The operation of an EF amplifier constitutes a form of negative feedback. The collector is generally connected directly to the power supply and the emitter is connected to another supply voltage - often ground - through an emitter resistor, R E. ![]() This is the sense in which the output voltage at the emitter “follows” the input. Unlike the CE amplifier, the output voltage and input voltage of an EF amplifier are in-phase with each other and of nearly the same magnitude. As with the CE amplifier, we will study the EF amplifier using a single transistor biased in a Class A configuration. “Emitter-follower” is more descriptive than “common-collector,” and will be used henceforth for this reason. The name “emitter-follower” originates from the fact that the output signal, taken at the emitter, follows the input signal, applied at the base, with nearly unity gain. ![]() Emitter-follower amplifiers are commonly used as output stages that are capable of driving low impedance loads due to their current gains and low output resistances. The emitter-follower, EF, also called common-collector, CC, amplifier provides nearly unity voltage gain, and current gain, which can be large, and low output resistance. It is WRONG.The common-emitter amplifier introduced in the “Class A NPN Common-Emitter Amplifier” lab provided voltage and current amplification, but suffered from a large output resistance that was equal to the equivalent collector resistance that was present for AC signals. Re' which you calculated is in series with a parallel combination of the bias resistors (if we assume the source Z to be high) so the number 25 ohms is a wonderful calculation but does not solve the physical schematic problem. Apart from fact that hybrid pi analysis really doesn't work well especially in terms of the accuracy of the r-pi resistor. This problem is why I came here for inspiration.īimpelrekkie - fundamental flaw in argument. That's a sanity check on the answer, it still has to be calculated. The parallel combination is 373 ohms, only a little less than RE. hoc is 117E-6 Siemens from LTspice simulation. I have just been doing this for a 2n 2369 cct where the Ic is 6.16 mA dc. You may need to use LTspice to plot the family of C for you because most dsheets are light on for this data/curve set. Hoc= hoe which is the slope of the Ic versus VCE curve at the bias current in question. Ic will control what hoc turns out to be so the bias resistors in the input circuit are involved in the output calculation as long as hoc is calculated. Note that the current controlled current source on OP has infinite small signal impedance because it drives the programmed current into the load regardless of the small signal load, within limits. INvert again and we have the output Z in resistance units. If we invert RE to convert an admittance and add to hoc (BUT Measured at the quiescent collector current of the circuit, not some arbitrary datasheet value) then we have the output admittance. ![]() ![]() Is in parallel with the parameter hoc which is normally in admittance units. He might be assuming an absolute value for hfc but this gives me pause.Īt the risk of stating the flaming obvious, the RE across which the OP signal is derived Hfc = - (1 hfe) which gives him a negative Ze at first glance. The formula or Ze above is ignoring a couple of things. You might also want to look at this derivation which does not use the \$h\$ parameters. At \$I_c\$=1mA that would give 25 ohms, which is generally much lower than the value of \$R_L\$ The output impedance is an indication of how much the voltage at \$V_\$. The trick with the CC circuit is that there's local feedback going on.
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