Passive Components

Rayat Shikshan Sanstha’s S.M. Joshi College Hadapsar-028 Department of Electronics Science Passive Components Presented by- Dr. Kakade K.P

Op-Amp • Introduction of Operation Amplifier (Op-Amp) • Block Diagram • Analysis of ideal Op-Amp applications • Comparison of ideal and non-ideal Op-Amp • Non-ideal Op-Amp consideration Operational Amplifier

BLOCK DIAGRAM Input Differential stage Gain stage Output Buffering stage Operational Amplifier

Stages • Input Differential Stage • Intermediate Signal-Ended High-Gain Stage • Output Buffering Stage • Current Source / Short Circuit Protection

Input Differential Stage The input stage consists of the transistors Q1 through Q7 with biasing performed by Q8, Q9, and Q10. Transistors Q1 and Q2 are emitter followers which causes input resistance to be high and deliver the differential input signal to the common base amplifier formed by Q3 and Q4. Transistors Q5, Q6, and Q7, and resistors R1, R2, and R3 form the load circuit of the input stage. This portion of the circuit provides a high resistance load. Transistors Q3 and Q4 also serve as protection for Q1 and Q2. The emitter-base junction of Q1 and Q2 breaks down at around 7V but the pnp transistors have breakdown voltages around 50V. So, having them in series with Q1 and Q2 protects Q1 and Q2 from an accidental connection between the input terminals.

Intermediate Single-Ended High-Gain Stage The second stage is composed of Q16, Q17, Q13B, and the resistors R8 and R9. Transistor Q16 acts as an emitter follower giving the second stage a high input resistance. Transistor Q17 is a common-emitter amplifier with a 100-Ώ resistor in the emitter. The load of this amplifier is composed of the output resistance of Q13B. This use of a transistor as a load resistance is called active load. The output of this amplifier (the collector of Q17) has a feedback loop through Cc. This capacitor causes the op-amp to have a pole at about 4Hz.

Output Buffering Stage The Output Stage consists of the complimentary pair Q14 and Q20, and a class AB output stage composed of Q18 and Q19. Q15 and Q21 give short circuit protection (described later) and Q13A supplies current to the output stage. The purpose of the Output Stage is to provide the amplifier with a low output resistance. Another requirement of the Output Stage is the ability to dissipate large load currents without dissipating large quantities of power. This is done through the class AB Out- put Stage.

Operational Amplifier (Op-Amp) • Very high differential gain • High input impedance • Low output impedance • Provide voltage changes (amplitude and polarity) • Used in oscillator, filter and instrumentation • Accumulate a very high gain by multiple stages Operational Amplifier

IC Product DIP-741 Dual op-amp 1458 device Operational Amplifier

Single-Ended Input • + terminal : Source • – terminal : Ground • 0o phase change • + terminal : Ground • – terminal : Source • 180o phase change Operational Amplifier

Double-Ended Input • Differential input • 0o phase shift change • between Vo and Vd Qu: What Vo should be if, Ans: (A or B) ? (A) (B) Operational Amplifier

Distortion The output voltage never excess the DC voltage supply of the Op-Amp Operational Amplifier

Common-Mode Operation • Same voltage source is applied • at both terminals • Ideally, two input are equally • amplified • Output voltage is ideally zero • due to differential voltage is • zero • Practically, a small output • signal can still be measured Note for differential circuits: Opposite inputs : highly amplified Common inputs : slightly amplified  Common-Mode Rejection Operational Amplifier

Common-Mode Rejection Ratio (CMRR) Differential voltage input : Common voltage input : Common-mode rejection ratio: Output voltage : Note: When Gd >> Gc or CMRR  Vo = GdVd Gd : Differential gain Gc : Common mode gain Operational Amplifier

CMRR Example What is the CMRR? Solution : (2) (1) NB: This method is Not work! Why? Operational Amplifier

Op-Amp Properties • Infinite Open Loop gain • The gain without feedback • Equal to differential gain • Zero common-mode gain • Pratically, Gd = 20,000 to 200,000 • (2) Infinite Input impedance • Input current ii ~0A • T- in high-grade op-amp • m-A input current in low-grade op-amp • (3) Zero Output Impedance • act as perfect internal voltage source • No internal resistance • Output impedance in series with load • Reducing output voltage to the load • Practically, Rout ~ 20-100  Operational Amplifier

Frequency-Gain Relation • Ideally, signals are amplified from DC to the highest AC frequency • Practically, bandwidth is limited • 741 family op-amp have an limit bandwidth of few KHz. 20log(0.707)=3dB • Unity Gain frequency f1: the gain at unity • Cutoff frequency fc: the gain drop by 3dB from dc gain Gd GB Product : f1 = Gdfc Operational Amplifier

? Hz 10MHz GB Product Example: Determine the cutoff frequency of an op-amp having a unit gain frequency f1 = 10 MHz and voltage differential gain Gd = 20V/mV • Sol: • Since f1 = 10 MHz • By using GB production equation • f1 = Gdfc • fc = f1 / Gd = 10 MHz / 20 V/mV • = 10  106 / 20  103 • = 500 Hz Operational Amplifier

Ideal Vs Practical Op-Amp Operational Amplifier

Ideal Op-Amp Applications Analysis Method : Two ideal Op-Amp Properties: • The voltage between V+ and V is zero V+ = V • The current into both V+ and V termainals is zero For ideal Op-Amp circuit: • Write the kirchhoff node equation at the noninverting terminal V+ • Write the kirchhoff node eqaution at the inverting terminal V • Set V+ = V and solve for the desired closed-loop gain Operational Amplifier

Noninverting Amplifier • Kirchhoff node equation at V+ yields, • Kirchhoff node equation at Vyields, • Setting V+ = V– yields • or Operational Amplifier

Noninverting amplifier Noninverting input with voltage divider Less than unity gain Voltage follower Operational Amplifier

Inverting Amplifier • Kirchhoff node equation at V+ yields, • Kirchhoff node equation at Vyields, • Setting V+ = V– yields Notice: The closed-loop gainVo/Vin is dependent upon the ratio of two resistors, and is independent of the open-loop gain. This is caused by the use of feedback output voltage to subtract from the input voltage. Operational Amplifier

Multiple Inputs • Kirchhoff node equation at V+ yields, • Kirchhoff node equation at Vyields, • Setting V+ = V– yields Operational Amplifier

Inverting Integrator • Now replace resistors Ra and Rf by complex components Za and Zf, respectively, therefore • Supposing • The feedback component is a capacitor C, i.e., • The input component is a resistor R, Za = R • Therefore, the closed-loop gain (Vo/Vin) become: • where • What happens if Za = 1/jC whereas, Zf = R? • Inverting differentiator Operational Amplifier

Op-Amp Integrator • Example: • Determine the rate of change • of the output voltage. • Draw the output waveform. Solution: (a) Rate of change of the output voltage (b) In 100 s, the voltage decrease Operational Amplifier

Op-Amp Differentiator Operational Amplifier

Non-ideal case (Inverting Amplifier)  Equivalent Circuit • 3 categories are considering • Close-Loop Voltage Gain • Input impedance • Output impedance Operational Amplifier

Close-Loop Gain Applied KCL at V– terminal, By using the open loop gain,   The Close-Loop Gain, Av Operational Amplifier

Close-Loop Gain When the open loop gain is very large, the above equation become, The close-loop gain now reduce to the same form as an ideal case Operational Amplifier

Input Impedance can be regarded as, where R is the equivalent impedance of the red box circuit, that is However, with the below circuit, Input Impedance Operational Amplifier

Input Impedance Finally, we find the input impedance as,  Since, , Rin become, Again with The op-amp can provide an impedance isolated from input to output Operational Amplifier

Output Impedance Only source-free output impedance would be considered, i.e. Vi is assumed to be 0 Firstly, with figure (a), By using KCL,io = i1+ i2 By substitute the equation from Fig. (a), R and A comparably large, Operational Amplifier

Passive Components