Simple Transistor Impedance Understanding - Page 1
Forum section: electronics / beginners

author: spitta
author_rank: Newbie
timestamp: February 23, 2019, 07:50:32 pm
content: I'm struggling understanding a fundamental transistor impedance equation.See pictured: common emitter amp with R1 and R2 divider at the baseWhen there is a voltage divider at the base of a transistor, they are considered "in parallel".Furthermore, the input signal sees R1 (110kΩ), R2 (10kΩ), and the impedance looking into the base in parallel. I've accepted this for years, but I don't truly understand it. WHY ARE R1 AND R2 CONSIDERED "IN PARALLEL"? They are not in parallel. In the picture, R1 is connected to 20V rail, while R2 is connected to ground. They only share one connection point. They are not in parallel... right?

author: billbyrd1945
author_rank: Regular Contributor
timestamp: February 23, 2019, 08:02:18 pm
content: The odds of my being able to answer any question on an electronics forum is near zero. However, I think I can answer the part about R1 and R2. They are separated by a node (actually two nodes). I'm pretty sure that that alone would constitute a parallel configuration.

author: spitta
author_rank: Newbie
timestamp: February 23, 2019, 08:09:08 pm
content: Well, series configurations also share a node. Parallel resistors must share both nodes.I think there is a fundamental assumption with basic transistor configuration equations that I am missing.

author: IanMacdonald
author_rank: Frequent Contributor
timestamp: February 23, 2019, 08:23:46 pm
content: R1 and R2 are in series as far as the DC supply is concerned. As far as an AC signal is concerned they are in parallel because R1 returns to ground via the supply decoupling capacitor. The fact that the top end of R1 is not at 0v makes no odds to the AC signal.

author: magic
author_rank: Super Contributor
timestamp: February 23, 2019, 08:54:07 pm
content: They aren't parallel. But the junction between them behaves exactly the same as if they were two parallel resistors connected to 1.66V.1.66V is the voltage at the junction if no current flows from the divider to anywhere outside.If an external load pulls the voltage down by 1mV, the top resistor will source 1mV/110k more current into the junction and the bottom will sink 1mV/10k less. So the junction will source 1mV/110k+1mV/10k into the load. Just like two resistors in parallel with 1mV across them.See Thevenin's theorem.

author: Zero999
author_rank: Super Contributor
timestamp: February 23, 2019, 09:53:19 pm
content: Quote from: spitta on February 23, 2019, 07:50:32 pmThe impedance seen at the node where they meet is equivalent to their values connected in parallel, both at DC and AC, because the power supply has an impedance of close to zero.Look up Thevenin's theorem.http://hyperphysics.phy-astr.gsu.edu/hbase/electric/thevenin.htmlProof:Work out the voltage, using the potential divider formula.VOUT= VIN*R1/(R1+R2)Calculate the output resistance, which is R1|R2.ROUT= (R1*R2)/(R1+R2)Note that when the output is shorted to 0V, the current is equal to VIN/R1 and VOUT/ROUT.Take another resistor, with the same value as ROUTand connect it to the output of the potential divider. Work out the output voltage and you'll find it will be halved.And please do some image processing on photographs of schematics. It makes them much easeir to read and saves bandwidth.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 23, 2019, 10:08:24 pm
content: To do an AC small signal analysis:- Calculate the DC operating point (this is used to figure out nonlinear (diodes, transistors..) elements' incremental resistance)- Short out all non-dependent voltage sources, and open all non-dependent current sources.- On the resulting network (which is linear RLC + dependent sources), solve for whatever values you're looking for, as you usually would (nodal analysis, superposition, whatever).Voltage sources link two nodes together into what is called a supernode, which means they can be taken as identical for purposes of AC analysis, or that certain DC analysis steps can be done (like splitting two parallel branches off, each with its own voltage source linked by ground, because, same voltage, same nodes, it's all one big supernode).Another way to think of it is, suppose you inject a small current into that node. Does it matter where the current comes from (GND or +20V)? The current is divided among the components draining that node: R1, R2 and the base. The current through R1 goes to +20V and returns through that voltage source to GND, the same place R2's current is returned. So, for that small current injection, what is the small change in voltage, and the ratio thereof, Ri = dV/dI? Exactly the parallel combination of the three.Tim

author: rstofer
author_rank: Super Contributor
timestamp: February 24, 2019, 07:52:00 am
content: Quote from: spitta on February 23, 2019, 07:50:32 pmThe short answer is that the power supply has an impedance of, say, 0 Ohms. So, draw a short circuit from the top of R1 to the bottom of R2 (Gnd). Now you can fold the circuit around until it is clear that the resistors are in parallel.

author: Zero999
author_rank: Super Contributor
timestamp: February 24, 2019, 09:01:35 am
content: Proof using LTSpice. Connecting a variable resistor to both the original circuit and its Thevenin's equivalent produces the same output voltage for a given load resistance.

author: RoGeorge
author_rank: Super Contributor
timestamp: February 24, 2019, 09:20:44 am
content: Quote from: spitta on February 23, 2019, 07:50:32 pmThe internal resistance of the 20V voltage source is zero ohms (ideally zero, in practice very very low, close enough to an ideal zero). So, there you have it: If you draw a shorting wire between ground and +20V (the voltage source acts like a short-circuit wire), the two input resistors (R1 and R2) are now connected in parallel.