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Lecture 01-02: Analog foundations

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PHYS 351 · Lecture 01-0201

Lectures 01–02

From Photodiodes to Mixers

Covers Labs 1 and 2
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PHYS 351 · Lecture 01-0202

Today, in one line

Sensor \rightarrow small current

Amplifier \rightarrow usable voltage

Multiply ++ filter \rightarrow new frequency

© Ran Yang, Ph.D.Advanced Instrumentation
PHYS 351 · Lecture 01-0203

Two ideas that run through the course

Feedback turns a vague device into a predictable one.

Multiplication moves information in frequency.

Labs 1 and 2 are entirely analog. From Lab 3 a computer takes over the decisions.

© Ran Yang, Ph.D.Advanced Instrumentation
PHYS 351 · Lecture 01-0204

Where each task points

TaskYou buildIdea
Lab 1, 1.1LED driver, 5.0 V max, current limitedseries resistor
Lab 1, 1.2the same driver with an op-ampfeedback current control
Lab 1, 2.1photodiode + resistor; current at 3.0-3.0 Vphotodiode model, sign
Lab 1, 2.2photodiode + op-amptransimpedance amplifier
Lab 1, 3analyse the given circuit: VaV_a, VbV_bbiased inverter + RC filter
Lab 2, 1AD633 product and squaring; quadratic fitmultiplier equation
Lab 2, 2offset adjustmentoffset, scale, nonlinearity
Lab 2, 3squarer in a loop \rightarrow square rootfeedback inverts a function
© Ran Yang, Ph.D.Advanced Instrumentation
PHYS 351 · Lecture 01-0205

Driving an LED

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PHYS 351 · Lecture 01-0206

LED driver: limit the current

An LED is not a safe load for a voltage source.

ILED=VsourceVFRI_{\text{LED}} = \frac{V_{\text{source}} - V_F}{R}

Design at the worst case: the top of the range.

© Ran Yang, Ph.D.Advanced Instrumentation
PHYS 351 · Lecture 01-0207

Op-amp driver: feedback holds the current

Sense the current across a resistor; the loop holds that voltage at a reference.

VFV_F drifts with temperature. A resistor and a reference do not.

Check three limits

output inside the rails \cdot LED current rating \cdot op-amp output current (often 20–25 mA)

© Ran Yang, Ph.D.Advanced Instrumentation
PHYS 351 · Lecture 01-0208

Detecting light

© Ran Yang, Ph.D.Advanced Instrumentation
PHYS 351 · Lecture 01-0209

Photodiode: the model to keep

Photodiode equivalent circuit: photocurrent source Iph is in parallel with diode resistance rd and capacitance Cd; series resistance rs connects that parallel network to the output terminal.

A light-dependent current source in parallel with a diode.

More light, more current. Linear over many decades.

A resistor turns the current into a voltage.

© Ran Yang, Ph.D.Advanced Instrumentation
PHYS 351 · Lecture 01-0210

Why reverse bias helps

Photodiode current-voltage curves shift with illumination. A second plot compares output current versus optical power for ideal linear behavior, reverse bias, and zero bias.
Light shifts the whole curve down. On the flat part IIphI \approx -I_{\text{ph}}, independent of the exact bias. The cost is dark current.

© Ran Yang, Ph.D.Advanced Instrumentation
PHYS 351 · Lecture 01-0211

Resistor detector: the DMM reads 3.0-3.0 V

I=VdetRdetI = \frac{V_{\text{det}}}{R_{\text{det}}}

Define the polarity and the current direction before the algebra.

A negative answer is a direction, not a mistake.

© Ran Yang, Ph.D.Advanced Instrumentation
PHYS 351 · Lecture 01-0212

Photodiode + op-amp: the TIA

Transimpedance amplifier block: input current Iin enters an amplifier with transimpedance Ztrans and produces output voltage Vout relative to ground.

Vout=IphotoRfV_{\text{out}} = -\,I_{\text{photo}}\,R_f

Inverting input held at virtual ground.

The photocurrent has nowhere to go but RfR_f.

The diode always sees the same bias.

© Ran Yang, Ph.D.Advanced Instrumentation
PHYS 351 · Lecture 01-0213

TIA: what limits it

Photoconductive transimpedance circuit: a reverse-biased photodiode feeds the op-amp inverting input. Feedback resistor Rf and parallel capacitor Cf connect output to that input; the noninverting input is grounded.

  • output swing inside the rails
  • op-amp output current
  • capacitance against RfR_f: ringing; a small CfC_f fixes it

Oscillating? This is why.

© Ran Yang, Ph.D.Advanced Instrumentation
PHYS 351 · Lecture 01-0214

Lab 1 Task 3: two blocks in series

1. Biased inverting amplifier, referenced to V+V_+, not ground.

2. RC low-pass, fc=1/(2πR5C1)f_c = 1/(2\pi R_5 C_1).

H=11+(f/fc)2φ=arctan(f/fc)|H| = \frac{1}{\sqrt{1 + (f/f_c)^2}} \qquad \varphi = -\arctan(f/f_c)

Show the lag in your sketch. Check nothing exceeds ±15\pm 15 V.

© Ran Yang, Ph.D.Advanced Instrumentation
PHYS 351 · Lecture 01-0215

Multiplying signals

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PHYS 351 · Lecture 01-0216

Multiply two tones

cosω1tcosω2t=14(ejω1t+ejω1t)(ejω2t+ejω2t)=12[cos(ω1ω2)t+cos(ω1+ω2)t]\begin{aligned} \cos\omega_1 t\,\cos\omega_2 t &= \tfrac14\Big(e^{j\omega_1 t}+e^{-j\omega_1 t}\Big)\Big(e^{j\omega_2 t}+e^{-j\omega_2 t}\Big)\\[2pt] &= \tfrac12\big[\cos(\omega_1-\omega_2)t+\cos(\omega_1+\omega_2)t\big] \end{aligned}

The inputs are gone. A difference tone and a sum tone remain.

© Ran Yang, Ph.D.Advanced Instrumentation
PHYS 351 · Lecture 01-0217

Try it: 10.710.7 MHz and 10.010.0 MHz

Ideal mixing spectrum: inputs at 10.0 MHz and 10.7 MHz produce output components at the difference, 0.7 MHz, and sum, 20.7 MHz.
A filter keeps the one you want. Any modulation rides along to the new frequency.
© Ran Yang, Ph.D.Advanced Instrumentation
PHYS 351 · Lecture 01-0218

Conversion loss

Each output tone has amplitude AB/2AB/2: half of ABAB.

Half the amplitude, a quarter of the power: 6-6 dB.

The product’s power splits between two lines. Keep one, lose half.

So a mixer is followed by gain at the intermediate frequency.

© Ran Yang, Ph.D.Advanced Instrumentation
PHYS 351 · Lecture 01-0219

AD633 analog multiplier

AD633 multiplier: differential inputs X1 minus X2 and Y1 minus Y2 feed a multiplier scaled by 1/10 V. The Z input is added to produce output W.

W=(X1X2)(Y1Y2)10 V+ZW=\frac{(X_1-X_2)(Y_1-Y_2)}{10\ \text{V}}+Z

Differential inputs: ground the unused one.

÷10\div 10 V keeps the output inside the rails.

0.1μ0.1\,\upmuF on each supply pin.

© Ran Yang, Ph.D.Advanced Instrumentation
PHYS 351 · Lecture 01-0220

Tie the inputs together: a squarer

AD633 squaring curve, W equals Vin squared divided by 10 V. Both input signs give nonnegative output; 1, 2, 5, and 10 V inputs give 0.1, 0.4, 2.5, and 10 V outputs.

W=Vin210 VW = \frac{V_{\text{in}}^2}{10\ \text{V}}

At Vin=1V_{\text{in}} = 1 V the output is only 100 mV.

A 20 mV offset is then a 20 % error.

© Ran Yang, Ph.D.Advanced Instrumentation
PHYS 351 · Lecture 01-0221

Fit the data, then read the coefficients

W=aVin2+bVin+cW = a\,V_{\text{in}}^2 + b\,V_{\text{in}} + c

aa should be 1/(10V)1/(10\,\text{V}): scale-factor error

b0b \neq 0: input offset, since (V+ε)2=V2+2εV+ε2(V+\varepsilon)^2 = V^2 + 2\varepsilon V + \varepsilon^2

c0c \neq 0: output offset

© Ran Yang, Ph.D.Advanced Instrumentation
PHYS 351 · Lecture 01-0222

Four error sources, four signatures

ErrorSignature in the fitFix
Input offsetlinear term bb; minimum away from 0trim at the unused input
Output offsetconstant ccopposing voltage at ZZ
Scale factora0.100V1a \neq 0.100\,\text{V}^{-1}calibrate, or report it
Nonlinearitya pattern in the residualsthe device floor

Plot the residuals. On the raw data every fit looks perfect.

© Ran Yang, Ph.D.Advanced Instrumentation
PHYS 351 · Lecture 01-0223

Square root by feedback

Put a block that computes f()f(\cdot) in the feedback path.

The loop finds the output that makes the fed-back value match the input: it computes f1()f^{-1}(\cdot).

One polarity only. Worst accuracy near zero, where the slope of V2V^2 vanishes.

Task 3 wants offsets under ±50\pm 50 mV first. Now you know why.

© Ran Yang, Ph.D.Advanced Instrumentation
PHYS 351 · Lecture 01-0224

Radio

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PHYS 351 · Lecture 01-0225

Inside a radio: the superhet chain

Superheterodyne receiver chain: antenna, RF filter and amplifier, mixer fed by a local oscillator, intermediate-frequency filter and amplifier, demodulator, audio amplifier, and speaker.
Same front end for AM and FM; only the demodulator differs.

© Ran Yang, Ph.D.Advanced Instrumentation
PHYS 351 · Lecture 01-0226

Tuning = set the LO

fRFfLO=fIF(455kHz AM, 10.7MHz FM)|f_{\text{RF}} - f_{\text{LO}}| = f_{\text{IF}}\qquad (455\,\text{kHz AM},\ 10.7\,\text{MHz FM})

All gain and selectivity live at one fixed IF.

The image at fLO±fIFf_{\text{LO}} \pm f_{\text{IF}} lands there too. The RF filter rejects it.

© Ran Yang, Ph.D.Advanced Instrumentation
PHYS 351 · Lecture 01-0227

AM vs FM

A slow message signal compared with amplitude and frequency modulation. In AM the carrier amplitude follows the message; in FM its amplitude stays constant while the spacing of cycles changes.

AM: message in the amplitude. Envelope detector.

FM: message in the frequency. Discriminator.

Constant envelope, so amplitude noise is ignored.

© Ran Yang, Ph.D.Advanced Instrumentation
PHYS 351 · Lecture 01-0228

Take away

Measure current across a known resistor.

Feedback turns a tiny current into a usable voltage, and inverts a function you cannot invert directly.

Multiply, then filter, to move a signal in frequency.

© Ran Yang, Ph.D.Advanced Instrumentation
PHYS 351 · Lecture 01-0229

Exit check

1. 2.0-2.0 V across 10kΩ10\kohm: current, and sign?

2. Why is Vout=IphotoRfV_{\text{out}}=-I_{\text{photo}}R_f?

3. 5.05.0 and 4.34.3 MHz multiplied: which frequencies? What selects one?

4. Both AD633 inputs at 3.0-3.0 V, Z=0Z=0: what is WW?

© Ran Yang, Ph.D.Advanced Instrumentation

Use ← → to move, Home / End to jump, and F for fullscreen.

Figure descriptions

Slide 9 · Photodiode: the model to keep

Photodiode equivalent circuit: photocurrent source Iph is in parallel with diode resistance rd and capacitance Cd; series resistance rs connects that parallel network to the output terminal.

Slide 10 · Why reverse bias helps

Photodiode current-voltage curves shift with illumination. A second plot compares output current versus optical power for ideal linear behavior, reverse bias, and zero bias.

Slide 12 · Photodiode + op-amp: the TIA

Transimpedance amplifier block: input current Iin enters an amplifier with transimpedance Ztrans and produces output voltage Vout relative to ground.

Slide 13 · TIA: what limits it

Photoconductive transimpedance circuit: a reverse-biased photodiode feeds the op-amp inverting input. Feedback resistor Rf and parallel capacitor Cf connect output to that input; the noninverting input is grounded.

Slide 17 · Try it: 10.7 MHz and 10.0 MHz

Ideal mixing spectrum: inputs at 10.0 MHz and 10.7 MHz produce output components at the difference, 0.7 MHz, and sum, 20.7 MHz.

Slide 19 · AD633 analog multiplier

AD633 multiplier: differential inputs X1 minus X2 and Y1 minus Y2 feed a multiplier scaled by 1/10 V. The Z input is added to produce output W.

Slide 20 · Tie the inputs together: a squarer

AD633 squaring curve, W equals Vin squared divided by 10 V. Both input signs give nonnegative output; 1, 2, 5, and 10 V inputs give 0.1, 0.4, 2.5, and 10 V outputs.

Slide 25 · Inside a radio: the superhet chain

Superheterodyne receiver chain: antenna, RF filter and amplifier, mixer fed by a local oscillator, intermediate-frequency filter and amplifier, demodulator, audio amplifier, and speaker.

Slide 27 · AM vs FM

A slow message signal compared with amplitude and frequency modulation. In AM the carrier amplitude follows the message; in FM its amplitude stays constant while the spacing of cycles changes.