Please explain this transistor buffer :) - Page 1
Forum section: electronics / beginners

author: Sigmoid
author_rank: Frequent Contributor
timestamp: March 09, 2015, 03:58:03 pm
content: Not really wanting to invest into a good triple-output lab supply, and neither wanting to buy a crap one, I decided to build a lab rail splitter so I can use my single-output one for breadboarding audio circuits with minimal extra boilerplate.In looking for designs, I found this here...I'm a bit confused as to the exact rationale behind the push-pull buffer consisting of four transistors. I'm familiar with totem pole output stages, and this looks like a variation on one...What is the advantage this provides over a simple totem pole layout with two transistors?

author: c4757p
author_rank: Super Contributor
timestamp: March 09, 2015, 04:04:03 pm
content: Q2 and Q4 are just diode-connected:It's a slightly unusual way of drawing them, so maybe it wasn't obvious.These provide two copies of IC1A's output signal, one shifted up by a VBEthreshold and one shifted down. This closes the gap that the naive push-pull circuit has where neither transistor is conducting, eliminating the crossover distortion.

author: Sigmoid
author_rank: Frequent Contributor
timestamp: March 09, 2015, 04:11:48 pm
content: Quote from: c4757p on March 09, 2015, 04:04:03 pmThanks, got it!

author: madires
author_rank: Super Contributor
timestamp: March 09, 2015, 04:16:09 pm
content: Looks like a current mirror with different transistors to get a higher output current.

author: Zero999
author_rank: Super Contributor
timestamp: March 09, 2015, 08:21:16 pm
content: Looks can be deceiving.Yes, the transistors are connected as diodes.You may ask why not just use diodes? Well you can. Quite often transistor B-E junctions have better characteristics than diodes. Another advantage is matching which I suspect is the case here. Is Q2 and Q4 are mounted in good thermal contact with Q1 and Q3 and are the same type of transistor, then their voltage drops will be similar.

author: Sigmoid
author_rank: Frequent Contributor
timestamp: March 09, 2015, 09:04:32 pm
content: Thanks, I've been wondering about that. It actually makes quite a lot of sense now that I think about it.

author: JohnnyBerg
author_rank: Frequent Contributor
timestamp: March 09, 2015, 09:09:47 pm
content: I agree with @madires: 2 mirrored current sources!http://en.wikipedia.org/wiki/Current_mirror

author: Paul Moir
author_rank: Frequent Contributor
timestamp: March 09, 2015, 09:15:03 pm
content: Would you mind pointing toward where you found it? I was planning on doing the same thing for the few times I needed it.

author: JohnnyBerg
author_rank: Frequent Contributor
timestamp: March 09, 2015, 09:30:56 pm
content: Quote from: Sigmoid on March 09, 2015, 03:58:03 pmI think the objective here is to create a virtual ground, that cansinkandsourcecurrent!Disadvantage of this circuit is that the transistors must be reasonable matched, otherwise sink current spec will differ from source current spec.

author: dannyf
author_rank: Super Contributor
timestamp: March 09, 2015, 09:53:28 pm
content: The diode connected transistor are there to provide bias and comparable temperature coefficients.Contrary to popular belief, they are not current mirrors.

author: krivx
author_rank: Frequent Contributor
timestamp: March 09, 2015, 10:08:02 pm
content: Wow... deleted posts?IMO calling those current mirrors really depends on the matching. If the bias transistors are small signal and the output transistors are large power devices then current is not exactly "mirrored".

author: dannyf
author_rank: Super Contributor
timestamp: March 09, 2015, 10:57:19 pm
content: QuoteEven if those transistors are perfectly matched, they are not current mirrors.

author: krivx
author_rank: Frequent Contributor
timestamp: March 09, 2015, 11:02:17 pm
content: Quote from: dannyf on March 09, 2015, 10:57:19 pmWhy? Because the emitters are not connected? Shouldn't the opamp keep them at the same voltage? Assuming matching that is.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 09, 2015, 11:02:33 pm
content: Specifically, the quiescent bias current is set thanks to mirror action. This depends on the transistors being matched, so don't expect too much drive power from it, or you'll thermally unmatch them, and wailing and gnashing of teeth will ensue.I suppose you could add another pair of transistors to make a Wilson-esque circuit, which would at least limit the voltage on the "sense" transistors, if not the current. Gives you another maybe 10x of current before runaway.Tim

author: dannyf
author_rank: Super Contributor
timestamp: March 09, 2015, 11:15:08 pm
content: A current mirror only holds when the two legs are "reasonably" symmetrical. That assumption is brutally violated here.Another way to look at it is to think about the goal of the circuit: a current amplifier. If the pairs worked like current mirrors, there would be no relief for the opamp: the opamp sees just as much load as the output emitter follower.All the diode-connected bjts do is to provide a "perfectly matched" Vbe, both in terms of Vbe levels and its temperature performance.

author: c4757p
author_rank: Super Contributor
timestamp: March 10, 2015, 03:23:15 am
content: Not a current mirror by far for practical purposes, only if these are transistors in SPICE-land... the output transistors are under significantly different conditions...Now, if these werethermallymatched pairs, then they might be closer. Still not quite what most people would consider a mirror.

author: dannyf
author_rank: Super Contributor
timestamp: March 10, 2015, 11:28:31 am
content: Since it is likely non-battery powered, you may also consider a diamond buffered here (either IC or discrete): it is much faster than the class B buffered used here.Another approach, if the opamp can provide a little bit of current, is to eliminate the diode connected bjt and use just a resistor to bias the output devices.

author: madires
author_rank: Super Contributor
timestamp: March 10, 2015, 11:57:28 am
content: Quote from: dannyf on March 09, 2015, 11:15:08 pmI disagree, a current mirror could also have other ratios than1:1. If both transistors are matched you'll get the 1:1 ratio, but if you select two different transistors (same technology, different area sizes for the emitter) the current mirror will have a ratio similar to the ratio of the area sizes.Quote from: dannyf on March 09, 2015, 11:15:08 pmGot it.

author: Sigmoid
author_rank: Frequent Contributor
timestamp: March 10, 2015, 01:56:58 pm
content: Quote from: T3sl4co1l on March 09, 2015, 11:02:33 pmHuh?! That sounds really weird. The whole point in having a transistor output stage is to have a crapload of current source / sink capacity. The output of the whole thing is fed back into the op amp. Isn't that supposed to circumvent any nonlinearities?The whole reason we invented op amps was so we don't need to worry about matching transistors.I'm pretty sure that c4757p's solution is the correct one, those transistors are wired as diodes, and the rest is a regular op-amp regulated push-pull.

author: krivx
author_rank: Frequent Contributor
timestamp: March 10, 2015, 02:05:02 pm
content: Quote from: Sigmoid on March 10, 2015, 01:56:58 pmI think most of us agree on this. It's just that under electrical and thermal matching conditions the transistor pairswouldfunction as current mirrors. Clearly there is no reason that this would actually be the case, as it would not be beneficial for this circuit.If the part numbers on the original circuit were legible this discussion would probably be shorted, I doubt the bias transistors are the same part as the outputs.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 10, 2015, 05:31:15 pm
content: Quote from: Sigmoid on March 10, 2015, 01:56:58 pmIs a mirror not simply "wired as diodes" too?Note I emphasizedquiescent. When there is load, the "mirror" is unbalanced and the emitters are at different voltages.Tim

author: dannyf
author_rank: Super Contributor
timestamp: March 10, 2015, 06:57:22 pm
content: QuotePutting aside Vbe differentials caused by Ib differentials, the emitters sit at substantially the same potential, regardless of load.