You have to keep the capabilities of the transistors in mind. The approximation of constant B-E voltage is no longer valid. With only 3 Ω on the emitter, the non-linear effects of the B-E junction will dominate. Just number them next time.) is so low that it might as well not be there. If you just want something like a microphone amplifier and the result is more important than the journey, just use a couple of opamps. Using NPN and PNP together, even in the same stage, can have advantages of impedance, stage gain, and distortion. Transistors are cheap and small, so there is no reason to economize the number of transistors. This allows good predictability and flat frequency response.Īnother thing to consider is to use both NPN and PNP transistors together. The feedback ratio, which defines that closed loop gain, can be set by two resistors. A better way is to make more open loop gain, then use negative feedback to set the closed loop gain to what you really want. That is more manageable, but still has problems.Įven with a gain of 17 per stage, the gain won't be all that well controlled. If the gain is balanced between stages, that would mean a gain of 17 per stage. You therefore should be expecting at least 3 stages. That is too much gain to expect a single-transistor stage to have. You want a gain of 5000, so at best that means a gain of 71 from each stage. To be clear and to protect from future edits to the question, here is the circuit you are asking about: However, there will be significant distortion as the output signal level rises due to the non-linearities of relying on \$r_E\$ as a gain limiting factor. This means that the gain of the first stage (with the 2nd stage connected and loading) is about 456/8.6 = 53.įor the 2nd stage, \$r_E\$ is 5.6 ohms and RC2 is 1,000 || 10,000 hence its gain is 909/5.6 = 162.
This is the dominat impedance loading the collector of the first stage and reduces RC1 to an equivalent of 456 ohms. Looking at the DS of the 2N3904 I make an estimate of hFE being about 150 so, reflected \$r_E\$ is about 840 ohms. Using the same calculation for \$r_E\$ we get the AC impedance at the emitter to be 5.6 ohms and this can be reflected to the base by transforming it by hFE.
However, the loading from the 2nd stage across RC1 will drop this somewhat. Add this tou your external AC resistor (RE1A) of 3 ohms and the gain of the first stage is:. your internal emitter resistance will be about 5.6 ohms. So, collector current is 4.7 mA and this gives an \$r_E\$ of 26 mV/4.7 mA i.e. This means emiiter current is about 4.7 mA and collector current is about the same. So, here's the first slight issue - you have 5.38 volts DC on the base and this sets the emitter at about 4.7 volts. This means that with a 1 kohm collector resistor you should design the collector current to be about 7.5 mA. Simulate this circuit – Schematic created using CircuitLabįor any decent amplifier you will want the quiescent voltage at the collector of stage 1 to be about 50% of the supply rail (7.5 volts). Here is log-log plot of signal_in and signal_out levels, to illustrate distortion Thus a whole nother world exists when you care about distortion. Scaled by gain of 5,000, the output is 50 volts peak. Thus for 0.1% distortion IM3, you need an input of about the size -10dBv IP - 3*10dB = -40dBv.
2N3904 BJT SERIES
Now the non-zero Rsource, and the non-zero Re1a, get included in your gain.īy the way, given the 2nd order and 3rd order Intermodulation Intercepts for a bipolar (at any bias current, assuming non-saturated) are both near -10dBv (+- some dBv, but we'll approximate both wrongly as -10dBv), that -10dBv being a peak number used in a Taylor Series model of the emitter diode nonlinearity, and -10dBv is 0.316 volts, and given the 3rd order products drop 20 dBc for every 10dB dropoff from the Intercept Point (that -10dBv), to have 0.1% distortion which is 3 factors of 20dB you need to be 3*10dB below -10dBv IP3. Gain now has dropped by 67% in each stage, We are still ignoring the discrete Re1a of 3 ohms. Finite beta, assume 100, and the low reac of 5 ohms, tells us the Rin is at most 5 * 100 = 500. Total maximum gain, ignoring Re of stage1, and the loading upon stage1 by stage2, etc, will be 200 * 200 = 40,000. The collector resistors are each 1Kohm, thus maximum gain is 1,000 / 5 = 200. Each transistor runs near 5mA, thus reac ( 1/gm) is near 5 ohms.
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