56. Combination FET and BJT Circuit (BIFET)

[@kionmahuermicio9860]

My left ear understood very well

[@janimker9518]

sweeet! I've been now looking on more of your videos, with basic learn topics like the CE and CC amplifiers, and all are great. One common suggestion for a more proper "finale" of each video: addition of a 5minute explanation where the particular circuit shines application wise (pro and cons inclusive). It is just a suggestion, not a complain, to give us a higher level view. Cheers. (Btw: I have seen a bit of H2 distortion in the output waveform - is this desired and is it implicit?)

[@vincentasare2013]

I have been trying to implement the same circuit but with a Vcc of 5V and Rload at 10 ohms to get a gain of 20V/V but it isn’t working. Is it the kind of circuit or is the voltage too small to get such a gain? I was also thinking it may be the load resistance that is too small. I need help please.

[@prakashreddyjaya9405]

I know answer: for example Vin:5v and gain:20 so that Vout:5*20= 100v practically this won't possible i.e our total always should be Vout

[@cardcaptorsakura97]

Is there any other way to solve for the I_D?

[@Jaycognito]

Looking at an old three stage amp in a R2R (P/B head pream) referred to as equalization amp has FET in middle. I assume for Z match and/or phase inversion, etc. In other words, the FET may be less for gain and more for matching or other effects. Us this commmon?

[@TheOffsetVolt]

Hi. I'n not sure about the phase inversion, that would depend on the surrounding resistor network. It could definitely be used for some type of matching. And thanks for watching.

[@chanyeolpark4724]

Can I ask what transistor you used need for our project? Hoping for your reply

[@khoulodmohamed3152]

I implement the same cuircut in multsuim and I add marker at vin and Vout to estimate the overall circuit gain but in stimulation it appear vin (sine wave) bigger than Vout and I don’t know the problem. How to solve it?

[@Cesar_III]

I took the time to simulate it in multisim and it yields a gain of 300 (without the emitter capacitor).

[@squadspud]

what are these capacitors values?

[@kionmahuermicio9860]

2nF aprox, it's more than enough

[@erickmiller728]

whay would be your preferred way to introduce feedback in this circuit?

[@TheOffsetVolt]

Hi, There are lot's of feedback methods and which one you use depends on the need. The simplest is to use resistors on the source and emitter as I've done here. Another option is to place a resistor and capacitor (in series) between Q2 collector and Q1 source (split R5 into two resistors and place the other end of the cap/resistor between the now split R5). I did a video on feedback that might help - kzbin.info/www/bejne/rKaymXWFlM-maqM

[@poincareelcartografo5406]

Maybe im wrong, but i think that the second stage is not beta independet

[@TheOffsetVolt]

Hello, The stage is largely beta independent, not entirely. The Q point is fixed by the base resistors and emitter resistors. The gain of the CE stage is fixed by (RC||RL)/RE1 so no beta there either. The current through RL is set by the Av. Beta is considered for the base resistance. Thanks for watching.

[@poincareelcartografo5406]

Thanks for answering. Your videos are very helpful for me

[@TheBornBaker]

what if we were given zin and av how do we know whst size resistors to use

[@TheOffsetVolt]

Hello Julian, Thanks for viewing my videos. I can't really answer your question without more detail. Since the transistor has the Av, you would change RCIIRL/RE1. Once you know RE1, multiply by Beta to get the approximate base resistance (use the lowest beta from the data sheet). You now have one resistance of the 3. From here you can use currents or rearrange the equation for parallel resistance (1/R?=1/Rth-1/R... I won't add more since this sounds like a homework problem :)

[@Rajuuuuuuuuuuu]

Is that E-mosfet?

[@TheOffsetVolt]

Hi Parithive - that's Jfet.