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with a collector resistor you can have half Vcc on emitter by splitting the other half Vcc between the transistor and the collector resistor so the sum makes the supply voltage

Hi, this is a canned formula that you have to accept. It is not something that you can derive. Hope this helps.

@@electronicsainteasy5253 Thank you very much.

Good question. I don t know I have never seen it explained.

It doesn t have to be 1 volt, really. What s important is that the 3 resistors, that is, the two in the voltage divider and the one in the emitter leg make the setup stable wr to beta, that is, changes to beta don t affect the bias voltages. I think i will makd a video of why it is so.

As long as your output signal doesn't clip, sure. Buffers like that are considered considered current source to drive low impedance loads like a speaker. If you are doing audio, i think it's better to use op-amps which have very high input impedance and low output impedance. They are just easier to deal with.

Thanks for your comment. No idea what it meams though

Thank you for your comment!

yes, this type of circuit boost the voltage. think of it as a pre-amplifier. To drive a speaker, you need to boost the current, not really the voltage. Commonly referred as power amplifiers. If you do audio, you may want to look at op amps for boosting the voltage to proper level before your power amplifier.

For 10 watts you need about 5 amps to drive a 4 ohm speaker. So you would need a power transistor. Google 10 watt power amp transistor and you should find proper transistor to use then look at the datasheet for that Q and you might also get some sample schematics on the ds. Instead of Q, you can use ic like the LM 1875.

@@electronicsainteasy5253thanks. Just looked it up. Playing with bjts was more for education purposes also got quite a few on hands including them beefy ones. I like the 1875 too, will pick it up next time I’m ordering parts. What about using mosfets(or any fets for that matter)? They have high input impedance too right? Are they used for audio?

Note that what the video describes is for maximum voltage transfer, not necessarily maximum power transfer as the 2 are mutually exclusive. In the case of a speaker as load, you actually want max power transfer, not max voltage transfer. To accomplish max power transfer, you need a buffer amplifier (a voltage follower, basically) with high input impedance and low output impedance. That way, having a relatively high impedance at the output of your circuit (feeding into the speaker) is not an issue anymore. You could use a transformer too to act as a buffer. I talk about that in one of the videos about tube amps.

I made another video where I go into the details of picking the right value for voltage divider setup that sets up the bias voltage. Title of video is something like "common emitter design (part 1)". the key is make the design not dependent too much on the beta of the transistor. This gives you an equation in R1 and R2, and since you know the ratio, it gives you exact values for R1 and R2. No guess work. Note that the choice of R1 and R2 will also dictate (I believe) the input impedance of the your amp. Something to consider as you want that input impedance to be high, rather than load. Otherwise, you are gonna load down your source. Another reason to use op-amps for audio stuff.

I made other videos where I go into much more details. Especially the one about common-emitter amp design (part 1 and 2). Theory is nice and all but I highly recommend using LTspice or similar software to try things out.

@@electronicsainteasy5253 yes I just got into LT spice. What a time saver. Theory got me close but then to fine tune it I just kept adjusting resistor values till I get the behaviour I want. Thanks for the tip on using op amps. I don’t have much expertise with those that’s why normally opted for bjts.

Sorry i went a bit fast there. You have two equations for two unknowns. Using first eqn, you can express one of the two unknowns as a fumction of the other and you plug that expression into the second equation. It was not done in my head but on paper. Hope that helps

Can you please explain a bit more as I am struggling to see this. Also if I take R1 // R2 = 10k using your values I get 9200 is this correct. @@electronicsainteasy5253

Thanks a lot for watching the vid and your kind comment!

Thank you for your comment. Well appreciated.

input of course. Maybe it's the earphone used as a mike that's confusing. Earphone = high impedance mike (piezo-electric kind).

Glad you like it!

Good point!

Great to hear!

Glad you like it!

Thanks for watching!

Thank you so much 🙂

Thank you so much for the comment.

The old days when you could still walk into a Radioshack and buy one of those. Now, only available in secondary market as far as I know. But you can get those things very cheaply on ebay and the likes. Elenco still makes those nice ones but they are the snap-on kind, which I don't think are as good as the ones with the spring thingies.

Thanks for watching. I guess I could go into that. The good thing about superheterodyne radios is that any book on electronics will have a good description of what's involved. Of course, it won't have all the details on how to design the whole thing but all the steps will be explained. Then, it's a matter of studying each stage and designing each stage. I was gonna go into a different direction, like designing a synthesizer, but designing a radio is cool too. Food for thought, as they say.

Thanks a lot for the comment!

Thanks!

He said the resistor Rc in part I video was chosen at a value because of the halfway of the load line emitter current to incorporate the AC signal with the DC signal to be optimal (he taught that in the last lecture). That was a great part in his last lecture I enjoyed since it is more necessary in MOSFETs designs where high frequencies are in the amplified voltages using them over BJTs.

Yes, it s not easy stuff. I highly recommend watching part one first since i explain the dc bias setup. And you are absolutely right, the operaring poing or quiescent point should be at the middle of the dc load line so that the ac swings around the dc op point stay in the linear region.

Thank you for your comment.

Thanks for the info, exactly what I needed. If I'm powering the circuit from the camera battery (2s liion, 6 to 8.4v) then it would be best to use a LM7833 to get constant gain, right?

In the electronics for inventors book, the author makes the same assumption, which i think is very reasonable.

Yes, correct for dc analysis. When looking at ac analysis, you need to consider Re ll Rload instead of just Re.