Transistor : Output voltage of the amplifier

[Jahnavi]

Homework Statement

Homework Equations

The Attempt at a Solution

This is a solved example given in my Physics textbook . I am not understanding a couple of things .

1) How is I_in calculated as V_in/R_in ?

Why is the book not taking potential difference between Base Emitter into consideration ?

Instead shouldn't Kirchoff's Voltage rule be applied in the Base Emitter loop ?

I think it should be I_in =( 1mV -V_BE )/2KΩ

2) Shouldn't the diagram be showing V_out across 100kΩ ?

I don't understand why book considers V_out to be V_CE ?

Please help me understand these issues ?

 

[Delta2]

$V_{BE}$ is equal to input voltage which probably is small, much smaller than 1mV, so $1mV-V_{BE}\approx1mV$.

 

[Jahnavi]

$V_{BE}$ is equal to input voltage which probably is small, much smaller than 1mV, so $1mV-V_{BE}\approx1mV$.

I don't think this is the case .

I think in a silicon PN junction potential difference between Base and Emitter is about 0.7 V .

This is quite large as compared to the bias battery .

 

[Delta2]

I don't think this is the case .

I think in a silicon PN junction potential difference between Base and Emitter is about 0.7 V .

This is quite large as compared to the bias battery .

You might be right , I am not an EE and don't know much about transistors, just said to share my thought on this.

 

[Jahnavi]

@gneill , @Charles Link

 

[Charles Link]

@gneill , @Charles Link

To me, it looks like a problem that is poorly presented. e.g., to forward bias $V_{BE}$ , you normally need $V_{BE}=.7$ volts. They also don't specify the collector bias voltage. Perhaps someone else has a solution, but my recommendation is to not spend much time on this one as is. This "common emitter" electronic configuration is a good one to know, but this problem looks to me to be a poor one to use to try to learn it.

 

[Jahnavi]

Thanks .

Suppose instead of 1mV battery there is a 1 V battery , then do you think what I did in the original post I_in=( 1V -V_BE)/2KΩ is correct ?

 

[Charles Link]

Thanks .

Suppose instead of 1mV battery there is a 1 V battery , then do you think what I did in the original post I_in=( 1V -V_BE)/2KΩ is correct ?

I looked at the problem in the "link" you had posted. For that one, with the book's example, a $V_{BE}$ of 18.5 Volts I think is totally inaccurate. I expect if you hooked it up in the lab, you might find that $I_B$ would be larger than what is computed using the value $\beta=200$, and you would find $V_{BE}=.7$ volts. And then I think you might have removed the "link": https://www.physicsforums.com/threads/find-the-state-of-transistor.935500/#post-5909999 $\\$ Edit: And on this one. a base voltage of 1 volt might work, but they still need to specify the collector voltage. Also $V_{in}$ (ac) usually gets coupled in with a capacitor, and $V_{out}$ (ac) is usually also coupled with a capacitor.

 

[Jahnavi]

Thank you .

Do you think post 7 is correct ?

 

[Charles Link]

Thank you .

Do you think post 7 is correct ?

That part works, but please see my edited additions to post 8.

 

[cnh1995]

Thanks .

Suppose instead of 1mV battery there is a 1 V battery , then do you think what I did in the original post I_in=( 1V -V_BE)/2KΩ is correct ?

I don't think this problem is practical and as Charles says, not much helpful in learning CE configuration.
The battery needs to be in series with the input ac source (theoretically). In this circuit, as per circuit theory, the battery can't dc-bias the transistor as the ac source is directly connected across the BE junction. Even if the battery were a 10V one, it couldn't dc-bias the transistor in this diagram.

 

[Charles Link]

I don't think this problem is practical and as Charles says, not much helpful in learning CE configuration.
The battery needs to be in series with the input ac source (theoretically). In this circuit, as per circuit theory, the battery can't dc-bias the transistor as the ac source is directly connected across the BE junction. Even if the battery were a 10V one, it couldn't dc-bias the transistor in this diagram.

See also my post 8. IMO, this textbook looks like it may contain numerous errors. I encountered that type of thing from time to time in my college days. The best solution would be for the OP @Jahnavi to learn the material from a book that presents it better and with fewer mistakes, if that is at all possible.

 

[Jahnavi]

The battery needs to be in series with the input ac source (theoretically).

@Charles Link , do you also think this is correct ?

 

[cnh1995]

See also my post 8. IMO, this textbook looks like it may contain numerous errors. I encountered that type of thing from time to time in my college days. The best solution would be for the OP @Jahnavi to learn the material from a book that presents it better and with fewer mistakes, if that is at all possible.

I have seen many junior college physics books in India showing wrong biasing circuit for CE amplifier. @Jahnavi, I believe that includes the NCERT textbook as well.
I agree that you should study amplifiers from some standard electronis books (or COP by H C Verma).

 

[Charles Link]

@Charles Link , do you also think this is correct ?

The way they show it, if you use $V=1$ volt, it would work, provided the $V_{in}$ was ac coupled (with a capacitor).

 

[cnh1995]

The way they show it, if you use $V=1$ volt would work, provided the $V_{in}$ was ac coupled (with a capacitor).

Yes, but I wasn't including a coulping capacitor in my reasoning. If there isn't one, the battery is ineffective.

 

[Jahnavi]

The way they show it, if you use $V=1$ volt, it would work, provided the $V_{in}$ was ac coupled (with a capacitor).

But then it would be something like an input AC signal placed parallel to a DC battery (1 V)

What would then be the net effective voltage applied across Base Emitter ?

This is so confusing .

 

[cnh1995]

then do you think what I did in the original post Iin=( 1V -VBE)/2KΩ is correct ?

That I_in wouldn't be the true I_in as it won't flow through the base. It only flows through the battery. The base current will be determined by the input ac voltage, which will be the same even if you remove the battery.

 

[Jahnavi]

That I_in wouldn't be the true I_in as it won't flow through the base. It only flows through the battery. The base current will be determined by the input ac voltage, which will be the same even if you remove the battery.

I was assuming AC input signal was not present .

 

[cnh1995]

But then it would be something like an input AC signal placed parallel to a DC battery (1 V)

The net BE voltage will be the input ac voltage. The battery is redundant as far as dc biasing is concerned.
Edit: assuming no coupling capacitor.

I was assuming AC input signal was not present .

Ok, but when you connect the ac source, I_in≠I_B.

 

[Charles Link]

Usually coupling capacitors are used on the ac signals, so that the bias voltage would not be short-circuited by the ac input connection. Meanwhile, oftentimes the ac signal can be a very small signal, sometimes from sensitive sensors where the signal needs to be amplified. A $V_{DC}=1$ volt could destroy some sensors. The ac coupling with a capacitor is quite necessary in these amplifiers.

 

[gneill]

I suspect that the given circuit diagram is flawed; Perhaps the original intention was to represent the bias network's essential small-signal equivalent circuit, but something went wrong after the illustrator got hold of it. The 1mV apparently should be associated with the input source, not a base bias battery. Or, that battery originally was meant to be the small signal source, and the "Input voltage" source is a mistaken evolution of what should have been just a text label to a labelled source.

Note that the way that input source is connected the base bias circuit as shown would have no effect: the base-emitter voltage would be entirely determined by that source. It would need capacitive coupling to isolate it from the bias network and superimpose the signal on the bias.

That said, the given solution clearly assumes that the 2 kΩ resistor is in series with a 1 mV input signal and the base of the transistor, since they use it to calculate the base current, $I_{in}$ . So I'd consider take the circuit to be a small signal simplification that should probably have looked more like this:

 

[Jahnavi]

That said, the given solution clearly assumes that the 2 kΩ resistor is in series with a 1 mV input signal and the base of the transistor, since they use it to calculate the base current, $I_{in}$ .

Thank you .

Even if that is the case , wouldn't applying KVL give I_B=( 1mV -V_BE )/2KΩ ?

 

[QuantumQuest]

Let me give my two cents recalling what I learned in Electronics some 30 years ago. For an NPN transistor of Si material the threshold in order to become conductive is $0.7 V$. Given the diagram, there is only $1 mV$ source and the input signal cannot pass through the junction before the sum of the voltages on the resistor of $2 K\Omega$ plus the $1 mV$ from the source gets over the threshold of the NP junction$^*$. Then the transistor will get conductive and the output will be a bigger signal regarding amplitude - as this is the basic purpose / application of a common emitter circuit i.e voltage amplifier. So, for 1) you have to take the total conductive voltage and divide it by $2 K\Omega$ in order to find $I_{in}$. For 2) $V_{out}$ is taken across the $100 k\Omega$ resistor - it is the amplified alternating signal. The DC voltage source gives the bias voltage between collector and ground.

EDIT: $^*$ As pointed out by qneill, a coupling capacitor in series is needed in order to get the sum of voltages of input signal plus $1 mV$ DC from the source, which is absent in the diagram provided. Apparently, there are various things missing / wrong with this diagram and the provided solution in OP.

 

[Jahnavi]

Usually coupling capacitors are used on the ac signals, so that the bias voltage would not be short-circuited by the ac input connection.

If no coupling capacitor is used just as in the original example , how does AC input signal short circuit the DC battery ?

Sorry , I am fairly new with AC and transistors .

I only understand "shorts" in a DC circuit .

 

[Charles Link]

If no coupling capacitor is used just as in the original example , how does AC input signal short circuit the DC battery ?

Sorry , I am fairly new with AC and transistors .

I only understand "shorts" in a DC circuit .

The input voltage is often a weak ac voltage or current source, e.g. a photodiode or a microphonic element for an audio (sound) system. If you attach it directly to the point $V_{BE}$, it could simply get fried, if it is a low resistance element. $V=1.0$ volts is often far too much for that sort of thing. $\\$ The purpose of amplifier circuits such as this is to amplify the (small) signals, so that the signal is large enough to easily work with, and use. Sometimes additional amplification is then employed, e.g. to drive speakers in a sound system.

 

[gneill]

Even if that is the case , wouldn't applying KVL give I_B=( 1mV -V_BE )/2KΩ ?

If the circuit is meant to represent an ac small signal equivalent, then the V_BE would be zero (it's part of the DC bias circuit).

If no coupling capacitor is used just as in the original example , how does AC input signal short circuit the DC battery ?

It doesn't short the battery, it shorts the branch containing the battery and its series 2 kΩ resistor. Applying an ideal source between two nodes fixes the potential difference between those nodes. In the given example the "input source" establishes the potential difference between the emitter and the base, rendering any bias circuit irrelevant.

 

[Charles Link]

If the circuit is meant to represent an ac small signal equivalent, then the V_BE would be zero (it's part of the DC bias circuit).

It doesn't short the battery, it shorts the branch containing the battery and its series 2 kΩ resistor. Applying an ideal source between two nodes fixes the potential difference between those nodes. In the given example the "input source" establishes the potential difference between the emitter and the base, rendering any bias circuit irrelevant.

I think the OP would do well to consider the recommendations of post 12 and 14. I myself have sometimes tried to work with books that weren't 100% in college. A good book can be worth its weight in gold, while a not-so-good book is worth about as much as you get get for it by recycling the paper.

 

[Jahnavi]

It doesn't short the battery, it shorts the branch containing the battery and its series 2 kΩ resistor. Applying an ideal source between two nodes fixes the potential difference between those nodes. In the given example the "input source" establishes the potential difference between the emitter and the base, rendering any bias circuit irrelevant.

Are you saying that potential difference between Base and Emitter is equal to the input signal voltage ( in the original problem where no coupling capacitor is present ) ?

 

[cnh1995]

Are you saying that potential difference between Base and Emitter is equal to the input signal voltage ( in the original problem where no coupling capacitor is present ) ?

Yes, and the battery is redundant.

 

[Jahnavi]

It would need capacitive coupling to isolate it from the bias network and superimpose the signal on the bias.

@gneill , please see the attached picture . This is from another reference book .Do you find it reasonable ?

 

[gneill]

Please see the attached picture . This is from another reference book .Do you find it reasonable ?

Yes. Note that capacitor $C_1$ DC-isolates the input signal from the bias network consisting of $V_{BB}$ and $R_B$. $R_B$ also contributes to the input impedance for the amplifier.

 

[Jahnavi]

Yes. Note that capacitor $C_1$ DC-isolates the input signal from the bias network consisting of $V_{BB}$ and $R_B$. $R_B$ also contributes to the input impedance for the amplifier.

OK.

Please help me understand this diagram .

Are you saying that the two voltages , DC from battery and applied AC input signal get added ?

Is the net input voltage in the Base still sinusoidal but the peak voltage now oscillating between V_BB+V_o to V_BB-V_o ?

 

[gneill]

Sorry, I need to retract my previous statement (post #32). In the circuit shown in post #31 the source $V_{BB}$ is going to clamp the potential at the end of $R_B$, so the input signal will have no effect. The input signal will be developed across the capacitor and the transistor will never see it.

I'm finding that your reference materials have rather dubious quality...

 

[Jahnavi]

In the circuit shown in post #31 the source $V_{BB}$ is going to clamp the potential at the end of $R_B$, so the input signal will have no effect. The input signal will be developed across the capacitor and the transistor will never see it.

Suppose the lower terminal of battery V_BB is earthed . Are you saying that the potential at left end of R_B is at a constant voltage V_BB irrespective of the input AC signal ?

 

[gneill]

Suppose the lower terminal of battery V_BB is earthed . Are you saying that the potential at left end of R_B (where the current is entering ) is at a constant voltage V_BB irrespective of the input AC signal ?

Yes, assuming that by "earthed" you mean that the bottom rail is taken to be the reference node.

 

[Delta2]

Ok sorry for me intervening out of nowhere BUT I think $V_{BB}$ will have some internal resistance so I think the transistor will see the input voltage after all...

 

[Jahnavi]

@gneill ,This picture is from another reference book . Here the author has put AC input signal in series with the DC battery . Does this look reasonable ?

 

[gneill]

@gneill ,This picture is from another reference book . Here the author has put AC input signal in series with the DC battery . Does this look reasonable ?

Yes, that's a better representation. Note though that it doesn't depict an input resistor for the base in this case. That might not be a problem with careful choice of $V_{BB}$.

 

[gneill]

Ok sorry for me intervening out of nowhere BUT I think $V_{BB}$ will have some internal resistance so I think the transistor will see the input voltage after all...

Unless the problem states that the components are not ideal, it's not advisable to make that assumption. Further, you'd have no idea what size of resistance to ascribe to that internal resistance, so analyzing the circuit would be problematical.

 

[Jahnavi]

Yes, that's a better representation. Note though that it doesn't depict an input resistor for the base in this case. That might not be a problem with careful choice of $V_{BB}$.

Ah ! Finally .

But the irony is that in the exams I need to stick with the diagram in post #31 however flawed it might be . ( It is the standard reference text )

Thank you for your patience .Please bear with me for some more time . I have few more relevant enquiries .

 

[Jahnavi]

Yes, that's a better representation. Note though that it doesn't depict an input resistor for the base in this case. That might not be a problem with careful choice of $V_{BB}$.

In the picture in post#38 , the DC voltage from battery and applied AC input signal get added ?

Is the net input voltage in the Base sinusoidal but the peak voltage now oscillating between VBB+Vo to VBB-Vo ?

 

[Delta2]

Ok @Jahnavi maybe in the text of the book mentions something like that all DC voltage sources are assumed to have a (small) internal resistance?

 

[gneill]

In the picture in post#38 , the DC voltage from battery and applied AC input signal get added ?

Yes.

Is the net input voltage in the Base sinusoidal but the peak voltage now oscillating between VBB+Vo to VBB-Vo ?

Yes, it's an AC signal with a DC offset. Mathematically you could write it as:

$V_{BE}(t) = V_{BB} + V_o sin(\omega t)$

 

[Jahnavi]

Yes, it's an AC signal with a DC offset. Mathematically you could write it as:

$V_{BE}(t) = V_{BB} + V_o sin(\omega t)$

Thanks .

What happens to the net input voltage in the original problem (post #1 where there is no coupling capacitor ) ?

I am referring to the case when AC input signal is applied across the branch having resistor and DC battery .

 

[cnh1995]

Thanks .

What happens to the net input voltage in the original problem (post #1 where there is no coupling capacitor ) ?

I am referring to the case when AC input signal is applied across the branch having resistor and DC battery .

Base-emitter voltage will be equal to the input ac voltage and the base current will have a half wave rectified waveform.
Plus, the battery will add a dc component in the ac source current.
Edit: I was assuming BE junction to be ideal (0V drop).If the BE junction is assumed to have 0.7V drop and if the ac source is of 1mV, then, as gneill said, it won't be able to turn the transistor on.

 

[gneill]

Thanks .

What happens to the net input voltage in the original problem (post #1 where there is no coupling capacitor ) ?

I am referring to the case when AC input signal is applied across the branch having resistor and DC battery .

The base DC bias circuit (battery and resistor branch) is rendered ineffective by the placement of the input source, so an equivalent circuit as far as the transistor is concerned would be:

Now, $1\;mV$ is not enough to forward bias the transistor, which needs about $700\;mV$ for a silicon transistor. Without a proper base bias to place the transistor operating point in the appropriate location on its $IV$ characteristic curves, the transistor would never turn on and no signal would pass.

 

[Jahnavi]

The base DC bias circuit (battery and resistor branch) is rendered ineffective by the placement of the input source,

Does that mean there would be no current in the branch consisting of the 2kΩ resistor and DC battery ?

Why does AC input voltage dominate over the DC battery so as to make it ineffective ?

 

[gneill]

Does that mean there would be no current in the branch consisting of the 2kΩ resistor and DC battery ?

No, the AC source will provide a path for it. KVL around the loop will show that the majority of the bias will be dropped across the resistor.

Why does AC input voltage dominate over the DC battery so as to make it ineffective ?

Any ideal source, AC or DC, would do the same. The point is, what ever source you place in that position will set the potential difference between the two ends of that "bias" branch.

 

[Jahnavi]

Thanks !