I did this in college many many years ago. I got my qualification and yet I did not understand the circuits properly . You have done a marvellous job in explaining it all. Thank you so much for your video and your sense of humour and humanity. :-)
I really like your videos. I've taken an Electronics class at a university, but I find your videos to be a really nice refresher. And some things you cover make me think about circuits in a different way. Thanks!
Remembering good times @ college... I always hated that configuration because you rarely get exactly what you calculated. Also temperature and replacing components with the same part number can change the performance, but it is still fun for academic usage. I'm wandering if you are going to talk about good old power amps? I'd suggest that you design one and just explain the basics on the current mirror, the differential amplifier and the always handy type B output stage... Maybe a video no longer than 20 mins? Still I think you are doing a great job on spreading technical knowledge and making it public and handy. As a teacher, your videos awesome because if they don't get me, they can always review your videos!
Too complicated for me. But judging by the positive comment section, you were on point with your explanation. I'll keep looking for my enlightenment in other videos. Keep up the good work!
Thanks for the great video. I would like to ask something though. You sayed that we shouldn't change the value of R1 and R2 because then we are messing up with the point that the load line crashes with the y-axis. But this point is Ic(sat) = Vcc/(Rc+Re). This is a value that is independent of R1 and R2. Actually I am reading a book right now that claims the same thing as you do. Am I missing sth here?????
Well, if you change the ratio of R1 and R2, you change Vbb, and since Vee = Vbb - 0.6; that changes Ie ( Vee/Re) and Ic, which changes Vcc (V - Ic*Rc). So, and Ic will change, and Vce. So the load line ( see 2:14 ) will change too.
I stayed awake and watched it twice. Could you correct the value for R1 for the final setup. You stated 1147 Ohms. 20 X 1147 / 1147+1000 = 10.68. I think you meant 11470 Ohms. What 3 resistors did you use to achieve 11470 ohms? Trying to build this example. Really great information. Thanks
Sorry Let me try this again, 1147 Ohms. 20 X 1147 / 1147+1000 = 10.68 is incorrect. It should be 20 X 1000/1000+1147 =9.31. But using 20 x 1000 /1000+ 11470 = 1.60
How does one know what a good input/output impedance? I've heard him say that the ~800ohm input impedance isn't good, but in another video he said that the low (~10ohm) impedance wasn't good either. Is there an ideal input/output impedance?
Lance Clark Without going into the possibility of considering a voltage or a current source or any of the four possible amplifier configurations as, Voltage to Voltage, Voltage to current, Current to voltage and current to current amplifiers let me try and give an analogy of input and output impedance. Output impedance. Assume a speed boat pulling a skier behind it, or a truck towing a car. Now the tow rope can have various stiffness and it could be elastic. If the tow rope can stretch then the load will not be at a constant position with respect to the pulling craft and the same happens when an signal source having a high impedance is used to drive a load, The voltage applied to the load will change with the load as when a truck pulls a car and the load changes when going uphill or on the level, or downhill really when no load can be assumed. Then elasticity of a tow rope may be considered as being analogous to the output impedance or a source or an amplifier. Input impedance. Well if a person is trying to lift a parcel with his outstretched hand, irrespective of how strong he is, the weight of the parcel will decide how far he can lift it. If the parcel is very light , then man can lift his hand at the end of his arm very high as the parcel is light. If the parcel is heavy, the man cannot lift his outstretched hand too high. Well the mass of the parcel may be looked upon as the input impedance of the parcel. If the parcel is very heavy, then then the input impedance is LOW, if the parcel is not so heavy, then then input impedance is HIGH. I shall not take this analogy any further as some amplifiers work with then " pressure " of the voltages applied and not the current flow, while there are other amplifiers which operate not pressure or voltages but with the rate of flow or current amount. In a water pipe one may have high pressure and no flow, or a lot of flow and little pressure. One can have constant flow systems and constant pressure systems in any hydraulic system all working with flexible/elastic or non flexible solid pipes. Basically all the effects of hydraulics using pressure or flow in solid or elastic pipes can be found in systems using voltage(pressure), current ( amount of flow) and solid or elastic pipes ( impedances or admittances) One should write a book on analogies to try and help people understand the behaviour of invisible behaviour in electrical systems. About your question how one does one know what is a good input and output impedance, well as in reality there is no perfection, youn can find the answer if you considering the problems in the TOWING INDUSTRY and consider what sort of truck one needs to pull or tow a particular broken car/ load and the choice of tow ropes youn can tollerate considering that even heavy trucks cannot pull all kinds of loads and choice of TOW ROPE IS also to be considered. Basically to conclude. A current input has a low impedance while a current output has a high impedance while a voltage input has a high impedance while a voltage output has a very low impedance, as they involve operating with either Pressure or the Flow. In hydraulics I find that many do not understand the difference between pressure meters and flow meters. Many students when asked what happens when a there is nan rate of change between a conductor and a magnetic field, most say, " A current flows" NO NO NO the answer is a voltage is generated (pressure). A current flow will only takes place if the conductor forms a closed circuit hence FLOW is possible. Again you can have pressure( voltage) without flow( current) such as an open circuit and one may have current (flow) with no pressure( voltage) such as a short circuit. A;ll this can happen in electrical sources and amplifiers.
Depnds on your source of signal and your load. Generally you do not want any attenuation as a rule of thumb, the output impedance should be 10 times smaller than your load impedance. This of-course is not true in some applications such as rf/transmission circuits where you match impedances of source and load for maximum power transfer.
I did this in college many many years ago. I got my qualification and yet I did not understand the circuits properly . You have done a marvellous job in explaining it all. Thank you so much for your video and your sense of humour and humanity. :-)
I really like your videos. I've taken an Electronics class at a university, but I find your videos to be a really nice refresher. And some things you cover make me think about circuits in a different way.
Thanks!
You never fail to entertain me while at the same time teach me something new. Thank you Pete!
In my memory the cutoff point for an amplifier is defined by the -3dB point. -3dB for voltage is 1/sqrt(2) not 1/2
Alright, I need to setup one of these. Haven't had a lesson on amplifier circuits since high school. Time to recall those memories. :)
i feel Johnny Depp teaching electronic
Another Great video from "According to Pete"
I learned a lot. Thanks.
Remembering good times @ college... I always hated that configuration because you rarely get exactly what you calculated. Also temperature and replacing components with the same part number can change the performance, but it is still fun for academic usage.
I'm wandering if you are going to talk about good old power amps? I'd suggest that you design one and just explain the basics on the current mirror, the differential amplifier and the always handy type B output stage... Maybe a video no longer than 20 mins?
Still I think you are doing a great job on spreading technical knowledge and making it public and handy. As a teacher, your videos awesome because if they don't get me, they can always review your videos!
Ahh,memories of glowing TO-92's. They only glow for a split second though. ;-)
Nice video, thats a cool function generator. Did you make it yourself?
Too complicated for me. But judging by the positive comment section, you were on point with your explanation. I'll keep looking for my enlightenment in other videos. Keep up the good work!
Do you have a video on the function generator that you use? I want to build one..
is there a video about the function generator used in this video?
Thanks for the great video. I would like to ask something though. You sayed that we shouldn't change the value of R1 and R2 because then we are messing up with the point that the load line crashes with the y-axis. But this point is Ic(sat) = Vcc/(Rc+Re). This is a value that is independent of R1 and R2. Actually I am reading a book right now that claims the same thing as you do. Am I missing sth here?????
Well, if you change the ratio of R1 and R2, you change Vbb, and since Vee = Vbb - 0.6; that changes Ie ( Vee/Re) and Ic, which changes Vcc (V - Ic*Rc). So, and Ic will change, and Vce. So the load line ( see 2:14 ) will change too.
I have built a powerfull self oscillating teslacoil with a very similar circuit using gate biasing to start oscillations
I subscribed! I'm gradually leaning...
I stayed awake and watched it twice. Could you correct the value for R1 for the final setup. You stated 1147 Ohms. 20 X 1147 / 1147+1000 = 10.68. I think you meant 11470 Ohms. What 3 resistors did you use to achieve 11470 ohms? Trying to build this example. Really great information. Thanks
what if we are using a current supply?
Cutoff is at -3dB. Which is at 10V -> 7.7V
Great tutorial... Thanks a lot for that.
How do you calculate Rbe?
How did he work Beta out to be 200?
why is the gain = Rc / Re?
Man I'm loosing the will to live
@5:10 Its like your a midget :D
Ps: Great lessons!
No link to part 1 in the description? Doh. If you want part 1 it's here th-cam.com/video/t0UOSIUve9E/w-d-xo.html
I fell asleep during the video, can not believe it!
Sorry Let me try this again, 1147 Ohms. 20 X 1147 / 1147+1000 = 10.68 is incorrect. It should be 20 X 1000/1000+1147 =9.31. But using 20 x 1000 /1000+ 11470 = 1.60
Thanks for using a working marker this time. Part1 was funny the first few times but ended up unwatchable.
How does one know what a good input/output impedance? I've heard him say that the ~800ohm input impedance isn't good, but in another video he said that the low (~10ohm) impedance wasn't good either. Is there an ideal input/output impedance?
Lance Clark Without going into the possibility of considering a voltage or a current source or any of the four possible amplifier configurations as, Voltage to Voltage, Voltage to current, Current to voltage and current to current amplifiers let me try and give an analogy of input and output impedance.
Output impedance. Assume a speed boat pulling a skier behind it, or a truck towing a car. Now the tow rope can have various stiffness and it could be elastic. If the tow rope can stretch then the load will not be at a constant position with respect to the pulling craft and the same happens when an signal source having a high impedance is used to drive a load, The voltage applied to the load will change with the load as when a truck pulls a car and the load changes when going uphill or on the level, or downhill really when no load can be assumed. Then elasticity of a tow rope may be considered as being analogous to the output impedance or a source or an amplifier.
Input impedance. Well if a person is trying to lift a parcel with his outstretched hand, irrespective of how strong he is, the weight of the parcel will decide how far he can lift it. If the parcel is very light , then man can lift his hand at the end of his arm very high as the parcel is light. If the parcel is heavy, the man cannot lift his outstretched hand too high. Well the mass of the parcel may be looked upon as the input impedance of the parcel. If the parcel is very heavy, then then the input impedance is LOW, if the parcel is not so heavy, then then input impedance is HIGH.
I shall not take this analogy any further as some amplifiers work with then " pressure " of the voltages applied and not the current flow, while there are other amplifiers which operate not pressure or voltages but with the rate of flow or current amount. In a water pipe one may have high pressure and no flow, or a lot of flow and little pressure. One can have constant flow systems and constant pressure systems in any hydraulic system all working with flexible/elastic or non flexible solid pipes. Basically all the effects of hydraulics using pressure or flow in solid or elastic pipes can be found in systems using voltage(pressure), current ( amount of flow) and solid or elastic pipes ( impedances or admittances)
One should write a book on analogies to try and help people understand the behaviour of invisible behaviour in electrical systems.
About your question how one does one know what is a good input and output impedance, well as in reality there is no perfection, youn can find the answer if you considering the problems in the TOWING INDUSTRY and consider what sort of truck one needs to pull or tow a particular broken car/ load and the choice of tow ropes youn can tollerate considering that even heavy trucks cannot pull all kinds of loads and choice of TOW ROPE IS also to be considered.
Basically to conclude. A current input has a low impedance while a current output has a high impedance while a voltage input has a high impedance while a voltage output has a very low impedance, as they involve operating with either Pressure or the Flow. In hydraulics I find that many do not understand the difference between pressure meters and flow meters.
Many students when asked what happens when a there is nan rate of change between a conductor and a magnetic field, most say, " A current flows" NO NO NO the answer is a voltage is generated (pressure). A current flow will only takes place if the conductor forms a closed circuit hence FLOW is possible. Again you can have pressure( voltage) without flow( current) such as an open circuit and one may have current (flow) with no pressure( voltage) such as a short circuit. A;ll this can happen in electrical sources and amplifiers.
Depnds on your source of signal and your load. Generally you do not want any attenuation as a rule of thumb, the output impedance should be 10 times smaller than your load impedance. This of-course is not true in some applications such as rf/transmission circuits where you match impedances of source and load for maximum power transfer.
This was messy :/