This video is part of the first course of our *ElectrONiX* massive open online course (MOOC) series. You can sign up for the course *ElectrONiX - Amplifiers* for free on the following links: English: imoox.at/mooc/local/landingpage/course.php?shortname=amps&lang=en German: imoox.at/course/amps Bosnian: imoox.at/mooc/local/landingpage/course.php?shortname=amps&lang=bs
Is it true that current-controlled transistors are better than voltage-controlled transistors by a factor of beta? At Texas Instruments I worked with I²L logic (Integrated Injection Logic). Beta was about 4, as opposed to discrete transistors with a beta of 200.
Thank you for your suggestion. We will put this on our loong list of planned tutorials. Do you have any concrete questions regarding the block diagrams at the moment?
You know that R1 =139k, Vin = 15v, Vout = 8.1v so use formula ; Vin x R2/(R1 + R2) = Vout. Transpose formula to get; Vin/Vout = (1+ R1/R2). Transpose further to get; Vin/Vout - 1 = R1/R2 again transpose to get R2 = R1/(Vin/Vout -1), which equals 163.5k. Select nearest listed value of 150K.
Thank your for your question. We suppose you are talking about the input impedance of R1||R2 at 10:15? R1 and R2 can be seen to be in parallel in the small-signal analysis as the positive and negative supplies are constant and are therefore both on small-signal ground. If you want to know more on small-signal analysis we recommend you to watch the following videos: Basics of small-signal analysis: ruclips.net/video/DKPqNQ3CZ9I/видео.html Similar example: ruclips.net/video/e067ySIG3gY/видео.html
I think that might be a mistake there. For voltage divider in general the conencted load shalll have at least 10x the resistance of the divider resistance. That means R1||R2
Don’t get it. You have an input open the valve but it’s not amplifying. The higher current is already there. I was expecting the low input current being amplified to produce higher output. Like a guitar signal in and a jimmy Hendrix out. But on your example this is not the case
This video is part of the first course of our *ElectrONiX* massive open online course (MOOC) series.
You can sign up for the course *ElectrONiX - Amplifiers* for free on the following links:
English: imoox.at/mooc/local/landingpage/course.php?shortname=amps&lang=en
German: imoox.at/course/amps
Bosnian: imoox.at/mooc/local/landingpage/course.php?shortname=amps&lang=bs
The world's best teacher thanks
Thank you, we will pass your feedback on to Michael.
Weary good explaining 😊
This is gold. Thank you
Thank you for the nice feedback. We are glad you like it :)
These are awesome tutorials- thank you!
Thank you for the nice feedback.
You're welcome ;)
Great video
Is it true that current-controlled transistors are better than voltage-controlled transistors by a factor of beta?
At Texas Instruments I worked with I²L logic (Integrated Injection Logic). Beta was about 4, as opposed to discrete transistors with a beta of 200.
hi thanks for sharing your knowledge,
could you please make a video how to read and learn block diagram
Thank you for your suggestion. We will put this on our loong list of planned tutorials.
Do you have any concrete questions regarding the block diagrams at the moment?
hey what we get by beeta times output impedance and 10 times output impedance?
You get a stupidly phrased question.
Sir R2 value how did you get it please explain , I didn't get it explained in the video
You know that R1 =139k, Vin = 15v, Vout = 8.1v so use formula ; Vin x R2/(R1 + R2) = Vout. Transpose formula to get; Vin/Vout = (1+ R1/R2). Transpose further to get; Vin/Vout - 1 = R1/R2 again transpose to get R2 = R1/(Vin/Vout -1), which equals 163.5k. Select nearest listed value of 150K.
@@georgeh1352how to get 1/1.17 value.can you please explain
At 7:30 is it Zin>=10*Zout?
Yes, you're right. We noticed that mistake too late...
This comment should be pinned or make an edit in the description
why is the R1 and R2 in parrelle for the impedence of 75K
Thank your for your question.
We suppose you are talking about the input impedance of R1||R2 at 10:15?
R1 and R2 can be seen to be in parallel in the small-signal analysis as the positive and negative supplies are constant and are therefore both on small-signal ground.
If you want to know more on small-signal analysis we recommend you to watch the following videos:
Basics of small-signal analysis:
ruclips.net/video/DKPqNQ3CZ9I/видео.html
Similar example:
ruclips.net/video/e067ySIG3gY/видео.html
@@ife.tugraz thank you
great
Ty
Fantastic
Thank you for the nice feedback :)
At 10:07 Rb value is 750k ohm but at 10:19, 10 times Rb is 75k ohm, why ?
I think that might be a mistake there. For voltage divider in general the conencted load shalll have at least 10x the resistance of the divider resistance. That means R1||R2
Yeah it should 10 * (R1 || R2)
nice
You're welcome.
Don’t get it. You have an input open the valve but it’s not amplifying. The higher current is already there. I was expecting the low input current being amplified to produce higher output. Like a guitar signal in and a jimmy Hendrix out. But on your example this is not the case