MRF 240 transistor with how much RF is exited at the base
arme but in datasheet I do not understand the specifications well so I ask if someone who is with this can help me I do not want to burn the transistor this that I do not know well is expensive to burn it so why yes
let's see some calculations
if the datasheet says, that the collector current can be in DC (continuous) up to 8A, and has a typical gain of 70, it means that the base current must be 8A / 70. Of course that will really depend on which part of the gain curve you are going to make it work, obviously that you are going to make it work just above the cutting current so that at rest there is almost no current, and that at most you will send it up to 8A / 70 = about 114 ma, but calculate a safety margin, I would give it 30%, I would not run it above 80ma at the base.
It has a good heat sink and cooler I also don't want to spend anything more capable at the entrance I'm asking it wrong
I assembled this RF circuit for the first time with this transistor I have as excite it up to 8wats in the input but I can burn it because there is a lot and only need 1wats to work
I am feeding it with almost 11 volts of power but at the entrance I want to know how many wats this transistor needs to work
When talking about transistors, talking about currents, transistors amplify currents.
I think in the tables I saw that there were examples of 1 2 and 4W, there I looked closely
What you are looking for is Figure 3 on page 3. Tagged as "Output power vs input power". horizontally the power applied to the base, and in vertical the power developed in the collector.
Of course, that table will depend on the correct choice of collector supply voltage. That will depend on the power to develop maximum desired.
Oh, and remember this .. that you should consider the maximum peak, with respect to the base ... the DC component + the amplitude of the RF at its upper peak (positive peak) .... that gives you an instantaneous value , q multiplied by the gain, would generate the collector current, and the collector current multiplied by the collector voltage results in the developed power.
(BEC + signal value current) * gain = IC * VCC = output W
If you consider a gain of 10dB, with 4W at the input it will give 40W at the output.
You must also take into account the loss of the entrance adaptation system so 4W will not reach the base and not burn it.
Greetings.
Ric.
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The vast majority of the necessary data is in the data sheet, not only must be cited but also read.
From to sheet Nº: 3
decibels is a logarithmic measure, if the theory in my head does not fail me, each decibel point means that the amplitude is twice that of the previous decibel.
Example
If 1db = original signal
then 2db = double
then 3db = the quadruple
... 4db = 8 times
... 5db = 16 times
... 6db = 32 times
... 7db = 64 times
... 8db = 128 times
... 9db = 256 times
and thus, it is a unit of logarithmic and nonlinear measurement.
Anyway, correct me if I'm wrong ...
Therefore speaking in terms of DB is not as simple as multiplication.
And of course the limit will be the VCC saturation
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