I have seen schematics with a resistor hooked up between the emitter and ground. In a book by Stan Gibilisco (I can't remember the name off hand) it stated that although the emitter was grounded from the signal that there was still a voltage due to the resistor. Is there a voltage because:
When the collector draws current from the emitter, that current passes through the resistor creating a voltage drop? I don't think that polarities are correct.
If this question doesn't even make sense to you guys, please let me know.
A bipolar resistor can be thought of as a current controlled switch. A small current flowing into (out of) the base of an NPN (PNP) transistor causes a larger current to flow in the collector - emitter. While the collector current is not quite equal to the emitter current (due to the base current) for easy analysis they are assumed to be since the difference is usually very small. The ratio between base current and the collector current is called the transistor Beta (current gain) and is usually specified as a minimum value according to the datasheet.
If on the resistor to which you are referring, one end connects to ground and the other end connects to the emitter, any current flowing through the emitter will flow through the resistor, bringing the emitter voltage above ground. In turn, the base voltage will be about .6V - .7V above the emitter voltage, assumig the transistor is on. If enough current is flowing, such that the collector voltage drops to about or below the base voltage, the transistor will enter saturation which means that an increase in base current will produce little, if any increase in collector - emitter current. In this mode of operation, the CE voltage is typically about .2V to .3V. This assumes, that there is a collector resistor, which is generating a voltage drop from the supply to the collector terminal. A circuit of this configuration will exhibit an electrical "gain" equivalent to Rc / Re. The idea is to design the circuit to be dependant on the resistors rather than the inherent properties of the transistor which can vary significantly
Sometimes a capacitor is placed in parallel with the emitter resistor. This is done so that at DC the emitter resistor is present to "bias" the transitor into a certain region of operation, while providing a high gain to AC signals. Note also, that the emitter has an inherent resistance associated with it that is proportional to the emitter current.
In this case, when collector current flows out the emitter and through R1, a voltage is generated across it. Assuming that the base voltage is high enough that Q1 doesn't otherwise turn off, this voltage moves the emitter voltage closer to the collector voltage.
But the usual 'view' of this is the other way around. One starts with the base voltage, drops it mentally by a presumed Vbe between the base and emitter (say, 0.7V, for example), assumes that this will be the voltage at the emitter, computes the current that would then flow through R1 based on that applied voltage, and then concludes that this is approximately the collector current -- less a small portion for the base current.
An emitter R simply results in a degree of local feedback. It wil help stabilise DC operating conditions and will also reduce AC voltage gain ( unless the emitter R is bypassed with a cap ).
That is exactly what happens. The emitter current is the sum of the collector current and the normally much smaller base current, so it is a way to produce a voltage related (mostly) to collector current that is applied to one of the transistor's input terminals.
Makes perfect sense. Just remember that the input signal that controls the transistor is the voltage between base and emitter. So as you apply a signal to the base, the emitter resistor applies an output (collector) current generated voltage that opposes it, acting as negative feedback. This makes the output current much more accurately proportional to the base voltage variations. It also raises the base impedance and the collector impedance. And while it lowers the gain (as all negative feedback does) it extends the frequency range that flat gain occurs. And it helps reduce the effects of temperature (increased current gain and lower base to emitter voltage) on the bias situation.
Thanks for all the replies, everyone. I will definitely check out the websites that have been suggested.
When you talk about current flowing through the emitter, is this current coming from ground? I think that current cannot flow from collector to emitter. If so, wouldn' t the voltage be higher at the bottom of the resistor (closer to ground)? That voltage got dropped across the resistor, leaving less voltage at the top.
I'm confused, but a little less so thanks to everyone.
Glenn
PS: the books I've been reading don't seem to address this very clearly.
Current will flow out of the base and into the emitter in a PNP transistor and yes, if the emitter were at ground, 0 volts, the base would be below this potential. PNP transistors have their place when the signals you want to work with are negative in amplitude such as an AC coupled amplifier, and when you wnat to switch something by bringing the voltage down rather than up, such as an active low signal.
The emitter current will be a combination of collector and base currents. The collector-emitter current concept defies normal / basic thinking and is a function of semiconductor physics. The application of base current injects charge carriers into the C-E junction called the depletion region. If the right type of charger carrier is injected the depletion region gets narrower an if the other type is injected it gets wider. When the depletion region is narrower and voltage is apllied to C-E pair charge carriers are able to be swept across the depletion region - hence current flows. The amount of current that is able to flow depends on the width of the depletion region, hence more base drive more current can flow in the C-E. The ratio of base (current) drive to collector - emitter drive is the transistors Beta.
You have to decide early in your career if you're going to use electron flow, which hams, techs, and other people who live in the real world use, or "conventional" flow, which academecians, scientists, and other visionaries use. ;-)
They're indistinguishable, except all of the signs are reversed.
Conventional flow flows in the direction of the arrow at PN junctions: one in a diode, and the emitter/base junction in BJTs (bipolar junction transistors). Electrons flow in the opposite direction to the arrow.
The conventional current in an electron beam goes upstream. ;-)
With a negative-ground supply, the electrons come out the ground, flow up through your circuit, and get sucked into the positive terminal of the battery, where they get circulated by the force of the chemical reaction.
All you have to do is change your point of view, and it's exactly the same as positive current flowing out of the positive terminal of the battery, down through the load, and to ground.
Conventional current uses the right-hand rule, and electron flow uses the left-hand rule, but save that for when you start to get into electromagnetics. ;-)
I have done my research and much of what you guys are saying makes sense!! Thanks!
BTW: How do you set the voltage at the emitter so that it is 0.7 v or so less than the voltage at the base? I know that the base voltage is set my some sort of potential divider resistor setup. But how do you ensure that the emitter will be less than the base?
Getting a bipolar transistor to do what you want (within its capabilities) is all about controlling b-e current, and there are numerous was to do it, so there is no one single answer to your question "how do you ensure that the emitter is less than the base"?
But there is an implication in your question that needs to be addressed. Your question implies that you may think the base must always be ~.7 volts higher than the emitter. You need to think of it a little differently.
The amount of b-e current controls the amount of c-e current. It could be that you want *no* c-e current, in which case you would not want the emitter to be ~.7 volts more negative than the base.
What you want to do is control the b-e current and not have the base *always* ~.7 volts higher than the emitter. The reason .7 volts is an important number is that there will be no c-e current until the base voltage is raised to about .7 volts more positive than the emitter voltage in the NPN transistor. In a PNP, the base would need to be about .7 volts more negative than the emitter for there to be c-e current.
First off, before I go any further with the above circuit, note that I added a (gnd) at the bottom. It doesn't mean anything except to note where my "reference point" is at. There are three nodes in the diagram other than (gnd) and these are (1), (2), and (3). If I talk about the voltage at (1) as being +5V, you should keep in mind that I am talking about the voltage at (1) compared with (gnd). Voltages are always measured with respect to somewhere. And by common agreement, we learn to designate one particular node as being special and calling it (gnd). Which one makes the most sense to label that way will depend on things and it will take some experience before you will always pick the same place to label that way that a professional might. But it doesn't change the circuit if you put the (gnd) label somewhere a professional wouldn't -- it just means that folks (including you) may be a bit confused when you talk about some voltage here or there. But I could just as well have labeled the node above called (2) as (gnd) and relabeled (gnd) as (15), for example. The labels are entirely arbitrary and mean nothing as far as the circuit itself goes. It's just inside our heads and nowhere else.
Okay. So trace out the above circuit in your mind until you agree with me that there are four places of common connections between components. We have 4 parts in the circuit; two batteries, one resistor, and one transistor. They are tied together with wire and since all places along a wire are considered to be at the same potential, we can mentally consider everywhere that only wire exists as being the same "point." I just labeled those points as (1), (2), (3), and (gnd). I hope you agree with that much.
Now, what is the potential at (1)? It's +5 (with respect to (gnd), of course.) That's because there is a battery there and its job is to make sure that one end of it is at a fixed potential away from the other end. In this case, the plus (+) side is tied to the base of Q1, so the base of Q1 _must_ be at +5V because the battery will make sure of this fact. That's its job.
Hopefully, we may agree that (1) will be at +5V (with respect to gnd.) This means Q1's base is at +5. Now think about the effective diode between Q1's base and Q1's emitter.. Ignoring the rest of the circuit, that looks about like this:
If you ignore R1 for a moment and think of it as just a wire, I think you can see that the diode will be forward biased. This means it will allow the flow of current, readily. Without R1 (with it shorted out by a wire), the voltage across this Q1's base-to-emitter diode would be the full 5V and a lot of current would flow. Way too much, and it is almost certain that Q1 would self-destruct. Now, if you re-imagine R1 included, what happens? Well, as more and more current is considered to be flowing through the diode _and_ R1, R1 starts dropping more and more voltage according to I*R1. Eventually, of course, all of the 5V would be used up by R1. In other words, if you imagine that "I" is large enough, the product of I*R1 would be exactly
5V. When that happens, there would be no voltage left to exist to forward bias the diode. And no current (not much) would be able to flow. So that would mean that that much "I" is too much to be reasonable. So we are sure that "I" must be less (
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