This is absolute legendary explanation. From my experience transistors have always been a very confusing topic considering so many variables and assumptions. Your videos are a treasure chest for this matter
Hello sir,I am from india, I am very big fan of your lectures sir, I took electronic and circuit course on edx, it's a mooch course, I really enjoyed your videos, I really liked the way of your teaching sir,keep it work, really you are best and master in this subject sir❤❤
Very informative playlists on your page! I think there is a correction in example 1. The base current is 4.3mA/100 = 0.043 mA and not 0.0043 mA. About 1%. Still negligible. 👍🏼
Don't you make a mistake at 32:45 when calculating Rin? Bre (beta times re) is 100.(2.8 ohms) = 280 ohms. That in parallel with 5k should give us around 265 ohms, isn't it? Or am I making a mistake somewere? In every case, I like your video's. Very nice and clear explanations.👍
If the base voltage is fixed to some Vb but the emitter is grounded, then there will be a nonlinear relationship between Ie and Vbe like that of a forward-biased diode. This can lead to it being a little unpredictable - the transistor might be in cutoff with Ie=0 (if Vbe is too low) or with Ie zooming up to some high transistor-burning number if Vbe is a little too high.
@@adannerThanks for the answer!! So it means that I must have a resistor there always? I have seen many (practical) circuits with a common emiter amplifier without a resistor at the emiter. Does it mean that it was wrong? My confusion is based on the problem I see when I try to understand some practical circuits (for example for radio control models, especially old ones from various books but not only) and I compare those circuits with theoretical circuits like here and it doesn't match sometimes. I feel like I am missing something here 🤔
@@grzesiek1x There could be a few reasons. For radio circuits they sometimes use Class C amplifiers for added efficiency (transistor is cutoff for part of each signal cycle) as the added distortion can be managed with a tuned load. The "gain" is a bit tricky though because the output signal then looks different than the input. The amplifier examples in this video are Class A (transistor on and burning power even when there is no signal) so in that sense they are linear.
@@solomioist (Voltage) gain in db = Adb = -20*log10(voltage gain in volts/volts) = -20*log10(vL/vB). If vL/vB=100 then Adb=-40, and if vL/vB=150 then Adb=-43.5. (If you want to express ratios of power in db, then you have to multiply the log10 of the ratio by 10. Make sure you don't forget the sign of the gain as appropriate.)
Amazing video as always. Great resource. New and very dedicated subsciber!
This is absolute legendary explanation. From my experience transistors have always been a very confusing topic considering so many variables and assumptions. Your videos are a treasure chest for this matter
Hello sir,I am from india, I am very big fan of your lectures sir, I took electronic and circuit course on edx, it's a mooch course, I really enjoyed your videos, I really liked the way of your teaching sir,keep it work, really you are best and master in this subject sir❤❤
Im also from india, and I watch these lectures everyday.
Also from India... Are you an engineer?? Or a hobbyist??
Professor Danner, thank you for explanation and examples.
Great lesson thanks
great video as always
Very informative playlists on your page! I think there is a correction in example 1. The base current is 4.3mA/100 = 0.043 mA and not 0.0043 mA. About 1%. Still negligible. 👍🏼
Good catch! Actually, more like about 8.6% of the current (from the voltage divider). Still small enough not to get too worried about.
👍👍👍👍👍👍👍👍👍👍
What about if I wanted a 50ohm input and output? Thanks
Don't you make a mistake at 32:45 when calculating Rin? Bre (beta times re) is 100.(2.8 ohms) = 280 ohms. That in parallel with 5k should give us around 265 ohms, isn't it? Or am I making a mistake somewere? In every case, I like your video's. Very nice and clear explanations.👍
I think you're right. Resistance in parallel is always less than all of its component resistances.
You're right.
@@adanner then the gain is -115? so the effect is quite different (around 26% smaller) from the calculation without minding r phi... is it right?
At 15:45 why is there a minus sign ? I know it needs to be there but that its not obvious from the AC equivalent circuit.
god's work
😍😍😍😍🥰🥰🥰🥰🥰
2:31 What if the emiter resistor is 0. Does it mean that I have to devide by zero??How can I obtain the emiter current if there is no emiter resistor?
If the base voltage is fixed to some Vb but the emitter is grounded, then there will be a nonlinear relationship between Ie and Vbe like that of a forward-biased diode. This can lead to it being a little unpredictable - the transistor might be in cutoff with Ie=0 (if Vbe is too low) or with Ie zooming up to some high transistor-burning number if Vbe is a little too high.
@@adannerThanks for the answer!! So it means that I must have a resistor there always? I have seen many (practical) circuits with a common emiter amplifier without a resistor at the emiter. Does it mean that it was wrong? My confusion is based on the problem I see when I try to understand some practical circuits (for example for radio control models, especially old ones from various books but not only) and I compare those circuits with theoretical circuits like here and it doesn't match sometimes. I feel like I am missing something here 🤔
@@grzesiek1x There could be a few reasons. For radio circuits they sometimes use Class C amplifiers for added efficiency (transistor is cutoff for part of each signal cycle) as the added distortion can be managed with a tuned load. The "gain" is a bit tricky though because the output signal then looks different than the input. The amplifier examples in this video are Class A (transistor on and burning power even when there is no signal) so in that sense they are linear.
Which unit has the calculated gain? If it is not dB, how can we convert it to dB?
If I may.. gain is a ratio of the output voltage over the input voltage. So it is unit-less.
@@rolfw2336 interesting. But what does the "gain" in db tell me with eg commercial amplifiers? So how can I derive it?
@@solomioist (Voltage) gain in db = Adb = -20*log10(voltage gain in volts/volts) = -20*log10(vL/vB). If vL/vB=100 then Adb=-40, and if vL/vB=150 then Adb=-43.5. (If you want to express ratios of power in db, then you have to multiply the log10 of the ratio by 10. Make sure you don't forget the sign of the gain as appropriate.)