Innovation Fellows Program · switch or amplify a signal. They typically consist of three...

Innovation Fellows Program

Analog Circuits Prof. Steven S. Saliterman, http://saliterman.umn.edu/ Department of Biomedical Engineering, University of Minnesota

Prof. Steven S. Saliterman

Course Workstations


Solderless Breadboards… Components and Wiring Top Side

Unseen Interconnections Bottom Side

DC Power Gnd or Neg. Rail DC Power Positive Rail

DC Power Gnd or Neg. Rail DC Power Positive Rail

Component Bus

Component Bus

Typical Use

22 gauge solid wire works best for interconnections.


Concepts to be Covered

Circuit Theory (Using National Instruments Multisim* Software) Kirchoff’s Voltage and Current Laws Voltage Divider Rule Resistance

Series and Parallel Resistors Thevenin’s Theorem

Capacitance Series and Parallel Capacitors Charging and Discharging a Capacitor

Semiconductors Operational amplifiers

* Multisim - Simulation Program with Integrated Circuit Emphasis - SPICE


Circuit Theory – Ohm’s Law

Ohm’s Law: 𝑉𝑉 = 𝐼𝐼𝐼𝐼

Where V = Voltage in volts, V. I = Current in amps, A. R = Resistance in ohms, Ω.

For example, if the voltage is 5 V (volts direct current), and the resistor is 220 Ω, the current flow would be:

𝐼𝐼 = 𝑉𝑉𝑅𝑅

= 5220

~.0227 𝐴𝐴 𝑜𝑜𝑜𝑜 22.7 𝑚𝑚𝐴𝐴 (𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚)

Image courtesy of Texas Instruments


Ohm’s Law Simulation…


Real World Measurement…

LRC meter showing .220 kΩ resistor.

(Inductance-Resistance-Capacitance)

Power supply set at 5 Vdc. (Voltage regulated - Current limited)

Multimeter in series with the power supply and resistor,

reading ~22 mA. (Volts AC & DC, Resistance, Diodes, Current)


Kirchoff’s Voltage Law Kirchoff’s Voltage Law:

∑𝑉𝑉𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 = ∑𝑉𝑉𝐷𝐷𝑆𝑆𝑆𝑆𝐷𝐷𝐷𝐷 The sum of the voltage drops in a series circuit

equals the sum of the voltage sources. 𝑉𝑉 = 𝑉𝑉𝑅𝑅1

+ 𝑉𝑉𝑅𝑅2

For series resistors, 𝐼𝐼𝑇𝑇𝑆𝑆𝑇𝑇𝑇𝑇𝑇𝑇 = 𝐼𝐼1 + 𝐼𝐼2 + ⋯+ 𝐼𝐼𝑛𝑛 For example, if V = 5 V, R1 = 1000 Ω, and R2 =

220 Ω, then the voltage drop across each resistor is determined as follows: Total resistance, R = 1220 Ω. ∴ 𝐼𝐼 = 𝑉𝑉

𝑅𝑅= 5

1220~4.1 𝑚𝑚𝐴𝐴

The voltage drop across R1 is: 𝑉𝑉𝑅𝑅1

= 𝐼𝐼𝐼𝐼 = .0041 𝑥𝑥 1000 = 4.1 𝑉𝑉 The voltage drop across R2 is:

𝑉𝑉𝑅𝑅2 = 𝐼𝐼𝐼𝐼 = .0041 𝑥𝑥 220 = .9 V Note that 4.1 V + .9 V = 5 V



Kirchoff’s Voltage Law Simulation…


Kirchoff’s Current Law

Kirchoff’s Current Law: ∑ 𝐼𝐼𝐼𝐼𝑛𝑛 = ∑ 𝐼𝐼𝑂𝑂𝑆𝑆𝑇𝑇 The sum of the currents

entering a junction equals the sum of the currents leaving the junction.

𝐼𝐼1 + 𝐼𝐼2 = 𝐼𝐼3 + 𝐼𝐼4



Voltage Divider Rule

Notice that the output voltage in this circuit is simply the voltage drop across R2.

𝑉𝑉𝑂𝑂𝑆𝑆𝑇𝑇 = 𝐼𝐼𝐼𝐼2 = 𝑉𝑉𝑅𝑅𝑇𝑇𝐼𝐼2 = 𝑉𝑉 𝑅𝑅2

𝑅𝑅1+𝑅𝑅2

Also notice that the output circuit will eventually provide additional pathways for the current to flow!



Parallel Resistors & Current Flow Notice that current will flow through both resistors (plus

any future additional output circuitry). Total resistance for parallel resistors is as follows:

1𝑅𝑅

= 1𝑅𝑅1

+ 1𝑅𝑅2

…+ 1𝑅𝑅𝑛𝑛

and 𝐼𝐼 = 𝐼𝐼1 + 𝐼𝐼2 = 𝑉𝑉𝑅𝑅

𝐼𝐼1 = 𝑉𝑉𝑅𝑅1

and 𝐼𝐼2 = 𝑉𝑉𝑅𝑅2

, also 𝐼𝐼1 = 𝐼𝐼 𝑅𝑅2𝑅𝑅1+𝑅𝑅2

and 𝐼𝐼2 = 𝐼𝐼 𝑅𝑅1𝑅𝑅1+𝑅𝑅2

For example, if V=5 V, R1=1000 Ω and R2=2200 Ω, then:

1𝑅𝑅

= 11000

+ 12200

,𝐼𝐼 = 𝑅𝑅1𝑅𝑅2𝑅𝑅1+𝑅𝑅2

~ 688 Ω

𝐼𝐼1 = 51000

= 5 𝑚𝑚𝐴𝐴

𝐼𝐼2 = 52200

= 2.27 𝑚𝑚𝐴𝐴

∴ 𝐼𝐼 = 𝐼𝐼1 + 𝐼𝐼2 = 5 + 2.27 = 7.27 mA

Confirm: 𝐼𝐼 = 𝑉𝑉𝑅𝑅

= 5688

~7.27mA


V

Math refresher: 1𝐼𝐼

=1𝐼𝐼1

+1𝐼𝐼2

=𝐼𝐼1𝐼𝐼1𝐼𝐼2

+𝐼𝐼2𝐼𝐼1𝐼𝐼2

=𝐼𝐼1 +𝐼𝐼2𝐼𝐼1𝐼𝐼2

∴ 𝐼𝐼 =𝐼𝐼1𝐼𝐼2𝐼𝐼1 +𝐼𝐼2


Parallel Resistor Current Simulation…


Series vs Parallel Resistor Simulation…


Example: Current Limiting an LED…

+ (longer lead) -


Resistor Color Code

Image courtesy of Electronix Express


Power…

The power through a resistor: 𝑃𝑃 = 𝐼𝐼2𝐼𝐼 = 𝑉𝑉𝐼𝐼 in watts.

Resistors must dissipate heat, and are rated based on watts – typically 1/8, ¼, ½ or a 1 watt resistor.

The resistor may be made of carbon composition, carbon film, metal film, metal oxide film, foil or even wire wound.


Capacitance Capacitors are made of two parallel plates, separated by a dielectric

material. These may be foil plates and rolled in a tubular shape or simply bonded as flat metal plates. The charge on a capacitor is: 𝑄𝑄 = 𝐶𝐶 𝑥𝑥 𝑉𝑉, where

Q is the charge in Coulombs, C is the capcitance in Farads and V is the voltage in Volts.

𝐶𝐶 = ε𝑆𝑆ε0𝐴𝐴𝑑𝑑 , where

C is the capacitance, in farads; A is the area of overlap of the two plates, in square meters; εr is the relative static permittivity (sometimes called the dielectric constant) of the

material between the plates (for a vacuum, εr = 1); ε0 is the electric constant ε0 ≈ 8.854×10−12 F/m; - farads per meter. d is the separation between the plates, in meters;

Image courtesy of Eric Schrader, Wikipedia


RC Charging of a Capacitor…

• Note the exponential rise in voltage across the capacitor as it charges.

• The oscilloscope is set at 1V/div and 50ms/div.

• The voltage across the resistor and current flow both fall as the capacitor charges.

• 𝑉𝑉 𝑡𝑡 = 𝑉𝑉0(1 − 𝑒𝑒−𝑡𝑡𝑅𝑅𝑅𝑅)

𝑄𝑄 = 𝐶𝐶𝑉𝑉0(1 − 𝑒𝑒−𝑡𝑡𝑅𝑅𝑅𝑅)

𝐼𝐼 = 𝑉𝑉0𝑅𝑅𝑒𝑒−

𝑡𝑡𝑅𝑅𝑅𝑅

Time

5V

Volts


Charging…

Image courtesy of hyperphysicis/phy-astr.gsu.edu

From previous example: R=750 Ω V0=5 Vdc C=100 µf RC=0.075s (i.e. t=.2s is 2.7 “RC”s Imax=0.00667 amps (Upon closing switch.) Qmax=500 µC


RC Discharging of a Capacitor…

Volts

Time

5 V

• Note the exponential fall in voltage across the capacitor as it discharges.

• The oscilloscope is set at 1V/div and 50ms/div.

• At the onset of discharge, voltage across the resistor is the capacitor voltage, and then rapidly decreases along with the current.

• 𝑉𝑉 𝑡𝑡 = 𝑉𝑉0𝑒𝑒− 𝑡𝑡𝑅𝑅𝑅𝑅

𝑄𝑄 = 𝐶𝐶𝑉𝑉0𝑒𝑒− 𝑡𝑡𝑅𝑅𝑅𝑅

𝐼𝐼 = 𝑉𝑉0𝑅𝑅𝑒𝑒−

𝑡𝑡𝑅𝑅𝑅𝑅


Discharging…

Image courtesy of hyperphysicis/phy-astr.gsu.edu

From the previous example: R=750Ω V0=5Vdc C=100µf RC=0.075s (i.e. t=.2s is 2.67 “RC”s Imax=0.00667 amps (Upon closing switch.) Qmax=500µC (the starting charge)


Diodes

Semiconductor devices that rectify current – allowing flow in only one direction.

They be used as switching devices, voltage-controlled capacitors (varactors) and voltage references (Zener diodes).

Typically a 2-terminal device – an anode and cathode. Ideally resistance is zero in one direction, and infinite in the other. Diodes allow current to flow in only one direction and can therefore be

used as simple solid state switches in AC circuits, being either open (not conducting) or closed (conducting).

A bridge rectifier, consisting of 4 diodes can be used to convert AC into DC, and typically followed filtering capacitors and/or Zener diode.


Half-Wave Rectifier

Scherz, P.& S. Monk. Practical Electronics for Inventors, McGraw Hill, New York, NY (2016).


Full-Wave Bridge Rectifier



Voltage Regulators – 7800/7900 Series

Fixed voltage integrated voltage regulator. Three terminal regulation. Output current to 1.5A Internal thermal overload protection. High-power dissipation capability with heatsink tab.


Simple Power Supply…



Transistors

Semiconductor devices that are used to switch or amplify a signal. They typically consist of three terminals. For an NPN or PNP bipolar

junction transistor these are called the collector, base and emitter. They may ay be a discrete component, or through nanofabrication may

number in the billions for a single CPU – e.g. 7.2 billion in Intel’s 22-core chip.

http://www.colorado.edu/physics/phys3330/phys3330_fa10/images/Transistors.JPG


Transistor Switch Simulation…


Transistor as an Amplifier

A transistor can be “off”, “on” or in a “linear state” where IB causes changes in IC based on hFE, the current gain factor: 𝐼𝐼𝐶𝐶 = 𝐼𝐼𝐵𝐵 𝑥𝑥 ℎ𝐹𝐹𝐹𝐹 Useful for example, for a common emitter amplifier.

In saturation, any changes in IB will not cause changes in IC.

When “off”, there is no base current applied.


Transistor Operating State…

Image courtesy of www.ermicro.com/blog


Simulation of an Amplifier…

Measured input voltage is 100mv pp and output voltage is 10v pp, which is a gain of 100!


Amplifier on a Breadboard Example…

Function Generator set at 1 kHz, 0.1 V pp (Actually measures 79.2 MV pp on oscilloscope)

Power Supply set at 20 VDC


Voltage Gain…

Amplifier Input 79.2 mV pp

Amplifier Output 7.6 V pp Gain is 7.6/.079 = 96.2


Operational Amplifier

Scherz, P. and S. Monk, Practical Electronics for Inventors, McGraw Hill, New York (2016)


Op Amp Closed Loop Configurations

A feedback loop allows for precise control of the voltage gain: Inverting Op Amp

𝑉𝑉𝑂𝑂𝑂𝑂𝑇𝑇 = −𝑉𝑉𝐼𝐼𝐼𝐼𝑅𝑅𝐹𝐹𝑅𝑅𝐺𝐺

𝑉𝑉𝑜𝑜𝑚𝑚𝑡𝑡𝑚𝑚𝑉𝑉𝑒𝑒 𝐺𝐺𝑚𝑚𝑚𝑚𝐺𝐺 = 𝑉𝑉𝑜𝑜𝑜𝑜𝑡𝑡𝑉𝑉𝑖𝑖𝑛𝑛

= −𝑅𝑅𝐹𝐹𝑅𝑅𝐺𝐺

Non-Inverting Op Amp 𝑉𝑉𝑂𝑂𝑂𝑂𝑇𝑇 = 𝑉𝑉𝐼𝐼𝐼𝐼

𝑅𝑅𝐺𝐺+𝑅𝑅𝐹𝐹𝑅𝑅𝐺𝐺

𝑉𝑉𝑜𝑜𝑚𝑚𝑡𝑡𝑚𝑚𝑉𝑉𝑒𝑒 𝐺𝐺𝑚𝑚𝑚𝑚𝐺𝐺 = 𝑉𝑉𝑜𝑜𝑜𝑜𝑡𝑡𝑉𝑉𝑖𝑖𝑛𝑛

= 1 + 𝑅𝑅𝐹𝐹𝑅𝑅𝐺𝐺

Carter, B. and R. Mancini, Op Amps for Everyone, Newnes & TI, Burlington, MA (2009)

a.

b.


Inverting Amplifier, Gain of 6.83

𝑉𝑉𝑆𝑆𝑆𝑆𝑇𝑇 = −𝑉𝑉𝑖𝑖𝑛𝑛𝐼𝐼2𝐼𝐼1

= 2 ×8212 = −13.66 𝑉𝑉𝐷𝐷𝐷𝐷 or −6.83 𝑉𝑉𝐷𝐷

𝑉𝑉𝑜𝑜𝑚𝑚𝑡𝑡𝑚𝑚𝑉𝑉𝑒𝑒 𝐺𝐺𝑚𝑚𝑚𝑚𝐺𝐺 =𝑉𝑉𝑆𝑆𝑆𝑆𝑇𝑇𝑉𝑉𝑖𝑖𝑛𝑛

=13.66

2 = 6.83 𝑜𝑜𝑜𝑜 −𝐼𝐼𝐹𝐹𝐼𝐼𝑖𝑖𝑛𝑛

= −8212 = −6.83


Non-Inverting Amplifier, Gain of 11

𝑉𝑉𝑆𝑆𝑆𝑆𝑇𝑇 = 𝑉𝑉𝑖𝑖𝑛𝑛𝐼𝐼1 + 𝐼𝐼2𝐼𝐼1

= 2 ×10 + 1

1 = 22 𝑉𝑉𝐷𝐷𝐷𝐷 𝑜𝑜𝑜𝑜 11 𝑉𝑉𝐷𝐷

𝑉𝑉𝑜𝑜𝑚𝑚𝑡𝑡𝑚𝑚𝑉𝑉𝑒𝑒 𝐺𝐺𝑚𝑚𝑚𝑚𝐺𝐺 =𝑉𝑉𝑆𝑆𝑆𝑆𝑇𝑇𝑉𝑉𝑖𝑖𝑛𝑛

=222 = 11 𝑜𝑜𝑜𝑜 1 +

𝐼𝐼2𝐼𝐼1

= 11


Voltage Follower…


Summary

Circuit Theory (Using National Instruments Multisim* Software)

Kirchoff’s Voltage and Current Laws Voltage Divider Rule Resistance

Series and Parallel Resistors Capacitance

Series and Parallel Capacitors Charging and Discharging a Capacitor

Semiconductors Operational Amplifiers Lab Exercise Equipment

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