Theory of Analog Electronics Vacuum Tube Triode How to invent Amplifier
Theory of Analog Electronics Vacuum Tube Triode How to invent Amplifier * In this presentation definitions and examples from Wikipedia, HowStaffWorks and some other sources were used ORT Braude Engineering College. Course: Theory of Analog Electronics 31401. Lecturer: Dr. Samuel Kosolapov [email protected] Quiz 04. 1 (before the start): Do You need this lecture ? Why EE needs DC and AC analysis to analyze amplifiers ? What EE wants to achieve while selection Q-Point ? What kind of capacitors are used in amplifiers ? 2 Items to be defined/refreshed/discussed • • • • • • Structure Amplification effect Practical Design Bias, Decoupling Capacitors DC and AC Analysis Why all this is still important 3 Vacuum Tube Triode : 1907 Anode ( A ) When positive, attract electrons Glass or metal Envelope Vacuum inside Grid ( G ) Positioned between cathode and anode DAC DGC Cathode ( C ) When heated creates cloud of electrons Heater 6.3V AC 4 Vacuum Tube Triode : Grid For simplicity of explanation Triode Geometric sizes are: DAC = 10 cm = 0.1 m DGC = 1 cm = 0.01 m E AC V AC V ; EGC GC D AC DGC 1 D AC V AC VGC E RESULTED E AC EGC D AC DGC E RESULTED 10 V AC 10VGC This means that influence of the Grid is 10 times more than influence of the anode. (Because Grid is 10 times closer to cathode) 5 Vacuum Tube Triode : Equation Modified Lengmure’s equation I ANODE k0 VGC k1VAC 1.5 PSpice default values are: k0 = 200E-6 ~ 1E-4 , k1=0.12 ~ 0.1 We have characteristic of the device by its physical and geometrical design. More Pins: A, G, C, heater. Problem: metal and glass for high temperature 6 Triode as a Concept Amplifier Anode (A) Grid Cathode (G) (C) Vac is positive. (Say 100V) Grid Influence: Case a: Grid potential is 0. Electrons run from cathode to anode according to Lengmure’s Law. Grid has no influence here. IA = F(VAC) LARGE CURRENT Case b: Grid potential is –10V. ETOTAL in that case is 0 No current (IA=0) SMALL Grid VOLTAGE REGULATES LARGE Anode CURRENT !!!! If –10 <VGC < 0 then change of VGC leads to change of IA 7 Let’s try to build Amplifier Suppose VAA = 200 V. I want VAC to be VAA /2 = 100 V. Then VR = 100 V. Let’s select proper RA value (Set Q-point) Ra I ANODE 0.00001* VGC 0.1VAC 1.5 VGC 0; VAC 100 I ANODE 0.0032 A 100 Ra 31250 30k 0.0032 8 Let’s try to build Amplifier Finally in this case: VGC = 0; VAC=100V VR = 100V IA = 3 mA 30 kW We know all currents and voltages (despite the fact that triode is NON-LINEAR device). 9 Amplification Effect Previous case: VGC = 0; VAC=100V VR = 100V IA = 3 mA Now, VGC = - 10V In that case: VGC = -10V IA = 0 (no anode current) VR = 0 (No current) VA = VAA = 200 V (Explain twice) 30 kW This means: VGC (INPUT VOLTAGE) was changed by 10 V (from 0 to –10V) VAC(OUTPUT WOLTAGE) was changed by 100V (from 100V to 200V) Then Voltage Gain is 100:10 = 10. (-10, phase change) We have some sort of Voltage Amplifier. (“Candidate for Amplifier”) We understand HOW each part of This Amplifier is functioning. 10 Better design needed Our “candidate for amplifier” works (increases pp amplitude of the input signal by factor ~ 10) in the following input voltage range: -10 < Vin < 0 (active region) Important: No amplification outside this active region. Only active region is interesting in the frames of this course 30 kW Problem #1. Our amplifier operates in case VGC change from 0 to negative value. We want it to change about ZERO. Problem #2. Output voltage changes in the range 100-200V. We want it to change (span) about ZERO. 11 Concepts: Bias, Decoupling capacitors DC and AC analysis VGC now is a superposition of two voltages: DC : VBIAS and AC : VSIG. Attention: Do not be baffled between AC (alternative current) and AC (anode-cathode) 12 Concepts: Bias, Decoupling capacitors DC and AC analysis The circuit in test is non-linear. We can write Voltage Node Equations Equation are non-linear in this case In most cases exact (“algebraic”) solution is not available We can solve non-linear equations only numerically (approximation) This is what is done in the simulation software In case we want to “analyze” the circuit we need parametric solution. Then trick: “DC / AC analysis” is used 13 DC & AC analysis DC analysis (short AC voltage sources): VSIG = 0 (short AC sources) VGC = -5V, VAC=AK ~ 150 (well, somewhere between 100 and 200 V : Non-linear Circuit) One may RECALCULATE Ra now for VAC to be exactly 150 in that case. {Set new Q-point in the “center” of “active” region} VLOAD = 0 in that case, because of C2 (no DC pass capacitor!) AC Analysis (short/ignore DC voltage sources): DVSIG is positive now (say +5V). Then VGC = +5-5=0 VAC = 100V DVLOAD is negative 100-150 = -50V. Voltage Gain is (100-150 ) / ( +5-0 ) = –10 DVSIG is negative now (say -5V). Then VGC = -5-5=-10V VAC = 200V DVLOAD is positive 200-150 = +50V. Voltage Gain is ( 200-150 ) / ( 0-(-5) ) = -10 Practically, VSIG must be SMALL signal, say 100 mV. Then VGC is changing in the small region about Q point Small signal AC technique can be used for “calculations”. 14 Concept: Self-bias (Auto-bias) Out amplifier works. (BTW, what Rin and Rout are ???? ) But we need 2 batteries: VAA and VBIAS. Bad (non-practical design). Solution: self-bias (automatic biasing) 15 Concept: Self-bias (Auto-bias) DC Analysis: Select VAA, RC and RA in such a way that VA(Q) will be (say) 150 V and that VGC(Q) will be –5V It is always possible for example by try and error method (set Variable Resistors and rotates until needed voltages will be achieved) VC(Q) then is POSITIVE, whereas VGC(Q) is negative: VGC(Q) = 0 – VC(Q) OK Explain need for RG: VG(Q) ~ 0, but VG(Q) != ZERO : small current from the Grid (explain: physics of the device bias) Explain LIMITS of the RG In this case AC analysis is the same as before Voltage gain is the same. 16 Concepts forever 17 Control Questions • What have I learned Outdated Device that most of EE will never use 18 Control Questions • Why did I learn it It is easy to grab idea of how amplifier works, What is the meaning of negative voltage gain Why do we need to set Q-point by using DC analysis Important example of how Q-point can be set. 19 Control Questions • How can I apply this We will reuse developed approach to analyze modern amplifiers based on FET and BJT 20 Literature to read 1. TBD 21
© Copyright 2024