The RF amplifier: circuit values, MOSFET ratings and operational conditions
Although it is possible to figure your RF values by an educated
trial and error (guess and check, using my values as a rough guide),
here are some rules of thumb that will help you get started. The values
are NOT that critical and the class E stage will actually operate with
component variations of 30% or more.
RF Amplifier Load Resistance
Start by figuring the load resistance of the RF amplifier (the load
line). This is slightly more than 1/2 (.57) the calculated DC
resistance (DC voltage / DC Current or DC Voltage squared / power). So,
for an amplifier operating with 40 volts @ 5 amps, the DC resistance is
8 ohms. Your load value is somewhat over 4 ohms (4.58 ohms).
The Shunt Capacitor
The function of the shunt capacitance is to reduce the peak voltage
across the MOSFET when the device is in the off state, and to spread the
width of the "off" pulse. The shunt capacitor is also part of the output
matching network total impedance.
The actual value of the shunt capacitor is important for a couple of
reasons. First, if the capacitance is too small, you will see a
very high RF peak voltage across your MOSFETs. If the value is too large,
the efficiency can suffer. This is not an extremely critical value, and your transmitter
will work over a wide range of values. For standard MOSFETs operating
on 160, 80 and 40 meters, the capacitive reactance
of the shunt capacitor should be roughly 6 to 12 times the load
resistance of the RF amplifier, depending on how the amplifier is driven.
The shunt capacitor value is correct if the peak RF voltage across the MOSFETs
during the "off" cycle is around 3.5 times the DC voltage applied to
the stage. If the voltage is higher than this value, you may need to increase
your shunt capacitor.
These calculations will work for class E amplifiers using sine wave drive. If
you are using square wave drive, you will most likely need to increase the
value of the shunt capacitor by 50% or more.
Using the 4 MOSFET RF amplifier described elsewhere in this document, the
load value is 4.58 ohms.
Figuring the actual
capacitance: Xc = 1/(6.28 x F x C) - or - C in pF = 1000000/(6.28 x
Fmhz x Xc). Solving the equation for a frequency of 3.9 mhz using Xc of
45.8 ohms (4.58 x 10) = 891 pF. The specified value for the amplifier is
750 ohms.
Don't be afraid to experiment a little with this value! MOSFETs
vary, and you may also find you have some reactance - inductive or
capacitive - reflected from your load, depending on tuning.
NOTE: It is possible to operate without this capacitor, if you have
PLENTY of voltage headroom. Without the external capacitor, you can
safely figure your peak voltage will be about 5 to 8 times the DC
applied voltage, depending on the internal capacitance of the MOSFETs,
and exactly how the RF amplifier is tuned. A 1000 volt MOSFET operating
at a *peak* applied voltage of 75 volts can be safely operated with no
shunt capacitor.
Make sure you choose a GOOD capacitor for the shunt
capacitor. The capacitor should have a low inductive component and
should be able to take plenty of RF current.
Figuring the resonant circuit Inductor and Capacitor values
In figuring the circuit values for the resonant circuit, it is very
desirable to use components that can be easily obtained. The circuit
will also be easier to build and adjust if the impedance of the circuit
is not too low or too high.
At resonance, the inductive and capacitive reactance of the input
inductor and the series capacitor will be approximately equal. Class E
does not operate at exact resonance, however it is close enough to use
for calculations. The inductive reactance of the input inductor should
be approximately 8 to 16 times the value of the load line of the class
E amplifier (as figured above). Note if you are going through a
transformer, you will transform this impedance by the impedance ratio
of your transformer. If you have a transformer with a 1:2 turns ratio,
the impedance transformation will be 1:4. Keep in mind, the leakage inductance
introduced by the transformer will also figure into your calculations.
As an example, the 6 MOSFET 75 meter transmitter described elsewhere
in this document has a load of approximately 3.2 ohms [DC values: 45
volts @ 8 Amperes = 5.625 ohms Load: 5.625 * .57 = 3.2 ohms].
There is a 4:1 impedance step up created by the
output transformer, which transforms the 3.2 ohm value to 12.8 ohms.
The approximate reactance of the input inductor and the series capacitor are
150 ohms, or about 11 or 12 times the 12.8 ohm load. Now, getting a value
for the inductor: Xl = 6.28 x F x L therefore L = Xl/(6.28 x F). Putting
in the values we obtained, and assuming a frequency of 3.9mHz (75 meters):
L = 150 / (6.28 * 3,900,000) which works out to 6.125uH.
The series capacitor value may be figured in the same way: Xc = 1/(6.28 x F x C), or
C = 1/(6.28 * F * Xc). Using our 150 ohm reactance:
C = 1/(6.28 x 3,900,000 x 150) = 272pF.
The actual values used in the circuit, of 7uH for the inductor and 250 to 300pF for the
capacitor are very reasonable and easy to obtain circuit values. In
this circuit, the output, or loading capacitor is a 4 section
variable capacitor, with a total maximum capacitance of around 1500.
Under normal operation, this capacitor is set to around 1000pF. Note,
that without the 4:1 impedance step up provided by the output
transformer, the reactance of the input inductor and series capacitor
would be around 40 ohms, which would make the series capacitor need to
be very large! When designing your class E output networks, use
impedance transformations to raise the impedance of your output
network, which will reduce the value and the cost of the variable
capacitors.
Tuning Capacitor Voltage Rating
The series tuning capacitor is subject to very high RF voltage, and
several thousand volts is not uncommon. Using a lower inductance and
higher capacitance in the resonant circuit will reduce the voltage
across the tuning capacitor somewhat. A useful rule of thumb for
figuring the tuning capacitor voltage rating is 2.5 to 3 times the peak
to peak RF voltage fed to the resonant network plus a safety factor. As
an example, you normally expect to see 500 volts peak across your class
E MOSFETs, and you are using a step-up transformer with a 1:2 ratio,
you will see 1000 volts peak to peak across the secondary of the
transformer, which is the input voltage to the resonant network. The
tuning capacitor should have a 2500V or better, a 3000V rating.
MOSFET Voltage Rating
The peak voltage across the MOSFETs is going to be a little less
than 4 x the DC applied voltage for a proper class E transmitter. This
will vary somewhat with tuning and your exact circuit. If you have very
low, or NO shunt capacitor, the ratio of peak RF voltage to applied DC
can be 6x or 8x the DC or MORE. For class e transmitters with proper shunt
capacitors, figure 4x the DC, plus a safety factor. If you're
running 40 @ 5 amps of carrier, and expect to modulate 150% positive,
your power supply voltage is going to be 100 volts. So, your MOSFETs
are going to see 400V at a minimum. Use AT LEAST a 600 volt
MOSFET, and I would personally use an 800V or 1000V MOSFET in this
application.
MOSFET Current Rating
The current rating of the MOSFET should be at least three times the Maximum
DC current expected. Again, if you're running 40 @ 5 amps of carrier,
and expect to modulate 150% positive, your power supply voltage is
going to be 100 volts. Base your calculations based on 100V. So, your
DC current at 150% positive modulation will be 12.5 amps. Figure your
total MOSFET current rating (all MOSFETs in parallel) should be at
least 36 amps.
Your efficiency will be better if you run less current, or use
MOSFETs with a higher current rating (or MOSFETs in parallel). The R
D-S on (Resistance Drain-Source when the MOSFET is on) becomes a big
factor, particularly as currents increase.
