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I'm designing an oscillator, I'm using as textbook Krauss one.

I would like to design the oscillator In function of the power it must give to the load and the fundamental frequency.

I have seen this expression in bjt amplifier design too.

According to this books, the power driven to the load (RL) is:

P = Icq2 × RL/8

So Krauss find the transistor's Q-point with this expression knowing the load and the power he wants.

I have try to deduce this equation and I have got this expression.

Let's assume the current (I) is the current across the load resistor (for example, using a simulator and measuring this signal).

P(rms) = IRMS2 × RL.

P(rms) = Ipeak2 × RL / 2.

P(rms) = Ip-p2 × RL / 8.

It's a similar expression, however it depends on the alternal current across the indictor, not Q-point.

How can I relation this expression with Q-point?


Also I have another question, let's say I have a voltage amplifier (CE). It has a gain of 2.

If the input signal is 1, I would have 2 volts in the output. So my RL with have this 2 volts signal across it, and this would generate a power.

Now I have 2 volts in the input, this will generate 4 volts in RL. So the power will increase.

According to this, the power depends on the input, the amplifier gain and the RL, not Q-point, so how does this expression makes valid ?

P(rms) = Icq2 × RL/8

schematic

simulate this circuit – Schematic created using CircuitLab

Thank you.

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  • \$\begingroup\$ What does 'ICQ' mean? (Explain it in your question, not in the comments.) \$\endgroup\$ – Transistor Mar 21 at 13:55
  • \$\begingroup\$ Q=Xc/R for series and R/Xc for parallel and Xc(fo)=XL(fo) at fo resonance. There must be some assumption for max power at some Q for stability or signal BW with feedback gain overall >1. \$\endgroup\$ – Tony Stewart EE75 Mar 21 at 13:57
  • \$\begingroup\$ Doesn't make all that much sense without a schematic. \$\endgroup\$ – user_1818839 Mar 21 at 14:42
  • \$\begingroup\$ Sorry for the semantic, I'm Spanish, we call ICQ to the Q-point of the transistor's (it's the same for DC and AC) \$\endgroup\$ – Sebastian Ospina Mar 21 at 19:48
  • \$\begingroup\$ then use Icq instead \$\endgroup\$ – Tony Stewart EE75 Mar 21 at 20:52

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