Voltage divider or op amp for output? - Page 1
Forum section: electronics / beginners

author: vmallet
author_rank: Regular Contributor
timestamp: September 21, 2020, 09:04:40 pm
content: I've been playing with a design (not mine) for an oscillator for an analog synth. It generates a sawtooth signal and shapes it in different ways (triangular, square, sinusoidal, pulse, etc). The various signals are then connected to the output terminals so they can be patched into the input stages of other modules (filters, mixers, etc).Some of these signals are taken straight off the output of the output stage op amp, so they will have a nice low impedance for the next module.However some other signals must have an amplitude that's too large and go through a voltage divider before being exposed. For example the square output goes from the output of a LM301 op amp (configured as a Schmitt Trigger) through a 2K2 / 2K divider, shrinking the output to 2K / (2K + 2K2) = 47%.The question: wouldn't it be wise to buffer the output of the voltage divider (with an op amp in a voltage follower configuration) to present a nice, low output impedance to the next module? Or, if we didn't care about phase, could we just use an inverting amplifier with say Rf = 51K1 and Rin = 107K (to have a gain of roughly -0.48) in place of the voltage divider to have a much stiffer output (is that the right term?) And if we cared about phase, add another inverting amplifier with a gain of -1 to shift it back in phase?In other words, if I have a signal swinging between +/-11V and I want to bring it to roughly +/-5V before the next stage, which one should I pick:a) voltage divider with nothing else, but not the lowest output impedanceb) voltage divider with bufferc) inverting amplifier with similar gain, but 180deg out of phased) inverting amplifier with similar gain, followed by another inverting amplifier with -1 gain.e) something else?Sorry about the newbieness of the question, I am just starting to discover the goodness of op amps and there is so much to learn there...

author: james_s
author_rank: Super Contributor
timestamp: September 21, 2020, 09:44:11 pm
content: It's probably not super critical, but I would use an op amp configured as a unity gain buffer and place that after the voltage divider. Op amps are cheap, at least general purpose ones like you'd probably want to use here are.

author: KT88
author_rank: Frequent Contributor
timestamp: September 21, 2020, 10:42:59 pm
content: The best solution depends on the length of the cable and it' capacitance. The resistive divider and the (parasitic) capacitance form a lowpass filter. If the capacitance is low and the resistance of the divider is low you might see no issues.With longer cables a buffer makes sense.An issue with opamps is their (more or less existing) inability to drive capacitive loads.The remedy for that is called Lead Compensation:https://www.analog.com/en/analog-dialogue/articles/ask-the-applications-engineer-25.html.Another possible issue with inverting amplifiers is noise gain that is inherently larger for the inverting configuration. As an example for an amplifier with a gain of -1 the noise gain is 2https://www.analog.com/en/analog-dialogue/raqs/raq-issue-91.html.Interestingly a higher noise gain cand help with stability at capacitive loads...CheersAndreas

author: Doctorandus_P
author_rank: Super Contributor
timestamp: September 21, 2020, 11:11:58 pm
content: All components you add, add noise to your system, even resistors.Bigger value resistors generate more noise. The root cause for this is the fixed charge of an electron. Each time an electron comes out of your resistor it gives a "bump". (Simplified explanatin).Audio circuits tend to have lots of resistors in the range of 1k to 10k, because those are the lowest values (less noise) that an regular opamp output can drive comfortably.Components also cost money, and ad complexity, you also have to solder them, they use up board space, they can fail and thus reduce reliability, make fault finding more difficult, etc.In the end it makes it all application specific. In lots of situations there is no need to add extra buffers.Even if the audio signal sags a bit because of output loading, your ears won't hear the difference, as they work sort of logarithmitically and small changes in level are insignificant.But don't believe me. Try it yourself.Build several variants of your audio circuit on a breadboard, do some measurements. Scope pictures often do not mean much for audio. Get yourself some decent headphones to listen to your circuits.If you don't have a separate headphone amplifier, there is a perfect opportunity to build one!

author: David Hess
author_rank: Super Contributor
timestamp: September 22, 2020, 10:00:56 pm
content: For applications which require a zero impedance source, then buffering is required, however this can be tricky to implement because of the capacitance of the cable as KT88 points out before my response to him below.The solution then becomes a "zero impedance output" which uses an operational amplifier with feedback taken *after* a series output resistor, and with appropriate frequency compensation to provide stability. Douglas Self discusses this in his book Small Signal Audio Design.In the past when I implemented the circuit he describes, (1) I deliberately included a shunt capacitor directly across the output to suppress RF trying to get into the output and to provide a consistent load. This limits bandwidth but provides a low AC impedance at high frequencies. This is the common "rail splitter" circuit using an operational amplifier configured to drive a heavy capacitance load. There are actually better rail splitter circuits so I do not use this one as such anymore but it does make a pretty good and tough audio line driver. Analog Devices shows this configuration at the last examplehereand describes the math behind it.Having said all of the above, usually zero impedance line drivers are not required and a source resistance of 10s to 100s of ohms are completely acceptable in audio applications.(1) I was not aware of Douglas Self or his book at the time or "zero impedance" drivers although I understood the concept just fine and implemented it on my own.Quote from: KT88 on September 21, 2020, 10:42:59 pmWhat you describe is only the case if the transmission line is not terminated. And it only has to be terminated at one end to provide full bandwidth.For example old Tektronix oscilloscopes have a vertical output with a source impedance of 950 ohms. When used to drive a 50 ohm cable, which presumably is connected to another high impedance input, then the cable capacitance limits bandwidth like you describe. But if the other end is terminated into 50 ohms, then full bandwidth is available, 10s to 100s of MHz, albeit with an attenuation of 20.

author: KT88
author_rank: Frequent Contributor
timestamp: September 22, 2020, 11:12:10 pm
content: I agree that every cable or PCB trace is a transmission line. In the above case I assumed that the connection won't be more than a meter (three feet) where transmission line properties are (usually) not an issue. Given that vmallet is talking about an analog synth, there won't be a reason to get into (esoteric) regions of (very) high-end audio or stage equipment that is based on 600-Ohms systems because of very long cables used...CheersAndreas