Class-B RF Amplifier

Hi All,



I wonder how to choose the resonance circuit values (L and C) to get the maximum gain of calss-B amplifier made for FM transmitter (100 Mhz).

I will use C2570 transistor.

Ampere flow is greater with higher C and smaller L.

Ampere flow is lesser with smaller C and larger L.

A large L:C ratio tends to create larger voltage swings.

If available current is small, then choose larger L. Calculate C accordingly.

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A suitable place to begin, is to choose L & C values so their reactive impedance is the same order of magnitude as neighboring resistance.

XC = 1 / ( 2 Pi f C )

XL = 2 Pi f L

where X is reactive impedance.

If you want maximum gain then remove R1, it reduces the gain and increases the input impedance so you get less Ib for the same amount of Vbe. or readjust the input circuit for better impedance matching.
Just using the LC circuit means that the collector output impedance is shunting it so it is damped also the load is damping the LC circuit.
You need to define your load impedance first, then match it to the output impedance of the transistor, which will be Vcc X2 /Ic . So there are several ways of doing this, tapping down the L, splitting the C into two parts and making a capacitive tap. Coupling a link coil over the L. . .
Once you have established the two impedances across the tuned circuit, you then need to put a figure in for the working Q, for the output of a transmitter, this is usually 10. If the working Q is higher then the losses associated with the tuned circuits will be higher. The dynamic impedance of the tuned circuit at resonance will be L/Cr, where r is the loss resistance of the unloaded tuned circuit (ωL/100 ?).
So now you have a handful of figures, just use some trial calculations to optimise your results.
Frank

This not Class-B amplifier, it's Class-C.
Q factor of a parallel resonance circuit is Q=(1/R)*SQRT(C/L)
Here R represents all parallel loads including the load,output impedance of the transistor,other L-C loss factors.
If you choose Q large, the amplifier will have narrowband ( it's preferable because Class-c amplifiers are very nonlinear amplifiers and they produce large harmonics.) or if you choose Q small, the amplifier will be wideband amplifier that is not very desired due to the reason mentioned above.
Q factor can be determined by playing around it, try ( simulate) and see..

I agree it operates in class-C because it is cutoff at idle because it has no bias.

Thanks all for the support.
I think you are right about it is class C not class B since there is no biasing to the transistor.

I think this should be my modifications.
1- no resistance at emitter
2- biasing to the transistor

I think this circuit has the same schematic that i think of :


in the above circuit the author stated to tune the VC2 to get maximum gain.

I don't actually understand why VC3 is needed and why should we tune it ?!

don't we need a matching impedance before the antenna to maximize the output power ?

mr_byte31 said:

Thanks all for the support.
I think you are right about it is class C not class B since there is no biasing to the transistor.

I think this should be my modifications.
1- no resistance at emitter
2- biasing to the transistor

I think this circuit has the same schematic that i think of :
View attachment 113659

in the above circuit the author stated to tune the VC2 to get maximum gain.

I don't actually understand why VC3 is needed and why should we tune it ?!

don't we need a matching impedance before the antenna to maximize the output power ?

Click to expand...

This is not same circuit because this one works as Class-A ( I presume the transistor has a right bias)
VC3 is used to tune antenna-circuit matching ( simple match ) and therefore it should be adjusted to get max. power AND less harmonic contents because VC3 is in a resonate with antenna effective reactance.( Probably inductive )

Your aerial will show a resistive impedance of about 75 ohms. If you PA transistor takes amps of Ic its output impedance will be in the order of ohms, if the PA takes a few milliamps, then its output impedance will be in the order of kilo ohms.
So if its the first case the transistor collector should be connected to a tap on the inductor near the Vcc end, if its the second case, the aerial should be taken to this tap instead. VC3 does not do a lot!
Frank

Hi Chuckey , I didnt get your point , sorry

If you PA transistor is taking , say 2A when the drive is taken to its base, then the impedance it will show in its collector circuit, would be 24/2 ~ 12 ohms, this should then be matched to the load (75 ohms) so the collector impedance needs to be stepped up. if you regard the L as a transformer, the impedance match must be 12/75 ~ .16, or .4 in terms of turns ratio. So if the coil has 10 turns , the PA collector must be connected 4 turns down from the Vcc end. This way the maximum amount of power will be fed to the aerial. The capacitor is connected across the L to tune it. This should be tuned for minimum collector current.
if you PA transistor is drawing , say 2mA with the drive on, its output impedance would be 24/2 K ohms, so the sums must be done , but this time the AERIAL would be connected to the tap on the coil.
Frank

Hi Frank,

I would try to make simulations on NI and check that, thanks for the support

Hmmmm,I built the circuit on NI , I want to simulate the Antenna to check the matching impedance!

shall i put it as resistor or how it should it look like ?

If the aerial is a quarter wave length long which it seems to be (F = 100 MHZ wave length = 3m, .75m = quarter wavelength) then it can be a 75 ohm resistor.
Frank