Will this Schematic work?

Started by Otwa05, December 21, 2021, 12:31:05 AM

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Otwa05

Gday all, first post here so I'm not sure if I'm doing it right but I'm just looking for someone knowledgeable to confirm if my schematic (link below) "works". It's just supposed to be a phase switchable boost that I plan on putting into a feedback looper to hopefully get something similar to the death by audio total sonic annihilation. Do I need a capacitor between the output of the opamps and the input of the transistor? Are the buffers wired correctly? Can I just use the one 430k resistor pulling up to 4.5v to bias before the transistor? Any help would be awesome!
Cheers 👍🏻



Keppy

Quote from: Otwa05 on December 21, 2021, 12:31:05 AM
Do I need a capacitor between the output of the opamps and the input of the transistor?
Yes, assuming that you don't want the transistor gate biased at 4.5v like the opamps.

Quote
Are the buffers wired correctly?
No.The non-inverting buffer (top) needs the ground connection on (-) input removed and the biasing resistor moved to the other side of the switch. The inverting buffer (bottom) needs resistors of equal value for input (switch to -in) and feedback (output to -in). Also, the outputs should be switched as well as the inputs. Opamp outputs directly connected will fight for supremacy, with undesirable results.

You might prefer a single-opamp design, like here: http://geofex.com/Article_Folders/polarity_reverser/polarity_reverser.htm

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Can I just use the one 430k resistor pulling up to 4.5v to bias before the transistor?
That would work for a transistor buffer, but not for a boost. The C/E resistor ratio is off too, compared to the bias resistors. Right now you're set up for less than a volt of bias on the gate (1/11 of 9v based on the bias resistors), which will put the emitter just above ground (gate voltage - .6v diode drop inherent to the B-E junction). Since the C/E resistors are the same value and the same current passes through both, the collector voltage will be exactly as far from 9v as the emitter voltage is from ground, putting it almost all the way at 9v. You want the collector closer to 4.5v for max headroom. Make sure you're working from a known-good boost circuit and then tweak from there. This schematic for the Linear Power Boost is similar to yours, but with resistor values known to work: http://beavisaudio.com/techpages/SchematicToReality/img/am1.gif
"Electrons go where I tell them to go." - wavley

antonis

"I'm getting older while being taught all the time" Solon the Athenian..
"I don't mind  being taught all the time but I do mind a lot getting old" Antonis the Thessalonian..

Otwa05

Quote from: Keppy on December 21, 2021, 01:36:04 AM
Quote from: Otwa05 on December 21, 2021, 12:31:05 AM
Do I need a capacitor between the output of the opamps and the input of the transistor?
Yes, assuming that you don't want the transistor gate biased at 4.5v like the opamps.

So if I was to leave the resistors out entirely would the transistor be biased from the previous op amp?
Also what benefit does moving the 1m bias resistor for the opamp to after the switch have?

Thankyou so much for your help so far 🙌🏻

antonis

Quote from: Otwa05 on December 21, 2021, 03:20:50 AM
So if I was to leave the resistors out entirely would the transistor be biased from the previous op amp?
Yes but not in a " conventional"  level.. :icon_wink:
Base on 4.5V means Emitter on 3.8 resulting into only +/- 2.6V Collector swing margin..

Place a 100nF or so cap, make 43k resistor 100k, 430k resistor 470k, 500k(pot..??) 1k and Collector resistor 4k7..
(just a suggestion for a happy CE working amp..) :icon_wink:
"I'm getting older while being taught all the time" Solon the Athenian..
"I don't mind  being taught all the time but I do mind a lot getting old" Antonis the Thessalonian..