ATAR Notes: Forum
VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Physics => Topic started by: Chavi on April 23, 2010, 05:38:38 pm
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What is meant by the term amplifier biasing?
In fact, what is biasing? What is a transistor? What are the significance of anodes and cathodes? How does amplification actually work?
What is the purpose of placing a diode in an AC circuit?
How do you read AC current / time graphs (they're just sinusoidal)?
How do you take a Vin/Vout graph and sketch Vin/time and Vout/time graphs?
What is the Vrms?
This entire section of the course isn't explained well be my textbook.
Any help would be appreciated.
Thanks in advance.
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yes i could do with some help as well. specifically regarding graphing amplification outputgraphs
is there a simple step-by-step procedure that you can follow that can help to simplify the questions, because currently i just look at the uinput graphs and even if i know the gain or whatever i just dont know where to start and what to do.
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thats trueeeee D: the book is pretty bad at explaining this section :(
amplifier biasing is baising the transistor from the middle of its linear range .. iono i got that from the book D:
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amplifier biasing is baising the transistor from the middle of its linear range .. iono i got that from the book D:
this can't be any more ambiguous . . .
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sorry i am a failwhale D:
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I think these pictures explain it pretty well..
(http://img260.imageshack.us/img260/3489/img0513og.jpg)
(http://img101.imageshack.us/img101/3210/img0514zi.jpg)
The wave going upwards in this one is the V-in signal
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that really does help. Thnx stroodle!
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Don't be intimidated by the Vin/Vout graph. Just read off the values and you'll see a pattern of sorts.
If your input was 5mV, and your graph at Vout at 2V, then your amplified wave will be at 2V where it was at 2mV. The clipping business occurs because you get the same Vout for more than one Vin value. So on your output signal graph you will have a flat section because where the input was a certain mV, it will be that V as per the graph, but will be for more values as well.
I don't think that makes sense... if you understand gain, just draw your amplified signal without clipping, then draw in horizontal lines where your min/max Vout occurs and it should be intuitive where to clip.
Vrms is Vp/sqrt(2). Vp is half of Vpp, and Vpp is the height of a wave on an oscilloscope. I was taught this as part of the elec/photonics, but from my brief look at the study design this is only in the detailed study?
A diode goes into an AC circuit usually as part of rectification. Again, I believe this to only be in the detailed study. One diode will give half-wave rectification (As it will only let the current go in one way through it - there will simply be no current going the other way so you cut off the -ve parts of the AC wave). Arrange 4 diodes in a certain way and you make a diode bridge - creating a DC or full wave rectifaction.
Hope that helps.
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Is biasing where the linear section of the transfer characteristic does not have its centre at the intersection of the x-axis and y-axis so that clipping occurs at different values for negative and positive Vin values? or is this something else?
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Is biasing where the linear section of the transfer characteristic does not have its centre at the intersection of the x-axis and y-axis so that clipping occurs at different values for negative and positive Vin values? or is this something else?
Clipping is when you try to amplify a device, but (somehow) certain values of V in (Above or below a certain point. This is specified by the V in graph) are unable to be "amplified" (I can't think of any other word, but you'll get what I mean by "amplified" when you go through a worked example). If you see a graph of V out that looks like a sin curve but the top or bottom are flattened out, the flat bits are the result of clipping. Clipping isn't wanted because it distorts "stuff", like a radio might sound retarded because of clipping.
Clipping is the 'problem'. Biasing is the 'answer' (to prevent clipping). I don't think we need to know how biasing works; just how to interpret biasing. Suppose the top of a V out curve is flattening out. To prevent that you'd would want to translate it down by biasing: 'negative biasing'.
I hope I made you less confused. Didn't quite understand what you were saying..had a long day..maybe you were right all along!
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Is biasing where the linear section of the transfer characteristic does not have its centre at the intersection of the x-axis and y-axis so that clipping occurs at different values for negative and positive Vin values? or is this something else?
Biasing is adjusting the input voltage (which makes it a biased voltage) so that it is in the centre of the linear section of the transfer characteristic - directly below the quiescent point. This way you can get the maximum output voltage without clipping. It's done using voltage dividers, but we don't have to be able to analyse these amplification circuits anymore.
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mm k. Thx. I was thinking biasing was a problem. its the same as what I was saying, just its a solution to the problem of clipping, not a cause of clipping... I think. thx again.