Clarification with biasing single supply op amps and what ground to use - Page 1
Forum section: electronics / beginners

author: ramonmaruko
author_rank: Contributor
timestamp: May 21, 2017, 11:28:00 am
content: Hello,I am trying to create a simple unity gain buffer for a guitar pedal. I'veattached the schematic for this oneSimpleBuffer.png.Since the op amp's non-inverting input has been biased withVcc/2(willbe calledVgndhereafter), should I set the circuit's output jack'sground toVgndso that the guitar amplifier or next guitar pedal in linewould correctly see a non strummed guitar signal as 0V (with reference to this circuit's signal out)?I am confused since some popular guitar buffer circuits (buffer1.gifandsab-buffer1.pdf) that I've seen in the Internet has the outputjack's ground connected toGND. Wouldn't that set a quiescent voltage ofVcc/2?

author: fcb
author_rank: Super Contributor
timestamp: May 21, 2017, 11:47:55 am
content: Input and output grounds should be the same as circuit/opamp ground.The capacitors at the input and output of the circuit block DC.

author: ramonmaruko
author_rank: Contributor
timestamp: May 21, 2017, 12:23:12 pm
content: I've tried simulating the circuit using LTspice with 0V Vin, and Vout is constantly outputting almostVcc/2.

author: fcb
author_rank: Super Contributor
timestamp: May 21, 2017, 01:30:45 pm
content: Please post the output of the simulation, and ensure you are reading after the output capacitor.

author: ramonmaruko
author_rank: Contributor
timestamp: May 21, 2017, 01:39:00 pm
content: Thank you for the replies. I've attached a screenshot of the circuit + graph. Noden001is the node connecting C2, R4, and R5.I ran the simulation using the trans analysis with Vin set to 0. The ac analysis line has been commented out.

author: mikerj
author_rank: Super Contributor
timestamp: May 21, 2017, 02:03:02 pm
content: 10uF and 2 Megohms gives a time constant of 20 seconds, your simulation is over in 100ms. 2 Megohms is way too high for this, both as an output load and for the biasing.

author: danadak
author_rank: Super Contributor
timestamp: May 21, 2017, 02:54:28 pm
content: QuoteThe input is AC coupled, so any value of DC as source on input, ignoring cap leakage,will result in output just reflecting the bias value.Regards, Dana.

author: rstofer
author_rank: Super Contributor
timestamp: May 21, 2017, 03:25:58 pm
content: Given the capacitive coupling on input and output, your output jack ground should be the same as your input jack ground and ignore the fact that the op amp is operating above ground.That's what the capacitors are for, removing the DC offset.

author: ramonmaruko
author_rank: Contributor
timestamp: May 21, 2017, 04:30:28 pm
content: After increasing the simulation time, I was able to seeVoutstabilizeto 0v. The reason I wanted for the high values for R1 and R2, was because Iwanted the circuit to have the same input impedance of my guitar amp (1M ohms).I didn't really know what I was doing with R3 and C3. I just followed thevalues found in the attached file (buffer1.gif) on my first post.How do I choose the "best" values for R3 and C3? I know that I just need it tohave a cutoff frequency set so that audible frequencies could pass through. Butthere is a very large possible combinations of R3 and C3 to achieve this. Howwould I know what combination to use for a given use case?

author: danadak
author_rank: Super Contributor
timestamp: May 21, 2017, 06:48:10 pm
content: You have a HPF with C3 R7, use this as a calculator -http://www.2pif.com/high-low-pass-filter.phpHowever R1, R2, R3, R4 affect the filter, essentially degenerating the effective value of C3.So do an ac spice analysis to get exact filter response.Also if C3 is an electrolytic they have very poor tolerance, so account for that accordinglyin cap selection.Regards, Dana.

author: mikerj
author_rank: Super Contributor
timestamp: May 27, 2017, 09:40:10 pm
content: Quote from: ramonmaruko on May 21, 2017, 04:30:28 pmYou don't need to match impedances in this case, you aren't dealing with a transmission line. A low output impedance is beneficial in terms of reducing noise pickup in the interconnects, and won't be the input capacitance of your amplifier or stray cable capacitances etc.

author: BrianHG
author_rank: Super Contributor
timestamp: May 27, 2017, 10:48:02 pm
content: If you are going to use a bass guitar, increase the value of c1, especially you shrink the 2meg input load. 10uf on the output isn't too bad for frequencies down below 40hz and above 20hz so long as you shrink the output R3 2meg resistor to something normal like 100k. Don't worry about the drive, that's the purpose of the op-amp, to drive the 100k without signal loss & to drive the cable and next external device stage even if it is something like a 50k, or 10k load.

author: ramonmaruko
author_rank: Contributor
timestamp: May 30, 2017, 04:47:01 pm
content: Thank you for the great replies everyone. Now I am starting to wonder why most of the distortion circuits I've look have such a high output resistors.

author: Zero999
author_rank: Super Contributor
timestamp: May 30, 2017, 05:47:11 pm
content: Quote from: ramonmaruko on May 21, 2017, 11:28:00 amYou told LTSpice to start up with all voltages at zero, so you had to wait for all the capacitors to charge. If you remove the 'startup' statement from the simulation command, it simulates perfectly, without having to wait for the capacitors to charge because LTSpice works out the initial operating conditions, i.e. the DC voltages and currents on capacitors and inductors respectively, before starting the simulation.