Hi MichaelHELLO
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
Hi Michaeli still dont find how to get desired voltage gain
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
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.
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.)
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.
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 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)
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