ElectronicsRobert R. Krchnavek

Spring, 2020

Review – Principles of Electric Circuit Analysis (PECA)

Major Concepts in PECA• Basic Elements – R, L, C, M

• Sources – independent and dependent

• Analysis – Node voltage, mesh current, loop current, CAD.

• Complex impedance

• H(s) vs h(t)

• Op amps

Minimum Required Skills• Recognize and solve a voltage divider. More detailed analysis, e.g. nodal

analysis, may be required but often is not.

• Handle impedances (R, L, and C) in series and parallel.

• Recognize and solve first-order systems (e.g., 1 pole, low-pass filter)

• transient response (t=0+).

• steady state response (t -> ∞)

• Solve op amp circuits: inverting, noninverting, differential, instrumentation, and others.

Op Amps Real vs Ideal?

Op Amps Ideal op amp

Power supplies: ignored Output current capability: Bias current: 0 Offset voltage: 0 Open loop gain: Bandwidth: Input impedance:

∞

∞∞∞

Op AmpsA comparison between two old op amps. Op amps keep getting better.

Bode Plots

Bode Plots

• Invented by Hendrik Wade Bode in the 1930s.

• A frequency response curve for a system.

• Usually includes both magnitude and phase.

• Various aspects of Bode plots allow one to easily account for the frequency aspects/requirements of a system.

Bode PlotsConsider the following RC circuit:

Vi(s) Vo(s)R

C

The steady-state (SS) response of this system is given by

H(s) =Vo(s)Vi(s)

=1

1 + sRC

Bode Plotsτ = RC

ω0 =1

τs = 𝚥ω

Substituting the following:

H(s) =Vo(s)Vi(s)

=1

1 + sRC=

1

1 + sω0

=1

1 +𝚥ωω0

Bode Plots

x-axis: logarithmic (base 10). y-axis: linear

Magnitude

Mag

nitu

de

0

0.25

0.5

0.75

1

1.25

Frequency (rad/s)1 10 100 1000 10000 100000 1000000

Bode Plots• Constant ⟶ constant in magnitude and phase angle

• Zero ⟶

• Simple zero ⟶

• Pole ⟶

• Simple pole ⟶

5 Basic Responses:

1 + s ⇒ 1 + 𝚥ωω0

s ⇒ 𝚥ωω0

1

1 + s⇒

1

1 + 𝚥 ωω0

1

s⇒

1

𝚥 ωω0

Bode PlotThe low-pass filter is an example of a pole:

1

1 + s⇒

1

1 + 𝚥 ωω0

https://www.electronics-tutorials.ws/filter/filter_2.html

Bode PlotThe high-pass filter is an example of a zero:

https://www.electronics-tutorials.ws/filter/filter_2.html

1 + s ⇒ 1 + 𝚥ωω0

Bode Plots• Constant

• Zero ⟶

• Simple zero ⟶

• Pole ⟶

• Simple pole ⟶

5 Basic Responses:

1 + s ⇒ 1 + 𝚥ωω0

s ⇒ 𝚥ωω0

1

1 + s⇒

1

1 + 𝚥 ωω0

1

s⇒

1

𝚥 ωω0

ω0

ω0

ω0

ω0

Bode PlotsExample 1 – Develop a “model” circuit using ideal components that models the frequency response of a real op amp.

Consider the AD712 op amp. Obtain the data sheet and look up the bandwidth (BW) and open loop gain of the op amp.

Bode Plots

Open loop gain in dB:

A(dB) = 20log10(150,000) ≈ 100

BW – 3 MHz. Open loop gain is 1. Unity gain or crossover frequency.

-20 dB/decade

Example 1 – Modeling the frequency response of an op amp

Bode Plots

R1

Vi

Vo

R2

R

C

Assumptions: Gain is 150,000 dB. Roll off starts at 100 Hz.

≈ 100

Vo

Vi= 1 +

R2

R1= 150,000

R1 = 1kΩR2 = 150MΩ

Let then

f0 = 100Hzω0 = 2πf0ω0 = 628(rad/s)

Let R = 1kΩ

RC = τ =1

ω0

1000C =1

628then

C = 1.6μF

Example 1 – Modeling the frequency response of an op amp

Bode PlotsExample 1 – Modeling the frequency response of an op amp

10 1000 106 f, Hz

ideal op amp

filter response

|A( !ω) |

Bode PlotsExample 1 – Modeling the frequency response of an op amp

10 1000 106 f, Hz

ideal op amp

filter response

total response

|A( !" ) |

Bode PlotsExample 1 – Modeling the frequency response of an op amp

10 1000 106 f, Hz

∠A(j!)

ideal op amp

filter response

total response

|A( !" ) |

-90

Bode PlotsExample 2 – Bass Boost Amplifier

2. Design a bass amplifier that provides gain between 25Hz and 100Hz.

K

25 20,000100 f, Hz

Bode Plots

K

25 20,000100 f, Hz

K HPF f1 = 25 Hz

LPF f2 = 100 Hz

Obtain the gain K with an ideal op amp circuit with gain.

Example 2 – Bass Boost Amplifier

Op Amps

Op Amps

• What are the differences between an ideal op amp and a real op amp?

• What causes these differences?

• How can you minimize their impact on circuit design?

Op Ampsv+

v+

vo

v-

v-

vo=A(v+- v-)

Ideal • • • • • • • • • •

Real • • • • • • • • • •

AOL = ∞AOL ≠ f( 𝚥ω)Rin = ∞Rout = 0

ignore power supplies

v+ − v− = vd = 0

common mode gain = 0

BW = 0slew rate = 0

drift = 0

Op Amps

Diodes

Diodes• A non-linear, two-terminal, passive device.

• Many types:

• Signal or Rectifier diodes, e.g, Si “pn” junction diodes. Also Ge, SiC.

• Schottky diodes, e.g., metal-semiconductor diodes.

• Light-emitting diode (LED), multiple colors available. High-brightness, blue LEDs (1993) led to the revolution in lighting we see today. See history in Wikipedia.

• Laser diodes. Used in optical communication systems.

• Zener

• PIN

• Gunn

• Tunnel

Diodes

ID = IDSS(eVDVt − 1)

Vt =kTq k = 1.38(10−23) J/K

T in Kelvin

q = 1.6(10−19) C

IDSS or IS is 10−6 → 10−14 A

Diodes

• Clamps

• Rectifier

• Peak Detector

• Doublers/Triplers

Applications

Diode – Applications

Vi, (v+-v-)1-1

V+

V-

Vo

v+

v+

vo

v-

v-

vo=A(v+- v-)

+V

-V

Diode – Applications

Vi, (v+-v-)1-1

V+

V-

Vo

v+

v+

vo

v-

v-

vo=A(v+- v-)

+V

-V

Never put a voltage on the input greater than .± V+

Diode – ApplicationsRectifiers

vs

vL

Diode – ApplicationsRectifiers

vs

vL

Diode – ApplicationsRectifiers

Diode – ApplicationsRectifiers

D1

C

D2

T1

CTvL < 0

What is vL?

Diode – ApplicationsRectifiers

D1

C

D2

T1

CTvL < 0

Diode – ApplicationsRipple and the size of the capacitor

vLC1

T1

D1

D2D4

D3

Load

iL

LOAD

Diode – ApplicationsRipple and the size of the capacitor

Equivalent circuit when the capacitor is discharging through the load:

C iL

iC

vLvC

Diode – ApplicationsCapacitor ESR

vLC1

T1

D1

D2D4

D3

Load

iL

LOAD

iL = 2A vripple = 1V

60 Hz

C= 16.7 mF

Find the following:

Assume transformer secondary is 12VAC and ideal diodes.

Ton Vs = 17VT = 16.7 msT/2 = 8.33 msT/4 = 4.17 msTon = 1ms

Diode – ApplicationsCapacitor ESR

vLC1

T1

D1

D2D4

D3

Load

iL

LOAD

iL = 2A vripple = 1V

60 Hz

C= 16.7 mF

Find the following:

Assume transformer secondary is 12VAC and ideal diodes.

Ton Vs = 17VT = 16.7 msT/2 = 8.33 msT/4 = 4.17 msTon = 1ms

Diode – Applications• Low loss requires a low ESR capacitor.

• Low ESR capacitors are usually expensive.

• The lifetime of a capacitor is usually inversely proportional to its temperature – hot capacitors die early.

• Another capacitor specification that must be met: WVDE - working volts, DC.

• Vpk on the capacitor should be 1.5 x Vpeak of your circuit. In our 12VAC supply, Vpeak is 17 so Vpk should be 25.

Capacitor ESR

NOTE: Similar details must be addressed for the transformer. Ipeak, cannot saturate the transformer, etc.

Diode – Applications• Vripple cannot be made low enough (mV) by only using a capacitor.

• A “regulator” is used to further lower Vripple. (Note: The capacitor is still required.)

• There are many categories of regulators: linear vs nonlinear, shunt vs series. LM317 is a popular linear, series regulator.

• We will consider a simple, linear, shunt regulator using a Zener diode.

Zener Diode• All diodes exhibit the reverse

“breakdown” shown in the curve.

• Most will fail in this region.

• Zener diodes are designed to work in this regime.

• A minimum current is required to achieve breakdown. See spec sheet. Typical is 1mA.

• Must consider the power dissipation in this region.

iDiZ

vD

vZ

VDVF

VZ

ID

Zener Regulatorunregulated

power supply

load

vLvunreg RLD1

Rlim

voltage regulator

Design the Zener regulator:

•Assume vunreg = 16V •vL = 5.0V •iL,max = 2A

•Find a Zener diode. •Calculate PD,max •Calculate Rlim

Precision Rectifier

RL

vi

vo

Find vo/vi.

Diode – ApplicationsWhat about the input to the supply?