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