Amp Circuit Analysis - Bassman v. JTM45 vs. Plexi vs. JCM800 | Pt. 1 

I love the smell of electrolytic capacitors in the morning. Smells like... victory.

I'm not proud of it, but just last week I had to suck off 1,000 Banshees while being spaghettified inside of a blackhole just to re=cap a 5150. Really would've preferred to just pick it up with my milk, at the local Groceletriceries-4-less

You noticed that to. it should be elementary electrical knowledge to know when explaining a circuit.

Yeah I noticed 😂

ps I really enjoy your channel, keep up the great work. I’m also a huge Uncle Doug, D-Lab, Blueglow and Guitologist fan

I see, so the voltage divider is not 68k / 68k + 1M as I explained. But it is just 68k / 68k. Thanks for the update.

@@KleyDeJong Yes Chris is correct.

Bemi-ton Yes absolutely! This long tail inverter is just a classic differential amplifier since the two amps share a cathode resistor. The input here goes to the top triode and the bottom triode of the pair just sees ground through a cap, and is biased on equal to input one through the 1 megs and the shared 470 ohm resistor. Because the current through the cathode resistor is constant, an inverted signal appears at the bottom triode's plate. So the two signals at the plates are inverted 180 degrees relative to each other. I don't really like the term grid leak resistor,... the 1 meg resistors set bias because they place the voltage at the grid below the cathode potential, partially reducing the current that would flow from Plate to Cathode without a voltage on the grid. One meg places the quiescent current at the mid point of the linear excursion, of a 12AX7 when the cathode resistor is 1.5 K. Different tubes require different bias,. As far as i know 12AU, AV, AX, and AY, work well with 1 meg to ground from the grid, with 1.5 K on the cathode. Changing the value of the grid resistor versus cathode resistor, move the bias point, so that there is more excursion above the quiescent point, or more excursion below the quiescent point. The Phase inverter uses 470 ohms as a common reference point on the Phase inverter so that the top half or bottom half of the signal get cut off in opposite directions. The top triode in the P I ends up passing more of the positive side of the signal, and the bottom triode passes more of the negative side of the signal. This allows for a greater overall excursion of the signal in the P I before the signals (now two inverted from each other ) reach the output tubes. It also means that for the same output half the drive current is needed for class B, (in which there is no overlap). In class A/B there is also a pretty significantly lower current draw for the same output at the speaker of a class A amp. This is why for the same output Class A/B is much more efficient in terms of the amount of power that is lost as just heat. I am quite sure that the so called imbalance of using 82K for the top half, and 100K for the bottom half, is actually a method for "balancing" the drive signal. I think there are two things that happen, number one in the top half, remember the plate signal and cathode signal are 180 degrees apart, add together for more excursion, and in the bottom triode they subtract, so the bottom needs more gain. Also one wants the center point of each plate offset, so using the same bias scheme, the difference in plate resistors change the voltage at each plate a little relative to each other, for the offset. Recently various guitar amps have tried to make the plate resistors equal which doesn't work so well, it seems to me.

Try either or both of theses links: www.valvewizard.co.uk/acltp.html www.aikenamps.com/index.php/designing-long-tail-pairs-the-load-line-approach On the Valve Wizard's site, first check out the link to the DC coupled LTPPI.

Thanks for that. So you still get the 50% voltage divider and with the lower input impedance you may experience some duller top end?

@@KleyDeJong The 1M resistor is shorted mechanically by the unused (stereo) input jack #1. Thus, the 2nd 68k resistor is shorted to ground and the voltage is divided by the two 68k resistors ("50%" as you called it). That is -6dB attenuation (quarter of the power) and that's why the signal is much quieter. Sound is duller because the input interacts with the passive guitar circuit (passive RLC circuit). This is called "loading effect. With a buffer it will be less pronounced. The input impedance is a mere 128k compared to the 1M on the 2nd input and to the output impedance of a guitar Cheers, Dror

Thanks, will fix.

Great suggestion, thanks!

Yeah that would work fine

Lol thanks. I believe it was a lid to a container my kid was playing with.

I believe that standby switches are not necessary. www.valvewizard.co.uk/standby.html

To answer your question, yes. With a modification described by the Valve Wizard in Kley's link, no standby switch would be necessary.

I believe it would send very high frequencies of opposing phases to cancel one another. It would help stabilize the amp. I imagine it is very high, possibly even beyond the 20khz range.

Kley De Jong: Thank you. You did a great job to go through all the details 👍

Send me an email at kleysquestions@gmail.com