Grid Driven Tetrodes
Grid Driven Tetrodes
Related page:
Screen Supplies
This page
explains how to
calculate resistors
for a griddriven
class AB1 amplifier. These are all class AB1 amplifiers, with grids never going
positive.
C1 is a low
RF impedance
capacitor, roughly 2
ohms reactance or
less on the lowest
frequency.
C2 is a very low
RF impedance
capacitor, roughly
Rg/25 at the lowest
frequency. Must have
a very low impedance
all the way to VHF!
L1 is a small RF
choke with a few
hundred ohms
impedance at the
lowest frequency. It
might be a good idea
to place a small
resistance across
L1.
Rg is the grid
swamping resistor
Rs is the series
resistor. This
resistor might not
always be used. If
Rg is higher than 50
ohms, you will have
to use a step up
transformer or L
network to match the
grid. If RG is 50
ohms or higher, Rs
is omitted.
Alternate
circuit. Not
generally as stable
as the first circuit because the choke is added. L1
should be shunted
with a 2k5k
resistor.
C1 must present a
low impedance all
the way to VHF.
Calculating Grid Resistance
(Please give me
credit for this if
you use it.)
The following
formulas apply to
class AB1
amplifiers.
Where:
Eb = bias voltage
Pd = drive power
in maximum PEP
Ipk = peak
current in grid
resistor
Eb²
/ 2Pd = Rg
For cases where grid
resistance is less
than 50 ohms, use
the following:
The first step finds
the peak resistor
current in a 50 ohm
load…..
(√ of Pd/50) *
1.41421356 =
Ipk
The next step finds
the grid resistor
value….
Eb/Ipk
= Rg
The final step finds
the series
resistance….
50Rg = Rs
For cases where grid
resistance
calculates to be
over 50 ohms, use
the following:
L and C are an L
network. Otherwise
you might be able to
use a broadband
transformer.
Where:
Eb = bias voltage
Pd = drive power
in maximum PEP
Ipk = peak
current in grid
resistor
Eb²
/ 2Pd = Rg
Examples:
A 4CX5000 has 300 volts of grid bias, and you have a ~100W PEP exciter. Eb=300 Pd=80 (to have headroom for variables). Eb²

next case:
We have a grid driven 6146B amplifier with 45 volts of control grid voltage and 2 watts PEP drive from a 12BY7A driver tube. Eb=45 Pd=2
Eb²

Another case:
A tetrode has 35 volts of bias and we have 100 watts PEP minimum drive at 50 ohms.
sqr (100/50) * 1.414 = 2 amps peak current into 50 ohms next: Eb/Ipk 50 – 17.5 = 32.5 ohms for Rs The exciter sees 17.5 + 32.5 = 50 ohms. With 100 watts current is 2*17.5=35 volts. This swings the grid from 70 volts down to zero

Also see Neutralizing Amplifiers