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Camenzind: Designing Analog Chips

Chapter 5: Current Sources

 

 

Vcc

Q1

Q2

 

 

 

I1

Iaux

 

Q5

 

 

10u

 

 

Q3

Q4

SUB

R1

R2

 

1.5k

1.5k

Fig. 5-14: Erdi current source.

used to make a current source, let me point out that the accuracy of this current source is of no great importance. A bulk (epi) pinch resistor will do or any of the lesser current sources discussed above.

Iaux is mirrored and split into two equal parts by Q1 and Q2; thus the operating currents for Q3 and Q4 are equal. Q4, however has 3 emitters, Q3 only one, thus there is a difference of about 29mV (at room temperature). Unbalanced, the collector voltage of Q3 rises until Q5 supplies enough current to make up the difference. This

current amounts to (deltaVBE)/R2.

Vcc can be as low as 1 Volt (or as high as breakdown voltages allow). Moving Vcc 20% changes I1 by 0.08%. The output impedance is 50MOhm. Temperature is strongly positive, a +25% change from 0 to 100oC. And the Monte Carlo variation is roughly that of R2, here assumed to be ± 25%.

 

18

 

 

 

 

 

 

 

 

 

 

 

16

 

 

 

 

 

 

 

 

 

 

 

14

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

vs. Supply Voltage

 

 

 

 

12

vs. Output Voltage

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

µA

10

 

 

 

 

 

 

 

 

 

 

I1 /

 

 

 

 

 

 

 

 

 

 

 

 

8

 

 

 

 

 

 

 

 

 

 

 

6

 

 

 

 

 

 

 

 

 

 

 

4

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

00

0.2

0.4

0 . 6

0.8

1

1.2

1.4

1.6

1 . 8

2

 

Voltage/V

 

 

 

 

 

 

 

200mV/div

Fig. 5-15: Performance of the Erdi current source.

CMOS Current Sources

None of the bipolar schemes work well for CMOS devices. They are based on VBE or delta-VBE , for which there are no equivalents. Trying to use circuits such as Figures 5-10 or 5-14 with CMOS width-ratios

 

 

 

 

 

 

 

 

 

 

 

 

 

I1

leads to inferior circuits. Also, due to the square-law

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

behavior of the gate voltage, the variations are roughly

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

M1

 

double those of bipolar designs.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Fortunately the CMOS transistor is a current

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Vref

 

 

 

 

 

 

 

 

 

 

 

 

source. If we simply apply a constant voltage to the

 

 

 

 

 

 

 

W=4.5u

 

1.2

 

 

 

 

 

 

L=20u

 

gate (such as a reference voltage) we can tailor the

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

width and length of the device to give us a certain

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

current.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Fig. 5-16: MOS

Using a rather exaggerated length, we can

transistor as a

minimize the channel-shortening effect. In the example

current source.

 

Preliminary Edition September 2004

5-7

All rights reserved

R1
5 k
Fig. 5-18: Accurate current source with an op-amp and external resistor.
-
W=5u
L=0.18u
100mV
M1
+
I1

Camenzind: Designing Analog Chips

Chapter 5: Current Sources

here the output current varies little with the applied voltage, amounting to an impedance of 38MOhm. But the variation (due to the uncertainty in the threshold voltage) is large: ± 39%. Add to this the change with temperature (0 to 100oC, -23%) and a variation in the reference voltage (± 3% causes a change of 10% in I1).

8

 

 

 

 

 

 

 

7

 

 

 

 

 

 

 

6

 

 

 

 

 

 

 

5

 

 

 

 

 

 

 

/µA

 

 

 

 

 

 

 

4

 

 

 

 

 

 

 

I1

 

 

 

 

 

 

 

3

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

00

0.5

1

1.5

2

2.5

3

3.5

Output Voltage/V

 

 

 

500mV/div

Fig. 5-17: I1 vs. voltage for fig. 5-16.

The Ideal Current Source

Sometimes a compromise is the best solution. If we allow just one component to be external to the IC and provide a pin for it, the performance of a current source improves dramatically. All other currents

within the IC can then be derived from it with current mirrors and are thus inherently

accurate.

In the last circuit an op-amp compares

the voltage across an external resistor with a

low internal reference (divided down, for

example, from a bandgap reference) and

drives the gate (or base) of an output device. Assuming no trimming, a 1% tolerance for the external resistor, 3% for the reference voltage and 2mV offset uncertainty for the op-amp, I1 will be within 6% at any voltage and any temperature.

Preliminary Edition September 2004

5-8

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Camenzind: Designing Analog Chips

Chapter 6: Analog Measures

6 Time Out: Analog Measures

dB

Analog scales tend to be very large. As an example, the hearing threshold of a young adult is 20 micro-Pascal; the maximum level without damaging the ear can be more than 20 Pascal, a ratio of one to 1 million. Particularly because of the widely varying sound levels a need for a logarithmic measure appeared early on in electronics. There are two of them: The Neper is based on the natural logarithm and named after John Napier, a 16th century Scottish mathematician who came up with the logarithmic table (and whose name was most likely spelled Neper in his time).

In the 1920's a measure based on logarithm with base 10 began to be used at Bell Laboratories. At first it was called the "transmission unit", then re-christened the Bel, after Alexander Graham Bell. The idea is simple, a Bel is the logarithm of two power levels:

Bel = log P1

P2

But the Bel turned out to be a bit course; one-tenth of that suited the Bell Labs people better, hence the decibel, or dB:

dB = 10 log P1

P2

This is the ratio of two power levels. Since power is related to the square-root of voltage (or current), we get:

dB = 20 log V1

V 2

Neper has more or less disappeared as a measure, dB proved to be more convenient. But, when using dB, always keep in mind there is a 2:1

Preliminary Edition October 2004

6-1

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Camenzind: Designing Analog Chips

Chapter 6: Analog Measures

difference between the ratios of power and those of voltages or currents (or pressure).

Even though a logarithmic ratio is very convenient, it is helpful to picture the (voltage) actual ratios:

-60dB

1/1000

-40dB

1/100

-20dB

1/10

-6dB

0.5 (exactly: 0.5012)

-3dB

0.707

0dB

1

20dB

10

40dB

100

60dB

1000

Fundamentally dB is relative, expressing only a ratio. But there are many modifications in which one of the two levels are absolute, among them:

dB(SPL)

Sound pressure level, where the hearing threshold (0dB)

 

is based on 20uPascal.

dBm

Power ratio where 0dB = 1mW. Originally this was based

 

on an impedance of 600 Ohms (that of a telephone line),

 

but now is used with any impedance (which is fine as long

 

as we calculate power ratios, not voltage ratios).

dBuV

Voltage ratio relative to 1uV.

RMS

RMS calculation was introduced by Charles Steinmetz more than 100 years ago. Steinmetz grew up in Breslau, Germany (now Poland), but shortly before he got his Ph.D. in mathematics and physics he had to flee to Switzerland because of his socialist activities. From there he emigrated to the U.S. and found a job as an assistant draftsman in Yonkers. The company fabricated hat-making machinery, but soon expanded into electrical motors. The year was 1889.

Steinmetz was a small man with a hunchback and one leg shorter than the other, a deformity he inherited from his father. Though he made the proverbial bad first impression, the people around him soon were in awe of his razor-sharp mind. No surprise then that the overqualified draftsman was

Preliminary Edition October 2004

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Camenzind: Designing Analog Chips

Chapter 6: Analog Measures

at the forefront in AC engineering within four years. Through mergers and acquisitions he found himself to be working for General Electric as head of the calculating department in Schenectady and teaching at Union College.

He led a bohemian life; afraid to marry because of his inherited deformity, he shared his house with an entire family and kept several crows, raccoons, eagles, owls squirrels dogs and alligators. He resumed his socialist activities, expounding his ideas in a book; he was against competition and advocated an industrial reorganization by the government. It is remarkable that he got along very well with his bosses at GE. In all respects he was a delightful man who and seemed to have a very happy life. Almost singlehandedly he moved electrical engineering from a craft to a profession.

Steinmetz found that few "electricians" used mathematics, his specialty. The first curriculum in electrical engineering had started at MIT in 1882 and very few people understood AC. George Prescott lamented in 1888: "It is a well-known fact that alternating currents do not follow Ohm's law, and nobody knows what law they follow."

For example, there were two things wrong with the "average" value meters of the time displayed: in the first place the true average of an AC waveform should have been zero; in the second place the product of the average current and voltage gave the wrong answer for power. Also, a phase shift between voltage and current left almost everyone perplexed.

In 1893 Steinmetz presented his first paper on the use of complex numbers in electrical engineering. It was heavy going, delivered in a thick German accent. But he kept at it in paper after paper and then a massive, three-volume text book. By 1901 he had it down pat and published a textbook that was finally easy to understand.

What he said was this: In order to calculate the power correctly, you need to square the voltage (or current), calculate its mean (average) and apply the square root. Hence RMS - root-mean-square. Or a bit more detailed: You divide the waveform into equal segments over one period, square each segment, add up the squared values, calculate the average and take the square root of that.

The power, by the way, is the average power. There is no such thing as RMS power, only RMS voltage and RMS current.

For a pure sine-wave this works out as:

Vrms = Vpeak = 0.707 Vpeak

2

Preliminary Edition October 2004

6-3

All rights reserved

Fig. 6-1: Arbitrary waveform for RMS calculation.

Camenzind: Designing Analog Chips

Here is a simple illustration of RMS calculation: Four time segments of 100usec each. First a voltage is at 5 Volts, then a zero, then at 2 Volts and finally at zero again.

Vrms =

52

+ 0

2 + 22 + 0

2

= 2.69V

 

 

4

 

 

 

 

 

 

Chapter 6: Analog Measures

 

5

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

V

3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Voltage/

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

50

100

150

200

250

300

350

 

 

 

Time/µSecs

 

 

 

 

 

 

 

 

 

 

50µSecs/div

But the RMS calculation has some limitations: it doesn't work with non-linear elements. In Steinmetz's

time there were no transistors, not even vacuum tubes. There were only linear elements (save perhaps for the occasionally saturating transformer), so he didn't consider what would happen if the impedance changes while you are measuring RMS voltage and current.

Take the case of a transistor stage, either linear or switching. You want to determine its power dissipation, so you measure the current through it and the voltage across it. But the impedance of the transistor constantly changes and Ohm's law doesn't hold. The product of RMS voltage and RMS current gives an absurdly wrong result for power. The only way you can determine the power is to integrate the instantaneous values of voltage times current. In a simulation, Spice does this very well.

But measurements aren't nearly so easy: "True RMS" instruments do indeed have a circuit element which measures RMS rather than average. But the inputs to almost all of them are capacitively coupled. If the waveform you are measuring has a DC component, it is ignored and the result reflect on the AC portion of it.

Noise

Imagine a current flowing through a wire connected between a negative terminal on the left and a positive terminal on the right. Through the wire are flowing millions of electrons from left to right.

Each electron carries a charge of 1.6e-19 Coulombs. Let's say we observe a current of 1uA, thus 7.8e12 electrons pass every second. If the interval between electron were the same, we would then see a ripple at 7800Ghz, like the teeth of a saw-blade moving at high speed.

Preliminary Edition October 2004

6-4

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Camenzind: Designing Analog Chips

Chapter 6: Analog Measures

But in a diode or a bipolar transistor, we see quite a different effect. Here the current is initiated by electrons and holes moving across a barrier and this movement is anything but smooth. Within any given time interval one electron or hole may cross the barrier, or 5, or 100, or none. The variation is so great that there is no discernable peak at 7800Ghz, or any other frequency. In fact, because of the very large number of electrons, this white noise or shot noise is so well distributed over the frequency range that it has a constant level over the entire spectrum:

 

Inoise(rms) = 2 × q × I × B

where

q = electron charge (1.6e-19 Coulombs)

 

I = dc current

 

B = bandwidth in Hz

There are two things you should notice here. First: Noise increases as the square-root of bandwidth. Second: When the current is decreased, noise becomes a larger fraction of it.

Let's illustrate the second part. With a bandwidth of 10kHz, 1mA dc produces 1.8nA(rms) of noise. That amounts to 0.00017% or -115dB. With 1uA of current and the same bandwidth the noise is 56pA(rms), i.e. 0.0056% or -85dB. At 1nA we get 1.8pA of noise, which amounts to 0.18% or -54dB. All of which shows that it is harder to design a low-noise circuit at low current levels.

There is also noise when no current flows at all. By the energy imparted by temperature, some electrons will suddenly leave an orbit and jump to another. The higher the temperature, the larger this irregularity becomes. Thus a resistor, doing nothing but lying on a bench actually has a noise voltage at its terminals:

 

Vnoise(rms ) = 4 × k ×T × R × B

where

k = Boltzmann constant (1.38e-23 Joules/T)

Thus a 1MOhm resistor always has a noise voltage of 13uVrms at room temperature, if measured over a bandwidth of 10kHz. This is called the Johnson Noise.

Whenever you mention a noise voltage or current, you also have to state the bandwidth. To avoid this, noise voltage is often expressed in nV/rtHz (nanovolts per root-Hertz). To get the real noise voltage you simply multiply this value with the square-root of the bandwidth.

Preliminary Edition October 2004

6-5

All rights reserved

Preliminary Edition October 2004
All rights reserved
FigTime/µSecs. 6-2:
Distorted sine-wave20µSecs/div.
0 20 40 60 80 100 120 140 160 180
-400
-200
0
D1-anode / mV
other, higher frequencies are created
which can be extracted in a Fourier series. There is an algorithm called "Fast
Fourier Transform", or FFT, which does
this. The name is somewhat meaningless,
there is no contrasting "slow" Fourier transform.
In our example here the positive half of a 5kHz sine-wave has been compressed. Converted into a frequency spectrum with the help of a Fourier transform we see the fundamental frequency of 5kHz and a series of
These two noise sources are fundamental, present in any current or resistor. But there is another one, not fundamental exactly, but always present. It is called 1/f noise or flicker noise.
Flicker noise is worst in an MOS transistor, which is a major reason why bipolar transistors are preferred in analog design. The silicon-oxide interface is capable of holding some electrons for a considerable period (seconds) and then releasing them in bunches. This increases noise at low frequencies far above the white-noise level; at 1Hz the noise level (in nV/rtHz) can be two orders of magnitude higher than at 1MHz.
Flicker noise is also present in bipolar devices, but to a lesser extent.
Fourier Analysis, Distortion
Jean Fourier was a mathematician who was active in the French revolution. He was arrested twice in the fight between the various factions but was spared the guillotine. In 1798 he joined Napoleon's army in the invasion of Egypt and then was appointed prefect in Grenoble; in 1809 Napoleon made him a baron.
In between his political and administrative duties he found time to not only publish a massive work on ancient Egypt but do mathematical research. He analyzed the flow of heat in mathematical terms, coming up with a novel expansion of functions as trigonometrical series. His memoir "On the Propagation of Heat in Solid Bodies" was read to the Paris Institute in 1807. His method, now called the Fourier series, was criticized by the leading French mathematicians and was not published until 1822 (and not translated into English until 54 years later). It turned out to have
applications in a wide range of areas, 400 including now electronics.
When a sine-wave is distorted
200
Camenzind: Designing Analog Chips
Chapter 6: Analog Measures

6-6

Camenzind: Designing Analog Chips

Chapter 6: Analog Measures

harmonics, multiples of the fundamental at 10kHz (second harmonic), 15kHz (third harmonic), 20kHz (fourth harmonic), 25kHz (fifth harmonic) etc., with gradually decreasing amplitudes. The square-root of the sum of the squares of all harmonics divided by the amplitude of the fundamental is the amount of distortion. You can usually disregard harmonics after the fourth or fifth, since their amplitudes become very small:

 

1

 

 

 

 

 

 

100m

 

 

 

 

 

 

10m

 

 

 

 

 

/ V

 

 

 

 

 

 

Amplitude

1m

 

 

 

 

 

 

 

 

 

 

 

 

100µ

 

 

 

 

 

 

10µ

 

 

 

 

 

 

1µ0

5

10

15

20

25

 

Frequency/kHertz

 

 

 

5kHertz/div

Fig. 6-3: Fast Fourier transform with low resolution.

Harmonics = 36mV 2 + 21mV 2 + 8.2mV 2 + 0.9mV 2 = 42.5mV

Fundamental = 550mV

Distortion = 42.4 = 0.077 = 7.7%

550

The peak at zero frequency shows the DC level, i.e. the asymmetry caused by the clipping.

Before you run a fast Fourier transform in Spice you need to choose two settings: how many samples should be taken over one period of the waveform and how many periods should be analyzed. In the example of Figure 6-3 there are in fact too few samples and periods, resulting in broad peaks.

As a general rule start with 25 samples and 50 periods. The first is determined by "maximum time -step" and "maximum print step" (set at 8usec in figure 6-4 for a 5kHz driving frequency) and the second by the total time in the transient analysis (1msec for 50 periods at 5kHz). You get the best results if both the number of samples per period and the number of periods are integers.

The fast Fourier transform has some flaws and limitations. For example, figure 6-4 shows peaks in between the harmonics, which, in

Preliminary Edition October 2004

6-7

All rights reserved

Camenzind: Designing Analog Chips

Chapter 6: Analog Measures

 

1

 

 

 

 

 

 

100m

 

 

 

 

 

 

10m

 

 

 

 

 

/ V

 

 

 

 

 

 

Amplitude

1m

 

 

 

 

 

 

 

 

 

 

 

 

100µ

 

 

 

 

 

 

10µ

 

 

 

 

 

 

1µ0

5

10

15

20

25

 

Frequency/kHertz

 

 

 

5kHertz/div

 

1

 

 

 

 

 

 

100m

 

 

 

 

 

 

10m

 

 

 

 

 

 

1m

 

 

 

 

 

/ V

100µ

 

 

 

 

 

Amplitude

 

 

 

 

 

10µ

 

 

 

 

 

 

 

 

 

 

 

 

1µ

 

 

 

 

 

 

100n

 

 

 

 

 

 

10n

 

 

 

 

 

 

1n0

5

10

15

20

25

 

Frequency/kHertz

 

 

 

5kHertz/div

Fig. 6-4: Fast Fourier transform showing

Fig. 6-5: Continuous Fourier transform

false peaks because of still insufficient

with high resolution.

resolution.

 

reality are not there. A superior method is the Continuous Fourier Transform, available in some analysis programs and shown in figure 6-5.

When you have a waveform which is symmetrical but not a sine-

 

1

 

 

 

 

 

 

 

 

 

 

 

 

0.8

 

 

 

 

 

 

 

 

 

 

 

0.6

 

 

 

 

 

 

 

 

 

 

 

0.4

 

 

 

 

 

 

 

 

 

 

/ V

0.2

 

 

 

 

 

 

 

 

 

/ V

Amplitude

-0.2

 

 

 

 

 

 

 

 

 

Amplitude

 

-0

 

 

 

 

 

 

 

 

 

 

 

 

-0.4

 

 

 

 

 

 

 

 

 

 

 

-0.6

 

 

 

 

 

 

 

 

 

 

 

-0.8

 

 

 

 

 

 

 

 

 

 

 

-1

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

 

 

Time/mSecs

 

 

 

 

 

200µSecs/div

1

 

 

 

 

 

 

 

 

 

 

400m

 

 

 

 

 

 

 

 

 

 

200m

 

 

 

 

 

 

 

 

 

 

100m

 

 

 

 

 

 

 

 

 

 

40m

 

 

 

 

 

 

 

 

 

 

20m

 

 

 

 

 

 

 

 

 

 

10m

 

 

 

 

 

 

 

 

 

 

4m

 

 

 

 

 

 

 

 

 

 

2m

 

 

 

 

 

 

 

 

 

 

1m

1

2

3

4

5

6

7

8

9

1 0

 

 

Frequency/kHertz

 

 

 

 

 

1kHertz/div

Fig. 6-6: Triangle wave.

Fig. 6-7: Fourier analysis of a triangle.

wave, you get only odd harmonics (i.e. the third, fifth, seventh etc.). Shown here is the example of a triangular wave. Total distortion (i.e. deviation from a sine-wave) is 12%.

When you have non-linearity in a circuit and two frequencies are present, you also get intermodulation distortion, i.e. not only harmonics are created, but differences as well.

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