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LW1DSE > TUBES 01.07.12 18:43l 166 Lines 7202 Bytes #999 (0) @ WW
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Read: DK3UZ GUEST DL1LCA DK3UD DG4IAK
Subj: The Cascode Circuit (CP437)
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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: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- IbÿRbÿ= 0 (3)
Now we solve equation (2) for Vgk2ÿand substitute into equation (3)
Vgk2ÿ= æ Vgk1ÿ+ Ibÿrpÿæ= 0 (4)
-æ Vgk1ÿ- Ibÿrpÿ- æ (æ Vgk1ÿ+ Ibÿrp) - Ibÿrpÿ- IbÿRbÿæ= 0 (5)
Substituting equation (1) into equation (5) gives
-æ Vinÿ- Ibÿrpÿ-æ (æ Vinÿ+ Ibÿrp) - Ibÿrpÿ- IbÿRbÿ= 0 (6)
If we multiply the parentheses through by æ, collect Ibÿterms and factor Ib
out as a negative quantity:
-æ Vinÿ- æ2ÿVinÿ-Ibÿ(rpÿ+ æ rpÿ+ rpÿ+ Rb) = 0 (7)
Now we factor out -Vin and at the same time factor out rpÿinside the
parentheses.
-Vinÿ(æ + æ2) -Ibÿ(rpÿ[æ + 2] + Rb) = 0 (8)
Let's factor out æ so we don't have to deal with that æ2ÿterm.
-Vinÿ(æ [æ + 1]) -Voÿ/ Rbÿ(rpÿ[æ + 2] + Rbæ) = 0 (9)
To get the output voltage into the equation and eliminate Ibÿwe write
equation (10) and substitute it into equation (9).
Voÿ= IbÿRb
Ibÿ= Voÿ/ Rbÿ (10)
Vo
-Vinÿ(æ [æ + 1]) = ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ (11)
Rbÿ(rpÿ[æ + 2] + Rb)
Now we divide through by -Vin, the parentheses on the right, and multiply
by Rb.
Voÿ Rbÿ(æ [æ + 1])
- ÄÄÄÄ = ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
Vinÿ (rpÿ[æ + 2] + Rb)
And finally we multiply through by -1 to obtain,
Vo -Rbÿ(æ [æ + 1])
Av = ÄÄÄÄÄ = ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ (12)
Vinÿ ÿ (rpÿ[æ + 2] + Rb)
The minus sign is because the amplifier inverts the signal and therefore the
gain is a negative number.
ÉÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍÍ»
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