# Gain equation of cascode common emitter amplifier with an LC tank

I'm trying to find the gain equation of the circuit presented below.

I know that for a cascode common emitter amplifier without resistor RE and with a current source instead of the LC tank, the gain is equal to:

$$A_V = \frac {V_{out}}{V_{in}} = -g_{m1} g_{m2} r_{o2} \cdot (r_{o1} || r_{\pi 2})$$

How does resistor RE and the LC tank modify this equation? And what is the purpose of resistor RE?

RE reduces and stabilises gain of the bottom transistor.In fact gm becomes the reciprocal of RE .Example RE = 1K ohm so gm is 1 milliamp per Volt which is the reciprocal of 1 Volt per milliamp .Now the LC tank has a high impedence at resonance which is dependant on Q .100K was my calculated tank impedence at resonance on a 455KHz If strip but your job is different.So the gm is multiplied by the tank impedence to get the voltage gain .Voltage gains of more than 1000 are possible .

What do Re, an RLC tank or a single resistor have in common ? They're all impedances. Some are frequency depedent, some are not (actually they all are but let's keep things simple).

Would you be able to find the gain of this circuit if the RLC tank at the collector of Q2 was a single resistor Rc ? Now replace Rc by Zc(s) which is the impedance of the RLC tank.

When Re is 0 (zero) what would be the transfer of input voltage to output current of the stage formed by Q1 ? Make the small signal equivalent circuit. Now add Re, what changes ?

For a small signal analysis, we ignore DC values and we say that Re has an AC voltage across it that is nearly equal to the AC voltage on the base. If the circuit were an emitter follower, it would have a voltage gain of about 0.99. We're not interested in the output being across the emitter resistor but, it's still important for what follows.

We make a reasonably accurate assumption that for AC signals, Vb = Ve

It therefore follows that the AC current thru the chain of transistors is the base AC voltage divided by the emitter resistance. This AC current flows thru to the collector load impedance (cascode or otherwise).

Because the AC current is the same thru the emitter as the collector you can say (with a reasonable degree of accuracy), that the gain of the amplifier is Rc/Re. But what is Rc or Zc?

Going back to the op's circuit, if Rd were zero ohms, the parallel LC AC impedance in the collector is infinite but it would be plain daft to say the gain was infinite because there are always losses so, work out what the impedance is at resonance and use that as equivalent to Rc in the gain equation mentioned above.

You might also take into account that a BJT in its active region might have a parallel resistance of about 20 kohm (generalism alert): -

Hint: the 20 kohm value is the flattest part of the characteristic shown above in the active region. On this particular graph it looks like 20V/1mA = 20 kohm.