OpenBCM V1.07b12 (Linux)

Packet Radio Mailbox

DB0FHN

[JN59NK Nuernberg]

 Login: GUEST





  
LW1DSE > TUBES    10.11.13 22:02l 190 Lines 7481 Bytes #-3840 (0) @ WW
BID : 7031-LW1DSE
Read: DK3UZ GUEST DL1LCA DG4IAK
Subj: The Cascode Tube Amplifier (CP437!)
Path: DB0FHN<HB9EAS<DB0ZAV<CX2SA<LU7ABF<LU1DBQ<LW8DJW
Sent: 131110/1916Z 32807@LW8DJW.#1824.BA.ARG.SA [Lanus Oeste] FBB7.00e $:7031-L

>From lw1dse%lw8djw.#1824.ba.arg.sa@lu7abf.ampr.org Sun Nov 10 16:34:24 2013
Received: from lu7abf.ampr.org by lu7abf.ampr.org (TCPIP-Net) with SMTP
	id AA126095 ; Sun, 10 Nov 2013 16:34:24 ART
Message-Id: <7031-lw1dse@lu1dbq.bbs>
>From: lw1dse@lw8djw.#1824.ba.arg.sa
X-JNOS-User-Port: Uplink   (LU1DBQ on port 70)  ->    Sending message

From: LW1DSE@LW8DJW.#1824.BA.ARG.SA
To  : TUBES@WW


[¯¯¯ TST HOST 1.43c, UTC diff:5, Local time: Sun Nov 10 08:08:16 2013 ®®®]

                             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                     ± Rb
                              ³  V2            R2   ±
                             ßßß                    ³
                            -----ÄÄÄÄÄÂÄÄÄÄÄÂÄı±±ÄÄ´
                            ÚÄÄÄ¿     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  ÚÄÄÄ¿ V1
                           ³      ³
                           ±      ÃÄÄÄ¿
                        Rg ±      ³   ³
                           ±      ±  ÄÁÄ
                           ³   Rk ±  ÄÂÄ C2
                           ³      ±   ³
                           ³      ³   ³
                          ÄÁÄ    ÄÁÄ ÄÁÄ
                          ///    /// ///

                  Figure 1:hematic 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 R1d R2 set so the
voltages across each triode are equal. This means that æ1æ2d rp1rp2.

                                  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:uivalent 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:

      Vgk1Vin                                                  (1)

Now, we start at ground and write the equation around the grid of the upper
triode:

      -Vgk2æ Vgk1Ib= 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 Vgk1Ibæ Vgk2gIbIb0               (3)

Now we solve equation (2) for Vgk2d substitute into equation (3)

       Vgk2æ Vgk1Ib                                   (4)

      -æ Vgk1Ibæ (æ Vgk1Ib) - IbIb= 0  (5)

Substituting equation (1) into equation (5) gives

      -æ VinIb (æ VinIb) - IbIb0       (6)

If we multiply the parentheses through by æ, collect Ibrms and factor Ib
out as a negative quantity:

      -æ Vinæýnbpæ rprpRb) = 0               (7)

Now we factor out -Vin and at the same time factor out rpside the
parentheses.

      -Vin + æý) -Ibp + 2] + Rb) = 0                     (8)

Let's factor out æ so we don't have to deal with that æýrm.

                                   Vo
      -Vin [æ + 1]) - ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ  = 0            (9)
                          Rbp + 2] + Rbæ)

To get the output voltage into the equation and eliminate Ib write
equation (10) and substitute it into equation (9).

      VoIb

            Vo
      IbÄÄÄÄ                                                   (10)
            Rb
                                   Vo
      -Vin [æ + 1]) = ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ                   (11)
                          Rbp + 2] + Rb)

Now we divide through by -Vin, the parentheses on the right, and multiply
by Rb.

         Vo   Rb [æ + 1])
      - ÄÄÄÄ = ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
         Vin (rp + 2] + Rb)

And finally we multiply through by -1 to obtain,

             Vo      -Rb [æ + 1])
      Av = ÄÄÄÄÄ = ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ                            (12)
            Vin  (rp + 2] + Rb)

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 220 k ohm plate resistor and a 470 k ohm resistor in the grid of the
following stage. For the 12AX7, æ = 100, and rp = 80 k ohms.

Solution:

The value of Rbc is 150 k ohms. This is the value to be used for Rb in
equation 4A.42.

       Vo      -Rb [æ + 1])
Av = ÄÄÄÄÄ = ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
      Vin  (rp + 2] + Rb)

         150Kê x 100 x 101
Av = - ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ = -182
        (80Kê x 102 + 150Kê)
ÉÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍ»       
º Osvaldo F. Zappacosta. Barrio Garay (GF05tf) Alte. Brown, Bs As, Argentina.º
º Mother UMC æPC:AMD486@120MHz, 16MbRAM HD SCSI 6.4Gb MSDOS 7.10 TSTHOST1.43 º
º                Bater¡a 12V 160AH. 9 paneles solares 10W.                   º
º                  lw1dse@yahoo.com ; lw1dse@gmail.com                       º
ÈÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍͼ



Read previous mail | Read next mail


 19.05.2024 02:32:23lGo back Go up