or I should be capable of connecting the input directly to the ground when there is no
input. So what do I do? I use another supply. So the dual supply concept comes if you
want to get rid of coupling the whole circuit by means of a capacitor.

In capacitor of things which have to be avoided in integrated circuit we are not developed
even though they are available with very small values. They have to be cautious. They
take up a lot of area, and therefore, this concept of biasing it by means of another supply.
How do I bias it? No, I can put a negative supply, but I just told you that it is easier to
supply a current into it and force the voltage. So I would rather apply a current here.

So using a current source what happens? Automatically, I can make sure that this can be
grounded. Now, when the signal is there not there, this can be grounded simply and
automatically. It gets biased by the required voltage in order to carry this current, IE. In
order minus this is the emitter current minus in order to, sort of, say that this current is
something that we are obtaining from another current source, we will call this as I0. Now,
I have biased it by means of a current, whereas biasing it by means of a voltage is a very
troublesome effect.

First of all, I have to get the exact voltage, which is very difficult, and even if I get the
voltage, it is going to drift because of temperature variation, whereas here, the current
emitter current is fixed, and even if the transistor based emitter voltage drift nothing
happens, as far as output current is concerned, cohescent current is concerned. So, biasing
by a transistor, by means of a current, is preferred to biasing the transistor by means of a
voltage. So, I would rather bias it by means of a current in this following manner. Next,
you will see that obviously the emitter has to be grounded, otherwise, it cannot be
common emitter amplifier.

How do I ground this?

For AC it should be a ground, or for signal it should be a ground. So normally, in a
discrete circuit, I would have put a capacitor, which is called a bypass capacitor. But, this
capacitor, for low-frequency operation, has to be very high in order to act as a short
circuit at that frequency. But in an integrated circuit, what should I do? So, in an
integrated circuit I should not use a capacitor, but I should use dynamic impedance which
will act as a short circuit for signal but it will not disturb my biasing current. I still know
what devising current it is, as long as I know what the biasing current is, that is called
stable biasing. If I am not able to tell you what the biasing current is, that is called
unstable biasing.

Suppose I ask you: Please tell me the beta of the transistor, only then I will come up with
the biasing current. Then it is unstable biasing. If I say, I know what the biasing current is
going to be irrespective of the transistor parameter. Then it is called stable biasing. In this
case, therefore, I have to put a device here. What will be the device that I will put?
Automatically, I will put a forward biased diode here. Please understand this, because a
forward biased diode is a short circuit for signal. But what will we do for this current?
Because of the IC property, the I0 is going to be split equally because bases are
connected, emitters are connected. So, these are identical transistors, and therefore, the
currents have to be equal.

The principle is that we depend upon so much in integrated circuits, that when the
transistors are equal, and if the base emitters are at the same potential then the current
should be equal, this is what is used for biasing it in a stable fashion. That means, the
biasing is not going to be disturbed, it is going to be I0 by 2, I am able to tell without any
hesitation. That means, it is a stable biasing. And apart from that, what have I achieved?
As far as the signal is concerned it is going to act as a near-short, not exactly a short.

What is the impedance from that emitter to ground? Incremental diode impedance which
is small RE itself and therefore this is going to still act as an amplifier which is very
similar to common emitter amplifier, with the gain now being reduced to half, because
instead of being a short circuit, it is also RE.

So this RE of a common emitter amplifier, and RE of this diode, will share the input
voltage. Earlier, input voltage was totally coming across based emitter junction of the
amplifier. Now, based emitter junction of the amplifier gets only half the input voltage,
because RE of this amplifier, as well as the diode, will be sharing the input voltage. When
the input voltage is fed here, Vi by 2 you can see here that the operating current is going
to be I0 by 2, I0 by 2. Then I say, in order to have the early effect also the same, if the
early effect what is the early effect? Collector base voltage is going to still determine
the collector current.

Therefore, if collector base voltage also happens to be the same, then I make sure that the
current deviation is equal, and therefore, what do I do?
I also connect this so that it also has the same collector, because otherwise I will have to
leave the collector hanging; instead, I will as well connect the collector resistance through
the same value with the resistance RC to the supply. What is this? This is nothing but the
difference. This is one way of presenting differential. It is not how this has developed.
This is one way of presenting the differential amplifier as something that has originated
out of necessity of getting rid of the disadvantages of discrete common emitter amplifier
when it is being adopted for IC version.

When you are adopting a common emitter amplifier for IC version we can say that we
want to get rid of the bypass capacitor. We do not want coupling capacitor, and this is
one way you can get rid of coupling capacitor, as well as bypass capacitor, in any discrete
circuit. This is one important powerful technique converting every discrete circuit into
integrated version.

Any discrete version of a circuit that you have can be easily converted getting rid of all
the bypass capacitors simply by making it differential. This is a powerful technique. So
you have to understand this technique, therefore, I brought it out. Now, once you say that
this is going to be differential, what happens here? This current is going to remain
constant I0.

If signal voltage gets dropped here as Vi by 2, and this will be minus Vi by 2

superimposed over the Iq, this will be plus minus, minus Vi by 2, and therefore, it will
develop a voltage which will be in phase opposition to this, but of the same magnitude,
when therefore, if this is minus gmRC, this will be plus gmRC. The gm in this case is 1 by
2RE and not 1 by RE because the input voltage is divided between two diodes. So, Vi by
2RE into RC is the single stage gain with the minus for the first stage, and for the other
stage it is plus RC by 2RE. Therefore if you take the differential output here as the output,
what will we get? One is going to be minus RC by 2RE, another is plus RC by 2RE.

(Refer Slide Time: 42:37)

If we take the differential output, it is RC by RE, or gm into RC. That means, you are
getting back the common emitter gain which you had thought you have lost because of
converting this stage into a differential stage. So you have not lost anything at all by
converting this single stage into a differential stage. It retains all the small signal
properties associated with the common emitter amplifier. The gain remains the same, but
it has advantages now. The output voltage can be made equal to 0 when input voltage is
0. That is because of the fact that we are using the dual supply. That is what brings in
dual supply in all analog circuits, and then we have got rid of the bypass capacitor and the
coupling capacitor.

We will discuss further about this important building block. This is the most important
building block again in analog integrated circuits next to current mirrors, next to
translinear principle. It is part of the translinear principle.