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LW1DSE > TODOS 11.06.19 17:30l 200 Lines 8156 Bytes #999 (0) @ WW
BID : 1361-LW1DSE
Read: GUEST
Subj: Handbukito(s) muy largos
Path: IW8PGT<LU4ECL<LW3DBH<LU7DQP
Sent: 190611/1434Z @:LU7DQP.#LAN.BA.ARG.SOAM #:179 [Lanus Oeste] FBB7.00i
From: LW1DSE@LU7DQP.#LAN.BA.ARG.SOAM
To : TODOS@WW
[¯¯¯ TST HOST 1.43c, UTC diff:5, Local time: Tue Jun 11 09:00:05 2019 ®®®]
Hola a todos. Como ya he manifestado, me intersa sumarme al proyecto
"Handbukito". Empero, estoy observando algunas peculiaridades que me
gustar¡a resaltar:
1) Veo qe son demasiado largos para leer, sin `eof' lo que hace que las
l¡neas de texto sean largu¡simas excediendo los 80 caracateres normales, y
que a£n con unpaginado de 30 lneas en un BBS, sea sumamente inc¢modo
leerlos online.
2) Se hace menci¢n a im genes en el texto que no aparecen en "plain text".
Sugerir¡a pasarlos a alguna forma de texto como el de "ACSII art" como hace
John G8MNY, y del que aprend¡ a hacerlos yo ambi‚n. De esa manera lo
podemos disfrutar todos, los "DOSeros", los WINderos y los "LINUXeros".
Cascode Amplifier
ÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍ
The very first stage of a VHF TV or FM receiver must amplify the small signal
and add as little noise as possible. The emission and travel of electrons in
the tube is a random process and randomness is noise. Pentodes give superior
performance at RF because they have inherently higher gain and the screen
grid acts as an electrostatic shield between the control grid and plate to
keep down oscillation. But those two extra grids give the electrons more
opportunities to run into something and produce more random noise. It would
be real nice if we could have a tube as quiet as the triode, and with the
gain and high input impedance of a pentode.
Enter stage left the cascode amplifier.
³³
ÚÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÂÄÄÄ´ÃÄÄÄo Vo
³ ³ ³³
³ ± C4
³ ± Rl
³ V2 R2 ±
ßßß ³
-----ÄÄÄÄÄÂÄÄÄÄÄÂÄı±±ÄÄ´
ÕÍÍ͸ ³ ³ ³
³ ÄÁÄ ± ³
³ ÄÂÄ ± R1 ³
³ ³ ± ³
³ ³ C3 ³ o +Ebb
³ ÄÁÄ ÄÁÄ
C1 ³ /// ///
³
³³ ßßß
Vi oÄÄÄÄ´ÃÄÂÄÄ-----
³³ ³ ÕÍÍ͸ V1
³ ³
± ÃÄÄÄ¿
Rg ± ³ ³
± ± ÄÁÄ
³ Rk ± ÄÂÄ C2
³ ± ³
³ ³ ³
ÄÁÄ ÄÁÄ ÄÁÄ
/// /// ///
Figure 1:ÿSchematic of Cascode Amplifier.
Cascode amplifiers are constructed using duo triodes such as the 12AT7 which
was designed for use in the tuners of early TV sets. The two sections are
closely matched and they are in series for DC so the quiescent plate currents
will be equal. The resistor network consisting of R1ÿand R2ÿis set so the
voltages across each triode are equal. This means that æ1ÿ= æ2ÿand rp1ÿ= rp2.
rp
G2 P2
ÚÄÄÄÄÄÄoÄ ÚÄÄÄı±±ÄÄÄÄÄoÄÄÄÂÄÄÄo
³ + ³ ³
³ ³ -->Ib ³
³ ³ + ±
³ (÷) -æVgk2 ±
³ ³ - ±
³ ³ ³
³ - ³ ³
³ oÄÄÄÄÁÄÄÄÄÄÄÄÄÄÄÄÄo ³
³ K2 ³K2 ³
³ ³ ³
³ rp ³ ³
³ G1 ³P1 ³
³ ÚÄÄÄoÄ ÚÄÄÄı±±ÄÄÄÄÄo ³
³ ³ + ³ ³
³ ³ ³ -->Ib ³
³ ³ + ³ + ³
³ (÷) Vin (÷) -æVgk1 ³
³ ³ - ³ - ³
³ ³ ³ ³
ÀÄÄÁÄÄÄoÄÄÄÄÅÄÄÄÄÄÄÄÄÄÄÄÄoÄÄÄÁÄÄÄo
K1 ³ K1
³
ÄÁÄ
///
Figure 2:ÿEquivalent Circuit of Cascode Amplifier
We don't need to make any distinction between æ and rg for the two triodes.
First, we will write the loop equation in the grid of the lower triode:
Vgk1ÿ= Vin (1)
Now, we start at ground and write the equation around the grid of the upper
triode:
-Vgk2ÿ+ æ Vgk1ÿ+ Ibÿrpÿæ = 0 (2)
The equation for the main plate circuit loop, given they are in series,
Ib1 = Ib2, so there insn't need for notate them separately.
-æ Vgk1ÿ- Ibÿrpÿ- æ Vgk2gÿ- Ibÿrpÿ- IbÿRlÿ= 0 (3)
Now we solve equation (2) for Vgk2ÿand substitute into equation (3)
Vgk2ÿ= æ Vgk1ÿ+ Ibÿrpÿæ (4)
-æ Vgk1ÿ- Ibÿrpÿ- æ (æ Vgk1ÿ+ Ibÿrp) - Ibÿrpÿ- IbÿRlÿæ = 0 (5)
Substituting equation (1) into equation (5) gives
-æ Vinÿ- Ibÿrpÿ-æ (æ Vinÿ+ Ibÿrp) - Ibÿrpÿ- IbÿRlÿ= 0 (6)
If we multiply the parentheses through by æ, collect Ibÿterms and factor Ib
out as a negative quantity:
-æ Vinÿ- æýÿVinÿ-Ibÿ(rpÿ+ æ rpÿ+ rpÿ+ Rl) = 0 (7)
Now we factor out -Vin and at the same time factor out rpÿinside the
parentheses.
-Vinÿ(æ + æý) -Ibÿ(rpÿ[æ + 2] + Rl) = 0 (8)
Let's factor out æ so we don't have to deal with that æýÿterm.
Vo
-Vinÿ(æ [æ + 1]) - ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ = 0 (9)
Rlÿ(rpÿ[æ + 2] + Rlæ)
To get the output voltage into the equation and eliminate Ibÿwe write
equation (10) and substitute it into equation (9).
Voÿ= IbÿRl
Vo
Ibÿ= ÄÄÄÄ (10)
Rl
Vo
-Vinÿ(æ [æ + 1]) = ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ (11)
Rlÿ(rpÿ[æ + 2] + Rl)
Now we divide through by -Vin, the parentheses on the right, and multiply
by Rl.
Voÿ Rlÿ(æ [æ + 1])
- ÄÄÄÄ = ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
Vinÿ (rpÿ[æ + 2] + Rl)
And finally we multiply through by -1 to obtain,
Vo -Rlÿ(æ [æ + 1])
Av = ÄÄÄÄÄ = ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ (12)
Vinÿ ÿ (rpÿ[æ + 2] + Rl)
The minus sign is because the amplifier inverts the signal and therefore the
gain is a negative number.
Example:
Calculate the gain of a resistance coupled cascode amplifier using a 12AX7
with a 220Kê plate resistor and a 470Kê resistor in the grid of the following
stage. For the 12AX7, æ = 100, and rp = 80Kê.
Solution:
The value of Rlc is 150Kê. This is the value to be used for Rl in the equation
(12):
Vo -Rlÿ(æ [æ + 1])
Av = ÄÄÄÄÄ = ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
Vinÿ ÿ (rpÿ[æ + 2] + Rl)
150Kê x 100 x 101
Av = - ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ = -182
(80Kê x 102 + 150Kê)
Mis mejores deseos para el proyecto y no se desanimen. Es mi opini¢n nada
m s.
ÉÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍ»
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