I came looking for answers as I try to decipher some very basic sold state guitar amplifiers. In my application, as I understand it, this increases the current, without amplifying the guitar signal. In doing so, it makes sure that the tone-stack does not take all the 'oomph' out of the signal on its way to the amplifier (LM386 in this case). Thanks for the clear explanation!
@byterock3 ай бұрын
Hmm in this case they are most likely not using that LM386 op-amp as a 'voltage-follower' or as 'buffer' which would be the analogous transistor circuit. The LM386 has a preset gain of 20DB, so it will always amplify the input by at least that. I can only guess at what they are doing but the LM386 is often used as a 'Bass Boost' in amps, perhaps this is what you are seeing. If there is a cap then resistor that can be traced from pin '5' (V-out) to pin '1' (GAIN) odds are that is what you are looking at. Thanks for you comment.
@byterock3 ай бұрын
Opps I think I got that dead wrong. I misread you comment. I would say yes there is a very good likelihood that there is a buffer into a tone stack before it goes to an LM386 though the gain of the LM386 can be as high as 200DBs so any lost 'oomph' could be easily added back in. ;)
@kensmith56944 ай бұрын
Cathode followers are also kind of nice in filter circuits. You can make a two or three pole filter or a notch with a known Q a little over 1.0 using one to do a positive feedback action. Also if you are making an oscillator, it is worth considering that you can make the LC tank circuit step up the amplitude from the cathode and feed it into the grid. You can then take the output at the plate. This is a nice thing to do because the tuned circuit is not connected to the output so the loading has almost no impact on the frequency.
@byterock4 ай бұрын
Yeah you could write a few book chapters on them unfortunately most of them are long out of print ;). I just stuck to the very simple example as it is a 'Short' video. I just started playing with an old Tek scope (545) and it looks like (to me) 1/2 of the tubes in that are using the cathode is some way. So much to relearn.
@kensmith56944 ай бұрын
@@byterock They will need to get an exact gain so cathodes will likely get doing something
@ludmilascoles11955 ай бұрын
Thanks I was wondering about that the other day😂
@byterock5 ай бұрын
Funny how that works out ;)
@davidluther39554 ай бұрын
THEIR SHOULD BE A GRID RESISITOR FROM GRID TO GROUND TO PROVIDE THE NEGATIVE VOLTAGE ON THE GRID.
@byterock4 ай бұрын
@@davidluther3955 Yes that is true . For simplicity I left all the 'Normal' bits off, like the coupling caps, out of the first example. A real circuit can be seen at the 4:00 mark, After all it is suppose to be a 'Short' video. Oh by the way you do not 'Need' a '-' voltage on the grid, you just need a '-' potential difference between the grid and the cathode. Thanks you gave me an idea for another short circuit 😉