PLEASE HELP voltage gain

[michael 1978]

HELLO

i need help about bjt amplifier

i know ce class a to bias, but i want a desired voltage to have
for example i want my ce amplifier to have av 50 voltage gain

wich steps i have to take, how to calculate if input is 4 ma, and i want a gain of 50
can somebody in similary way explain me

THNX

 

[goldsmith]

HELLO

i need help about bjt amplifier

i know ce class a to bias, but i want a desired voltage to have
for example i want my ce amplifier to have av 50 voltage gain

wich steps i have to take, how to calculate if input is 4 ma, and i want a gain of 50
can somebody in similary way explain me

THNX

Hi Michael
A suggestion ! take a look below , please :
http://ellabedu.physics.upatras.gr/Pages/Single stage amps/Common Emitter/CE amp.htm
Or perhaps this one :
http://www.physics.ohio-state.edu/~durkin/phys617/commonemitter.pdf
Best Wishes
Goldsmith

 

[michael 1978]

Hi Michael
A suggestion ! take a look below , please :
http://ellabedu.physics.upatras.gr/Pages/Single stage amps/Common Emitter/CE amp.htm
Or perhaps this one :
http://www.physics.ohio-state.edu/~durkin/phys617/commonemitter.pdf
Best Wishes
Goldsmith

hey goldsmith

i still dont find how to get desired voltage gain

for example you make a biasing ce class a
you make a amplifier and you calculate and you get a voltage gain
but not a desired voltage
do you understand what i want

 

[goldsmith]

i still dont find how to get desired voltage gain

Hi Michael
i.e : AV with classical method is RCllRL/RE+re hence you can change AV with changing un bypassed resistor .
You understand what i mean ?


Best Wishes
Goldsmith

 

[LvW]

for example i want my ce amplifier to have av 50 voltage gain
wich steps i have to take, how to calculate if input is 4 ma,
THNX

Hello Michael,

I think, we should try to find a systematic approach to your task.
And this requires to clarify first some basic questions.
At first, what about your input signal? Is it really 4mA ? Or is it a typo and you mean 4 mV ?
What is the frequency of the input signal? Do you have to consider any load resistance?
What about power supply? Any choice?

Next, what do you know about a common emitter stage? You have mentioned that you know about biasing. OK - do you have any gain formula?
Do you know something about operating point stabilization using an emitter resistance?
Do you know that you can (and perhaps should) split Re into two parts - and one bypassed by a capacitor?
(That`s what Goldsmith was referring to).

 

[BradtheRad]

In operation the amount of bias has the net effect of changing the internal resistance of the transistor through C-E.

Say you put a resistor with value 100 ohms in the collector leg.

Then your bias causes the internal resistance to go up and down. It forms a voltage divider. The result is voltage gain at the output. This is a common way to use a transistor.

I'm not sure how easy it is to calculate what values will work, based on the amplitude of the incoming signal.

 

[LvW]

Then your bias causes the internal resistance to go up and down. It forms a voltage divider. The result is voltage gain at the output. This is a common way to use a transistor.

I am not quite sure if this "explanation" really helps. How can a voltage divider provide gain?
This attempt to explain the transistor operation completely ignores the fact that the transistor resembles a controlled current source.

 

[BradtheRad]

I am not quite sure if this "explanation" really helps. How can a voltage divider provide gain?
This attempt to explain the transistor operation completely ignores the fact that the transistor resembles a controlled current source.

True, it is indirectly-obtained voltage gain.

I was thinking particularly of a class A amplifier and an AC signal. The crest might cause the transistor to go to 30 ohms resistance. The trough might cause it to go to 3300 ohms. This makes a dynamic resistive divider, with the collector resistor (100 ohms in my example). It creates up-and-down swing in the output voltage.

If we think of the transistor as having a variable resistor inside it, I believe that can be as useful in aiding our understanding, as to call it a current source.

Sometimes we ask the transistor to provide current gain and we are not interested in voltage gain.
Sometimes we ask the transistor to provide voltage gain and we are not interested in current gain.

I also was not sure how to take the 4 mA input figure. Sometimes a question comes in an uncertain form, and we must connect the dots. I can't always be sure that my reply addresses the question that was intended.

 

[LvW]

I was thinking particularly of a class A amplifier and an AC signal. The crest might cause the transistor to go to 30 ohms resistance. The trough might cause it to go to 3300 ohms. This makes a dynamic resistive divider, with the collector resistor (100 ohms in my example). It creates up-and-down swing in the output voltage.
If we think of the transistor as having a variable resistor inside it, I believe that can be as useful in aiding our understanding, as to call it a current source.

Hi BradtheRad,

I could imagine what you mean and, of course, I know that you are well aware about the real function of a BJT.
However, I still have severe doubts if introducing a voltage divider concept really helps "in aiding our understanding" of the transistor principle.
This "voltage divider" would consist of a collector resistor Rc and the (variable) C-E path of the transistor - driven by the supply voltage Vcc.
In this case, any change of Vcc would also change the current Ic through this series connection - a severe (and for my opinion: a catastrophic) misunderstanding of BJT operation.
As you know - exact the opposite is true, because Vcc does NOT determine the current Ic (neglecting the Early voltage influence.)

 

[michael 1978]

oh sorry i mean for example take 4mv input signal

the power suply lets take 10V

frequency min 20KH

Rc 10K

how you get now desired voltage can you show example

of you mean by input imedance divide by output impdeance you get desired voltage gain

thnx for reply

Hi BradtheRad,

I could imagine what you mean and, of course, I know that you are well aware about the real function of a BJT.
However, I still have severe doubts if introducing a voltage divider concept really helps "in aiding our understanding" of the transistor principle.
This "voltage divider" would consist of a collector resistor Rc and the (variable) C-E path of the transistor - driven by the supply voltage Vcc.
In this case, any change of Vcc would also change the current Ic through this series connection - a severe (and for my opinion: a catastrophic) misunderstanding of BJT operation.
As you know - exact the opposite is true, because Vcc does NOT determine the current Ic (neglecting the Early voltage influence.)

 

[BradtheRad]

Hi BradtheRad,

I could imagine what you mean and, of course, I know that you are well aware about the real function of a BJT.
However, I still have severe doubts if introducing a voltage divider concept really helps "in aiding our understanding" of the transistor principle.
This "voltage divider" would consist of a collector resistor Rc and the (variable) C-E path of the transistor - driven by the supply voltage Vcc.
In this case, any change of Vcc would also change the current Ic through this series connection - a severe (and for my opinion: a catastrophic) misunderstanding of BJT operation.
As you know - exact the opposite is true, because Vcc does NOT determine the current Ic (neglecting the Early voltage influence.)

Yes, my way of explaining it would be misleading to the extent it ignores the current control ability (which is a chief defining characteristic of a transistor).

I hope never to be the cause of any catastrophes.

To consider the change of supply voltage Vcc...

Start with supply 10V and load 5 ohms. A transistor will assume 5 ohms internal resistance when biased so it delivers 1 A. Then we raise the supply to 15 V. My voltage divider explanation above breaks down if we stick with the simple concept of a voltage divider. Instead however, we know the transistor will continue to provide the same 1A, by assuming 10 ohms internal resistance.

For me it's one of those mental constructs we must build in order to explain the operation of something that's not easy to grasp.

For instance, when we speak of a transistor amplifier providing different amounts of voltage gain in its three basic configurations. (Common emitter & common base, high voltage gain. Common collector, less than 1.) The voltage gain depends somewhat on the transistor, and some of it depends on the configuration.

Whether we are a novice or experienced, we might be working on a project, and ask 'Why won't that transistor pull that wire up to supply level? Does it need more gain?' Often it is not a matter of transistor gain but rather to do with neighboring components and their values. So in a case like this it can help us solve a problem if we picture the transistor being part of a resistive divider, as much as thinking of it as a current source.

 

[michael 1978]

hi everybody

i want to say i dont understand nothing pfffffff

wat can i do i am a beggginer i try to learn by my self

but i whait my book common emitter amplifier configuration

i goona seeee maybe i can learn

thnx for reply to everybody

Yes, my way of explaining it would be misleading to the extent it ignores the current control ability (which is a chief defining characteristic of a transistor).

I hope never to be the cause of any catastrophes.

To consider the change of supply voltage Vcc...

Start with supply 10V and load 5 ohms. A transistor will assume 5 ohms internal resistance when biased so it delivers 1 A. Then we raise the supply to 15 V. My voltage divider explanation above breaks down if we stick with the simple concept of a voltage divider. Instead however, we know the transistor will continue to provide the same 1A, by assuming 10 ohms internal resistance.

For me it's one of those mental constructs we must build in order to explain the operation of something that's not easy to grasp.

For instance, when we speak of a transistor amplifier providing different amounts of voltage gain in its three basic configurations. (Common emitter & common base, high voltage gain. Common collector, less than 1.) The voltage gain depends somewhat on the transistor, and some of it depends on the configuration.

Whether we are a novice or experienced, we might be working on a project, and ask 'Why won't that transistor pull that wire up to supply level? Does it need more gain?' Often it is not a matter of transistor gain but rather to do with neighboring components and their values. So in a case like this it can help us solve a problem if we picture the transistor being part of a resistive divider, as much as thinking of it as a current source.

 

[LvW]

Hi BradtheRad,

I appreciate your desire to simplify things in order to facilitate the understanding of operation principles for parts, which are slightly more complex than a resistor.
However, it`s a critical path and there is always a danger to oversimplify things.
According to my experience, it is of great importance for beginners to learn that voltage-current relationships in electronic circuits - in particular, if they contain semiconductors - are non-linear.
That means - not all parts behave according to Ohms law. This is very important to realize - and this is my key argument against a "mental construct" like "A transistor will assume 5 ohms internal resistance when biased so it delivers 1 A.
I am afraid, such a thinking can (will) lead a novice into a false direction.
More than that, as a by-product one can learn about the characteristic properties of linear and non-linear elements.
Example: Frequently, I have heard that capacitors and inductors are "non-linear" (because of frequency dependence)

- - - Updated - - -

i want to say i dont understand nothing pfffffff
wat can i do i am a beggginer i try to learn by my self
but i whait my book common emitter amplifier configuration
i goona seeee maybe i can learn
thnx for reply to everybody

Hi Michael,
it seems we couldn`t help you - and I am very sorry about that.
However, my recommendation to you: Try to learn how to ask questions and how to provide all information to the forum, which are necessary to help you.

Here is an example:
In post#5 I have started my help with some questions to you (what do you know about a common emitter stage? You have mentioned that you know about biasing. OK - do you have any gain formula?
Do you know something about operating point stabilization using an emitter resistance?)

Why didn`t you answer these questions? So I don`t know how and at which level I should start with my explanations. You see what I mean?
Anyway, it is a good decision to consult a textbook.

Good luck.

 

[BradtheRad]

Hi BradtheRad,

I appreciate your desire to simplify things in order to facilitate the understanding of operation principles for parts, which are slightly more complex than a resistor.
However, it`s a critical path and there is always a danger to oversimplify things.
According to my experience, it is of great importance for beginners to learn that voltage-current relationships in electronic circuits - in particular, if they contain semiconductors - are non-linear.
That means - not all parts behave according to Ohms law. This is very important to realize - and this is my key argument against a "mental construct" like "A transistor will assume 5 ohms internal resistance when biased so it delivers 1 A.
I am afraid, such a thinking can (will) lead a novice into a false direction.
More than that, as a by-product one can learn about the characteristic properties of linear and non-linear elements.
Example: Frequently, I have heard that capacitors and inductors are "non-linear" (because of frequency dependence)

Yes, and I recognize it's important to be grounded with a right understanding of the basics. I'm thinking about Feynman's story telling how appalled he was at seeing the inaccuracies in science textbooks which he reviewed.



As for transistor principles, every primer I've seen speaks of majority and minority carriers, emitter region, base region, etc. I suppose any other electronics student has seen the same.

I did not realize my post #6 was addressing the basics of transistor operation. After all my phrase was 'net effect'.

I acknowledge that my reference to a transistor acting as part of a voltage divider, would not appear in chapter 1 about transistor usage.

However such a discussion might appear in a later chapter.

May I provide these additional arguments as a follow-up to your argument?

1.

Chapter 2 (or 3 or 4, take your choice) tells about the basic configuration of transistor amplifiers. Common base, common emitter, common collector.

Charts compare each types parameters. Such as voltage gain, input impedance, output impedance.

I used to look at these specs, but I did not grasp where they came from. A transistor was supposed to be a current source. Such specs were obscure theory until I tried to devise model transistor behavior for my own homebrew simulator.

I had to calculate the transistor's internal resistance, after first figuring out the maximum current it could draw depending on bias current. I recognize that I must first evaluate it as a current source, secondly as a variable resistor.

2.

As a teacher you have just explained to a class about the operating principles of transistors. You are careful to point out that it is a controlled current source, not a variable resistor.

There is bound to be an alert student who will ask:

'Where is Ohm's law in this? Every day we hear how Ohm's law governs electrical behavior. How does the transistor comply with Ohm's low?'

The answer will have to be about the transistor changing its own internal resistance, in order to act as it does.

3.

Transistors and mosfets are known to overlap in their usage. Among the chief specs for mosfets is minimum-On resistance.

It's natural to extend that to transistors as well. Because when using either a transistor or mosfet to switch high current, its On-resistance becomes a prime concern.

4.

A retired electronics teacher showed me a quiz he used to give his students. It took a schematic of an ordinary transistor amplifier, and duplicated it several times. Each image had volt readings at nodes.

In each case the question was 'What fault in the transistor would cause these unworkable volt readings?'

To solve the questions, I had to picture how the transistor 'pulls the collector terminal low or high', or 'shows hi resistance to bias current', etc.

The troubleshooting process required that I imagine the transistor as having invisible resistors inside.

 

[LvW]

Hi BradtheRad,

wow - what a detailed answer. Nevertheless, thanks for it.

My only intention (post#9 and post#13) was to tell about my experience - nothing else. I am sure you have made other experiences, of course.
As you have seen, the questioner (michael 1978) seems to be a real novice in transistor electronics - and I had (and still have) some doubts if we really help him (and guide him towards the right direction of thinking) by simply stating that the transistor "forms a voltage divider" and that this would be the "common way to use a transistor" (both quotations from post#6).

Just one short comment to section (2) of your previous posting
(There is bound to be an alert student who will ask: 'Where is Ohm's law in this? Every day we hear how Ohm's law governs electrical behavior. How does the transistor comply with Ohm's low?'
The answer will have to be about the transistor changing its own internal resistance, in order to act as it does.)

In contrary, my answer would be: "Perhaps it`s surprising for you, but you should know that there are some parts which do not comply with Ohms law".
And my simple explanation would be: "The voltage across the C-E path does not determine the height of the current that flows".

Regards