# Grid Driven Tetrodes

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explains how to
calculate resistors
for a grid-driven
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 2k-5k
resistor.

C1 must present a
low impedance all
the way to VHF.

### Calculating Grid Resistance

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

(√ of Pd/50) *
1.41421356  =
Ipk

The next step finds
the grid resistor
value….

Eb/Ipk
= Rg

The final step finds
the series
resistance….

50-Rg = 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
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 / 2Pd = Rg     so  300^2/ 160 =   562 ohms for Rg.

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 / 2Pd = Rg  so   45^2 / 2 = 506 ohms  This means the grid impedance would be about 500 ohms. The 12BY7 would have to match that impedance.

Another case:

 A tetrode has 35 volts of bias and we have 100 watts PEP minimum drive at 50 ohms. (√ of Pd/50) * 1.41421356  = Ipk sqr (100/50) * 1.414 = 2 amps peak current into 50 ohms next: Eb/Ipk = Rg   = 35/2 = 17.5 ohms grid 50 – 17.5 = 32.5 ohms for Rs The exciter sees 17.5 + 32.5 = 50 ohms. With 100 watts current is 1.414 amps RMS or 2 amps peak, so peak grid voltage is 2*17.5=35 volts. This swings the grid from -70 volts down to zero volts, the grid never goes positive. Also see Neutralizing Amplifiers