Voltage at common node in transistor

This is a rather elementary question but I think my fundamentals are not clear on this. What is the voltage at the common node? Firstly, I thought voltage was taken across something i.e it's taken with a reference to some other voltage. How can we say what is the voltage at the node? My professor claims that the voltage at that node will be Vcc since the negative of the battery Vee has zero voltage and the positive of Vcc is Vcc.

He also claims that voltage Vcc should be greater than Vee to keep the Emitter Base Junction in forward bias and Collector Base in Reverse Bias. Why is this?

I am also perplexed in other bjt configurations- Common Emmitter. What is the voltage at the common node here and how do I know if they are properly biased in their active region?

• You're correct that voltage is across something; if one end isn't specified, it's assumed to be to circuit ground. Nether of those circuits has an obvious or specified ground! Also, all but one of the 'cells' symbols are "upside down", which makes the diagrams far more confusing than they need to be. – pjc50 Jan 21 '13 at 12:58

Voltage is indeed taken with reference to something. It is a potential difference between two points. So you cannot just say "this node is at 5 volts", it has to be "this node is at 5 volts relative to this node".
Usually, to make things easier we define a circuit reference point, e.g. circuit ground so we can measure all voltages relative to this. Then in this case you will hear "this node is at 5 volts" and can assume it is relative to the reference point.

So in your first circuit above, you need a reference point to determine the voltage at the node. If your professor said "the negative of the battery has zero voltage", this would be incorrect. You could say that the negative of the battery is your zero volt reference, or circuit ground, in which case the node pointed to is at Vcc.

Usually you will have a symbol representing circuit ground, such as this one: We can see it used at the bottom of this circuit, and the voltages in reference to it : There are different ground symbols, depending on what the actual reference point is (e.g. the earth underfoot, a metal chassis, or a local circuit ground on a PCB): Your last example is drawn confusingly (as Phil says, you rarely have two batteries in a circuit like this) It's actually a common collector circuit, and the common point is at 5V here (relative to the circuit ground symbol): • I'd say that in the US at least, the particular meanings of those ground symbols is not so clear. Schematics tend to use the "earth ground" symbol by default, and use the other symbols when there is a need to specify a different ground. The specific meaning of that ground is probably best not assumed, and should be in the schematic text somewhere. – Phil Frost Jan 21 '13 at 12:43
• I agree, there are standards for these symbols but many people don't follow them. As long as you know what's what in a particular circuit (i.e. notes as you say) then all should be okay. I do actually use them as shown, with the unfilled triangle for my circuit grounds. Just found the standard - it's IEEE Std 315-1975, Section 3.9: Circuit return (see Wiki Ground). – Oli Glaser Jan 21 '13 at 12:54
• I think the problem is that there are standards, not a standard, and they don't all agree :) – Phil Frost Jan 21 '13 at 13:14
• I agree with Phil. While all three symbols clearly mean "ground" of some sort, in my experience the earth and chassis grounds are often reversed as you show them. Sometimes people get sloppy and use your signal ground symbol, or something like it, for everything. +1 for a good explanation though. – Olin Lathrop Jan 21 '13 at 13:24
• Ok, now I get to take a node voltage with reference to ground. However, I'm still confused- What will be the node voltage for the common node in the common base transistor (the first image)? Another thing I don't understand is why does Vcc in the Common Base circuit need to be greater than Vee? – Prabhpreet Jan 22 '13 at 5:47

A first estimate, and a practical one, is to consider that the base-emitter junction is simply a diode. As long as the voltage difference between base and emitter is enough to forward-bias the diode (about 0.6V for silicon devices), then the voltage will be 0.6V.

This approximation is valid enough in most circumstances, because in this region, current goes up exponentially with voltage. By the time you get the voltage up to 0.7V, there is a whole ton of base current and you may be approaching the rated limits of the transistor.

Your common-emitter schematic is backwards from how a normal person would draw it, with higher voltages at the top, signal flow from left to right. Usually it would be like this: And typically, you wouldn't have two batteries in the circuit, either. Personally, I find these contrived textbook examples to be more confusing than helpful. Here's a real example: Let's assume the transistor is on. The collector current will be about $V_{B1} / R_2 = 5V/50\Omega = 100mA$ (by Ohm's law). Actually a bit less, since the voltage from emitter to collector of a transistor can go down to only about 0.2V, but it doesn't make much difference.

The base current will be $(V_{B1} - V_{Q1\_be}) / R_1 = (5V - 0.6V) / 1000\Omega = 4.4 mA$. Ohm's law again, but here the voltage drop of the base-emitter junction is taken into account.

The voltage as the common node (emitter) is 0V because we have called it ground. The voltage at the base is about 0.6V and the voltage at the collector about 0.2V, because this is what the physics of a fully on silicon transistor require.

In a real application, R1 will be connected not to the battery but to the output of something that can change, like a microcontroller output. And R2 will be something that does more useful work, like a light, motor, etc.