# Understanding NPN

#### TheTaoOfPhil

Mar 10, 2013
3
Hi,

I am working with several different electronics books to understand the basics of NPN transistors. The only discussion that I can make sense out of is that of Delton Horn, Basic Electronics Theory, pp. 278ff. Since this book is notoriously replete with errors, I want to make sure that what I'm taking away from his discussion is accurate.

I am including a diagram of the NPN configuration addressed by his common-emitter example. From his discussion, it seems that:

1) As per the voltage divider R1, R2, and R3, the voltage levels seen by the collector, base, and emitter must descend in polarity respectively. In other words, the collector voltage must be greater than the base voltage which must be greater than the emitter voltage. Other authors focus on the base-emitter polarity only.

2) The AC signal is applied to the base on top of this arrangement for biasing. That is, the DC bias maintains the emitter-collector flow, but the signal level at the base relative to emitter ground drives the amplified output at the collector.

Is all of this correct?

Final question: He says that R3 is necessary to drop the voltage seen by the emitter to below that of the base. The implication is that the emitter sees the ground to which R3 is connected. However, according to the schematic, the battery is also connected to ground. Shouldn't R3 be directly wired to the emitter so that the emitter sees it instead of seeing the battery ground? (I'm sure my question is wildly naive, but this is how one learns )

Many thanks!

--Phil

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#### john monks

Mar 9, 2012
693
1. For practical purposes this is mostly true.
2. Just for simplicity lets say that R2 is for biasing.
3. Lets just say that R3 helps prevent the base from becoming too positive thereby causing the transistor from turning on more that it should.

This schematic is very confusing because it does not show any values so if you could post a different schematic with values I will be happy to go through it completely.

#### davenn

Moderator
Sep 5, 2009
14,011
Final question: He says that R3 is necessary to drop the voltage seen by the emitter to below that of the base.

To me that doesnt really make sense

I prefer John's response in his points 2 and 3 as a much better description

The implication is that the emitter sees the ground to which R3 is connected. However, according to the schematic, the battery is also connected to ground. Shouldn't R3 be directly wired to the emitter so that the emitter sees it instead of seeing the battery ground?

As they are both connected to ground, its irrelevent, its a common node, the resistor IS connected between the base and emitter

Dave

#### KrisBlueNZ

##### Sadly passed away in 2015
Nov 28, 2011
8,393
Hi Phil and welcome to electronics point

1) As per the voltage divider R1, R2, and R3, the voltage levels seen by the collector, base, and emitter must descend in polarity respectively. In other words, the collector voltage must be greater than the base voltage which must be greater than the emitter voltage. Other authors focus on the base-emitter polarity only.
Yes, those three points do descend in voltage. (They are all the same polarity - positive with respect to the 0V rail, which is represented by the earth symbols.)

R1 is a bit of a red herring. It's not needed as part of the circuit, and won't be present in most practical circuits. I would pretend it's just a short circuit, and the collector load resistor RL is just connected straight from the collector to the positive supply (the battery).

2) The AC signal is applied to the base on top of this arrangement for biasing. That is, the DC bias maintains the emitter-collector flow, but the signal level at the base relative to emitter ground drives the amplified output at the collector. Is all of this correct?
Yes, that's right. R2 and R3 set the DC (static; quiescent) conditions, aka the bias, for the transistor, and have little bearing on the signal being amplified. They put the transistor into a useful part of its "transfer curve", by making it conduct some current through its collector-emitter path, so that it's able to amplify a signal, when one is applied. This bias also determines the DC voltage at the collector; the amplified signal is superimposed onto that DC voltage.

Final question: He says that R3 is necessary to drop the voltage seen by the emitter to below that of the base. The implication is that the emitter sees the ground to which R3 is connected. However, according to the schematic, the battery is also connected to ground. Shouldn't R3 be directly wired to the emitter so that the emitter sees it instead of seeing the battery ground? (I'm sure my question is wildly naive, but this is how one learns )
Well, as Dave says, all ground points are connected together and can be considered roughly the same in a low-frequency, low-current circuit like this one. But I wouldn't say that R3 drops the emitter voltage lower than the base! The emitter voltage is zero, because it's grounded. So it's not being "dropped below" anything; it is tied to a fixed potential.

The base voltage is around 0.7V. This is mostly determined by current flowing through R2 into the base-emitter junction of the transistor. R3's function is to stabilise the base-emitter voltage.

There are several ways of biasing a common-emitter stage, and this is not the best one. For better stability against variations in transistor parameters (current gain, mainly), a resistor is inserted between the emitter and ground, often with a capacitor connected across it. I hope this will be described later in your book, because it is widely used in simple discrete transistor amplifiers and illustrates an important principle.

#### TheTaoOfPhil

Mar 10, 2013
3
Hi everyone and thanks for your answers. I feel like I'm starting to get it.

To answer John's question: The author gives the example where the battery is at 9 V and the resistors are of equal value, such that each resistor is subtracting 3 V.

R1 is a bit of a red herring. It's not needed as part of the circuit, and won't be present in most practical circuits. I would pretend it's just a short circuit, and the collector load resistor RL is just connected straight from the collector to the positive supply (the battery).

Yeah that's what I thought.

Well, as Dave says, all ground points are connected together and can be considered roughly the same in a low-frequency, low-current circuit like this one. But I wouldn't say that R3 drops the emitter voltage lower than the base! The emitter voltage is zero, because it's grounded. So it's not being "dropped below" anything; it is tied to a fixed potential.

The author of this book says: "Resistor R3 subtracts another 3 V from the voltage allowed to reach the emitter." How confusing is that? What you say makes perfect sense, and I am not sure what the author means here.

The base voltage is around 0.7V. This is mostly determined by current flowing through R2 into the base-emitter junction of the transistor. R3's function is to stabilise the base-emitter voltage.

I don't understand how R3 stabilize's the base-emitter voltage. Can you elaborate?

There are several ways of biasing a common-emitter stage, and this is not the best one. For better stability against variations in transistor parameters (current gain, mainly), a resistor is inserted between the emitter and ground, often with a capacitor connected across it. I hope this will be described later in your book, because it is widely used in simple discrete transistor amplifiers and illustrates an important principle.

Yes, he does explain that later. I'm not sure I understand that part either. But the main point for me at the moment is to understand the distinction between the biasing voltage and the signal.

Many, _many_ thanks for your help.

#### BobK

Jan 5, 2010
7,682
The author gives the example where the battery is at 9 V and the resistors are of equal value, such that each resistor is subtracting 3 V.
That statement is just plain false. The load resistor and the transistor are in parallel to R2 and R3. And the base emiiter junction is parallel to R3. So they are not a simple voltage divider, and his asseriton that each drops 3V will not be true (he also seems to be assuming that R1, R2 and R3 are the same value, otherwise they would not divide the voltage equally.

If he was correct, the voltage betwen the base and emitter would be 3V, which cannot happen. It would cause basically unlimited current to flow between the base and emitter, destroying the transitor.

Bob

#### KrisBlueNZ

##### Sadly passed away in 2015
Nov 28, 2011
8,393
The author of this book says: "Resistor R3 subtracts another 3 V from the voltage allowed to reach the emitter." How confusing is that? What you say makes perfect sense, and I am not sure what the author means here.
Yes, as BobK says, that statement is simply wrong. There is no way you can have 3V across a base-emitter junction (unless the transistor has failed). I can't understand why the author would say this, as it's very obviously wrong.
I don't understand how R3 stabilize's the base-emitter voltage. Can you elaborate?
R2 and R3 form a voltage divider that's designed to present a bias voltage close to 0.7V at the base. Actually, unless there's an emitter degeneration resistor, I'm not sure that R3 actually improves the stability of the amplifier. The important function of biasing is to reduce variations in the collector voltage that are caused by differences in current gain, because for any given transistor type (component number), the current gain can fall anywhere within a range of between 2:1 to 4:1, sometimes even more. In other words, the current gain is a poorly controlled parameter for bipolar junction transistors.

The simplest biasing arrangement is just R2, without R3. This causes a base bias current that's equal to the voltage across R2 (which is the supply voltage minus about 0.7V) divided by the resistance of R2, according to Ohm's Law applied to R2. For example if the supply rail is 9V and R2 is 820k, considering R2 on its own, V=8.3 and R=820,000 so the current will be about 10 uA (microamps). The base current is pretty stable.

But if the transistor's current gain could range from 75 to 300 (a fairly typical range for a BJT), the collector current could range from 750 uA to 3 mA. If RL is 2k, for example, that range of collector currents will cause the voltage across RL to be anything from 1.5V to 6V, so the collector voltage could be anywhere from 3V to 7.5V. Since you want the collector voltage to be close to half the supply voltage, for maximum output voltage swing without clipping, this variation is a problem. It is the reason why emitter degeneration is used.

Yes, he does explain that later. I'm not sure I understand that part either. But the main point for me at the moment is to understand the distinction between the biasing voltage and the signal.
Yes, you seem to understand that pretty well.

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,505
Actually, unless there's an emitter degeneration resistor, I'm not sure that R3 actually improves the stability of the amplifier.

Spot on!