Need help determining polarity of a circuit - Page 1
Forum section: electronics / beginners

author: GeoNOregon
author_rank: Newbie
timestamp: March 07, 2015, 09:41:54 pm
content: Hi,I working on a programmed learning text covering the basics of electronics. So far the book has covered:Intro to ElectronicsPrinciples of Magnetism & ElectricityResistors & Resistive CircuitsInductance & TransformersCapacitance & CapacitorsResonance & FiltersI am now on Diodes & Transistors and have come up against a learning barrier. I'm at a discussion of bipolar transistors and forward or reverse bias in their circuitry. Below is a diagram of the circuits that are causing me trouble.In order to determine the circuit's bias, you have to know the polarity of the Emitter, Base and Collector legs. It's obvious the Emitter leg is positive and the Collector leg is negative, but I've found no information in the previous chapters of the book, nor is there any info in the current chapter about how to determine the polarity of the Base leg when one circuit joining the Base leg is positive and the other circuit joining the Base leg is negative.I haven't been able to find any rhyme or reason to the answers the book gives on these circuits and I AM stuck.This circuit is an introductory circuit to the concepts, further in the book are circuits with values for the voltages & resistors, but I still can't consistently get the Base leg polarity correct.Anybody have an explanation, rule of thumb, etc that can get me past this problem?Thanks, in advance.GeoD

author: Anand
author_rank: Contributor
timestamp: March 07, 2015, 10:04:04 pm
content: I think you're going too fast with your reading.My advice to you is to slow down and at every chapter, do some circuit simulations, using:http://www.falstad.com/circuit/http://everycircuit.com/https://www.circuitlab.com/Everycircuit has an app for android. The first two are interactive simulators, you'll be able to see effects in realtime when you modify your circuit.Start with that and I promise that you'll get better.

author: JohnnyBerg
author_rank: Frequent Contributor
timestamp: March 07, 2015, 10:44:15 pm
content: Your drawing is somewhat unconventionalWith the proper values of R1 and R2 the base voltage is 0,7V below the emitor voltage.

author: GeoNOregon
author_rank: Newbie
timestamp: March 09, 2015, 03:39:30 am
content: @ Anand: Trust me on this, I couldn't be going any slower or methodical with this. The book I'm working through right now is a programmed learning text. The format is something like this:There is a paragraph of 2-3 sentences ending with a fill in the blank statement.The beginning of the next paragraph gives the answers to the previous paragraph's blanks.Some of the fill in the blanks require computations/calculations and some are just concepts or terminology.At the end of each chapter there are 20-30 questions about the material with the answers in the back of the book.There isn't really any way to 'go too fast' with the reading.The point I'm stuck at, I can't move on to where I would have values to actually run a simulation. I tried forging ahead to later in the chapter and the chapter ending questions and I didn't figure it out.Thanks for the info on the simulators, though, I'll check them out. Maybe using them I can't figure out how to determine the polarity.@ JohnnyBerg: Well, I'm actually glad to hear you say the drawing is 'somewhat unconventional', I copied it straight out of the text! I thought it was unconventional to write an 'Intro to...' programmed text and then assume knowledge of the learners.The 0.7V difference between the emitter and the base is something that's been covered & I'm familiar with it, but I don't have this chapter down yet, because of this polarity issue. So, while what you mention is familiar, I'm not conversant with it.Thanks..

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 09, 2015, 05:50:00 am
content: If you look at the transistor as two separate PN junctions, then:Emitter is forward biased (+ to P, - to N)Collector is reverse biased (- to N)Base is just what's common between them.In practical terms, voltages are usually referenced to the emitter, so while it may be correct to say the above, it's more conventional to say "the base is forward biased" instead of the emitter.In normal operation, the collector is always reverse biased.The tricky part about a transistor is, it's not simply two diodes glued together; magic happens when the junctions are brought together, very closely. On the order of microns. Current flow in one junction induces current flow through the other, even if the one is reverse biased.The transistor is apparently palindromic, so you could interchange C and E; this is called inverted operation. However, it doesn't usually work very well, because transistors can be constructed asymmetrically (and usually are -- typical of the double and triple-diffused processes). Some special purpose transistors are made symmetrically (or nearly so), as well as most antique transistors (which are typically "alloy junction" germanium types). "Not very well" means: lower breakdown voltage (e.g., for 2N3904, Vebo = 7V, versus Vcbo = 60V!) and much lower hFE (typically under 5).Tim

author: GeoNOregon
author_rank: Newbie
timestamp: March 17, 2015, 03:28:57 pm
content: Hi Tim,Thanks. Could you expand a bit on "Base is just what's common between them."This is all so new to me, so I have no frames of reference. As goofy as the schematic might be, it's a common format through-out the course book. The author uses these simplistic drawings to introduce concepts, then moves to the same simplified schematics, but adds values for components and voltages, then to conventional schematics.The method he is teaching for figuring out trans bias on paper is to first determine the polarity of the leg, then based on whether the B, E, or C is anode or cathode, the bias is determined.Specifically, what I'm stuck on is the leg coming from R1 to the junction point is negative and the leg coming from R2 to the junction point is positive. They meet at the junction point with the Base leg and... Which wins out, negative or positive?The author gives no values for the resistors or voltage supplies, so higher/lower resistance or higher/lower voltage can't be one of the determining factors.I'm sure when it finally clicks for me it will be a 'slap of the palm to the forehead', but right now, it's a blank.GeoD

author: TimFox
author_rank: Super Contributor
timestamp: March 17, 2015, 03:56:55 pm
content: The base is a region between the emitter region and the collector region.When "majority carriers", positive-charged holes for the PNP example here, are injected into the base region by emitter current (the emitter-base junction is forward-biased), they feel the strong field from the reverse-biased base-collector junction and mostly flow into the collector region, where they can do useful things in the overall circuit. The small fraction that do not flow into the collector region form the base current. The ratio of the collector current divided by the base current is larger than one, and is called the "current gain" or "beta" of the bipolar device.In the common-emitter circuit (most popular), there is a small current and small voltage at the base-emitter connections, and a higher current at a higher voltage at the base-collector connections, so the device is useful as an amplifier (more power out than power in).

author: keithu
author_rank: Contributor
timestamp: March 17, 2015, 10:02:17 pm
content: Quote from: GeoNOregon on March 17, 2015, 03:28:57 pmThis is where you're going wrong - you haven't fully grasped what voltages are. When we talk about voltage we really mean Potential Difference i.e. The difference in electrical potential between two points. It's easy to forget this fact because we generally talk about voltages in absolute terms rather than relative terms, as a kind of shorthand.Although we might say that a certain point in a circuit is at 3V, for instance, what we really mean is that there is a potential difference of 3V between that point and the 0V point (often called ground).So what's so special about the 0V point? Absolutely nothing! It's just a point in the circuit that we have arbitrarily defined as 0V. We could define some other point as 0V and it wouldn't make the slightest bit of difference as long as we stuck to that definition. All our calculations would work out exactly the same.Just remember: All voltages are relative. No exceptions.Let's look at your circuit. You're asking "is B positive or negative". I say "relative to what?We can see that there is a battery between C and B with the positive terminal towards B. SoB is positive relative to C.We can see that there is a battery between E and B with the positive terminal towards E. SoB is negative relative to E.That's the only answer there is to your question.Looking at the transistor in this circuit, the emitter is the most positive terminal, the collector is the most negative terminal, and the base is somewhere in between. This is a typical biasing arrangement for a PNP device.