How can I calculate the input impedance of an LM3886? - Page 1
Forum section: electronics / beginners

author: hummusdude
author_rank: Regular Contributor
timestamp: May 14, 2021, 04:48:02 am
content: I built a gain clone amp based on the LM3886 that I found on circuitbasics.com and it sounds really good. But I'm struggling with how to calculate the input impedance. I'm making a preamp for the project and I'm trying to make sure the two stages are close enough. I've scoured the datasheet but TI doesn't give an explicit value. They show a typical application circuit, which the fine folks at circuitbasics.com based their design on. And they give a short explanation about how Rin relates but the I'm not sure how to interpret their explanation. To make it more confusing, circuitbasics.com added some input filtering to set the bandwidth and it seems like that could be impacting the impedance as well.And something else that I find confusing...TI says in their notes that the input impedance is determined by Rin but I've read over and over again that op amps are designed with very high input impedances (>1M) in order to keep the current very low.I could really use some help sorting the wheat from the chaff!

author: ledtester
author_rank: Super Contributor
timestamp: May 14, 2021, 05:20:13 am
content: QuoteYes, the inputs to the LM3886 are very high impedance -- basically no current enters them. Adding Rin actually decreases the input impedance of the system as a whole. The resistance to ground is Rin in parallel with a really, really large resistance which is basically Rin.If you post the schematic of the amplifier I'm sure we can determine the input impedance of the configuration.QuoteMy understanding is that you want the output impedance of the pre-amp to be a fraction (like 1/10th) of the input impedance of the amp.

author: hummusdude
author_rank: Regular Contributor
timestamp: May 14, 2021, 06:01:03 am
content: sorry...I meant to add these pics with the original post.

author: ledtester
author_rank: Super Contributor
timestamp: May 14, 2021, 06:37:59 am
content: My guess is that the input impedance in that schematic is set by the 20K resistor.As for the discussion of R_IN, this is what I think they are talking about:The input impedance is 110K and the pot will allow you to decrease the input volume to 0. Of course, the input signal will be attenuated by at least a factor of 11 even on the highest volume setting but perhaps that can be accounted for in the LM3886 stage.

author: hummusdude
author_rank: Regular Contributor
timestamp: May 14, 2021, 11:20:17 pm
content: QuoteThis makes perfect sense. Funny how easy it is to stare at a circuit and not see something so basic...ha ha! Thanks for pointing that out. I didn't know about the 1/10th thing. That seems curious in that I would expect for max power transfer you'd want to match impedances. But maybe there some other overriding consideration. I guess in my case it's easy to get that because I'm driving the LM3886 with another op amp which I'm guessing is pretty low output impedance by design.

author: ledtester
author_rank: Super Contributor
timestamp: May 14, 2021, 11:56:51 pm
content: QuoteThat's true for power transfer to the speakers, but between a signal source and the input of an amplifier you want to maximize voltage transfer.