5

... I need a more concrete definition showing also how it is obtained Since the base of the BJT is nailed down hard (zero impedance voltage source), the dynamic output impedance is (you can find the starting equation at this Wiki page on the BJT and the Ebers-Moll model): $$\begin{align*} \operatorname{D}\,I_\text{E}&=\operatorname{D}\left[I_\text{ES}\...


3

For the purposes of this qualitative explanation, you can replace the emitter resistor Re with the varying load R(L). Here is my intuitive explanation: In the circuit of the emitter follower, the input base-emitter voltage of the transistor is formed as a difference between two single-ended voltages - the input voltage VIN and the output voltage VOUT. The ...


2

I am assuming that since you say you are observing the S-Parameters you have access to a Vector Network Analyzer (VNA) or at least a Scalar Network Analyzer (SNA). Your musings about S-parameters are not all correct. There are several things you should be aware of. When you're using the "a" and "b" coefficient convention instead of voltage ratios \$a_1\$ is ...


2

Just use a MOSFET that has a rated RDson at your drive voltage, NOT the gate threshold voltage (which is when it just barely starts conducting). In this case, even the gate threshold voltage is far above 3.3V. So it seems like you just weren't looking when you picked the MOSFET. And if you're willing to put that much effort into the method you've dreamed up,...


1

Rout = Re//re Where re = the intrinsic emitter resistance of the transistor = 25mV/Ie (This simplification ignores any out output impedance of the battery, which will be small). The output impedance,Rout forms a potential divider with your variable resistor,R. This potential divider is between the emitter voltage (VB-0.7V) and ground.


1

simulate this circuit – Schematic created using CircuitLab Figure 1. The internal layout of the output voltage measurement instrument and the output impedance, RS. The impedance of the output is always 50Ω. For a load impedance of 50Ω, the displayed voltage is half the output voltage to reflect the voltage seen at the load. I suspect that they are ...


1

It depends on what you're driving the circuit with. Your noninverting arm has infinite input impedance, and your inverting arm has relatively small input impedance. This could be problematic, but not in every situation. For example, if your input is some sort of piezo-device, you will be very unhappy. The input impedance of the inverting input is \$330R \...


1

Are such cables able to be cut and connected (soldered) to the connectors pictured, and used without effect on their performance. Can't tell you, because we don't have a spec sheet to these connectors. USB2 is a (pseudo)differential bidirectional bus, and it relies on the cables being of defined impedance. If your connectors are designed to keep that ...


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