Transistor impedance in circuit - Page 1
Forum section: electronics / beginners

author: AxleD
author_rank: Regular Contributor
timestamp: March 09, 2015, 07:48:01 pm
content: Hi,I was wondering if someone could assist me in understanding what's happening with the attached circuit. It's modeled in LT spice and i am surprised to see that the transistor configuration has no impact on the current flow to the load which is R5 in this case. I would have thought that the transistor configuration had an impedance that would affect the circuit in some way.Thanks

author: max666
author_rank: Frequent Contributor
timestamp: March 09, 2015, 08:01:37 pm
content: You haven't specified a series resistance for the voltage source, so in your simulation the voltage source is an ideal voltage source with zero serial resistance.This means no matter how much load you put on the voltage source, it will always output 8 V. Specify a series resistance (right click on the voltage source), and the current through R5 will vary depending on how much current flows through the transistor.I hope this was what you were asking?

author: Zero999
author_rank: Super Contributor
timestamp: March 09, 2015, 08:10:49 pm
content: Yes, it's a perfect voltage source and the transistor is not in series with the load so it will not change the current through the load.Try connecting a 33R resistor in series with V1.

author: AxleD
author_rank: Regular Contributor
timestamp: March 09, 2015, 08:52:04 pm
content: Thank you both.I am aware that a non perfect voltage source will effectively have a series resistance with the source. But even including that series resistance I was wondering more on the impact of the emitter follower on current flow through to R5.If i assume its in parallel with R5 then shouldn't it impact the total resistance in some way? Or is the resistance of the transistor configuration just so large that the effective parallel resistance of R5 and the transistor configuration is always close to R5?

author: Zero999
author_rank: Super Contributor
timestamp: March 09, 2015, 09:05:38 pm
content: No, the transistor has no control over the current through R5 which solely depends on the power supply voltage.All components connected in parallel will have the same potential difference across them, just as all components connected in series will have the same current flowing through them.If you look at the current flowing into Q1's base you'll also find it's equal to IR3-IR2.Look up Kirchhoff's circuit laws.http://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws

author: max666
author_rank: Frequent Contributor
timestamp: March 09, 2015, 09:18:12 pm
content: Like I said, the voltage source you specified is a perfect one, it will output 8 V no matter how big the load. Which of course every real life voltage source can't do, because they have a series resistance. This is why you're having problems imagining this I guess, because a ideal voltage source is counterintuitive, but very useful for calculating things.The node you ladled ".op" will always be at 8 V. So R5 will always have 8 V across and will therefore always have the same current.It is only when V1 has a series resistance, that the voltage on node ".op" will drop depending on how much load you put on, which in turns will depends on how big the parallel resistance of R5 and the resistance of the transistor path is.

author: krivx
author_rank: Frequent Contributor
timestamp: March 09, 2015, 09:21:53 pm
content: OP's line of thinking is correct for a current source - the current is split between the two paths depending on their resistances.

author: AG6QR
author_rank: Frequent Contributor
timestamp: March 09, 2015, 09:44:12 pm
content: Quote from: AxleD on March 09, 2015, 08:52:04 pmYou don't have an emitter follower, at least not in the sense I've heard the term used.You have connected the emitter directly to the source voltage. So the voltage on the emitter will always be the source voltage.With an emitter follower, the emitter is connected to Vcc through a resistor. This allows the voltage at the emitter to change, and it can "follow" the voltage on the base.Put a resistor on the emitter if you want an emitter follower.

author: krivx
author_rank: Frequent Contributor
timestamp: March 09, 2015, 09:56:13 pm
content: Quote from: AG6QR on March 09, 2015, 09:44:12 pmThis is an NPN - the voltage source is connected to the collector, not the emitter. It could be called an emitter follower (the emitter will be follow the base voltage minus Vbe).

author: AG6QR
author_rank: Frequent Contributor
timestamp: March 09, 2015, 10:44:08 pm
content: Quote from: krivx on March 09, 2015, 09:56:13 pmSorry, my bad, you're right. Forget what I said.If R5 had been connected to the emitter, then the emitter follower would have a strong influence on the current through R5.But since R5 is connected directly to Vcc...

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 09, 2015, 10:55:30 pm
content: There is a theorem which states:Anywhere you have multiple branches connecting to a voltage source, each branch can be powered independently from an identical voltage source.Or, identical supernode voltages (chains of voltage sources which sum to the same nodal voltages) can be merged together under one source.(Conversely, current sources in parallel add, and current sources through multiple nodes in series can be split up into isolated meshes with separate (but equal) current sources, and so on.)So, your bias resistors can be supplied from one voltage source, the transistor collector from another, and R5 from a third, without losing any truth in the circuit whatsoever. Or you can split the grounds instead of the supplies, to the same end.Tim

author: AxleD
author_rank: Regular Contributor
timestamp: March 10, 2015, 05:58:03 am
content: That clears up my issues.The source resistance was the missing link and after adding it to the circuit things became clearer. At last the resistance for the transistor configuration of (Beta +1)*RE does get you to the correct values so I'm happyVery useful tips all round on how to think about voltages around a circuit and makes life allot easier.Cheers