How do you start selecting feedback value for CSA? - Page 1
Forum section: electronics / beginners

author: LoveLaika
author_rank: Frequent Contributor
timestamp: October 26, 2021, 06:34:26 pm
content: I'm looking through the datasheet for theOPA691, and I'm a bit lost on how to select and adjust optimal parameters for an inverting circuit. I was wondering if I could get some feedback on my thought process.I'm more or less trying to follow the example shown in Figure 9, expanding it out to multiple inputs and keep a gain of -2. If done correctly, I feel that I should be able to get a pretty high bandwidth of ~200 MHz. To help illustrate things, I've attached a picture which is more or less the same as Figure 9. The only thing to note is that when I expand it to accommodate N inputs, the box on the left is replicated in parallel to account for multiple inputs.So, to start off, the gain of each input, A, is written as [-RFB/RG]. To match the source impedance, [RG||RM] should be close to 50-ohms. Now, the noise gain (NG) is written as [1+ RFB/(((RG+{RM|| 50}))/N)], where the denominator is the sum of RGand the parallel combo of the source impedance and RM. N is the number of inputs in the inverting summer. Once you figure out the noise gain, you can use the chart in Figure 8 to approximate a feedback resistor value; from there, you follow the equation shown in Equation 4 to get a sum close to 472 for optimization.From all this, unless my reasoning is wrong, most of the terms are dependent on NG and RFBfor their calculations. However, you can't solve for NG without picking a value for RFB. However, at the end, RFBis determined by NG. This feels slightly circular. How would you approach this design. Do you start by picking a feedback resistor without taking into account impedance matching? From there, using this value, you then account for impedance matching? Sorry if this sounds confusing, but I'm just kind of lost.

author: magic
author_rank: Super Contributor
timestamp: October 26, 2021, 07:32:04 pm
content: Noise gain of inverting 2x configuration is 3 regardless of resistor values. With N inputs it becomes N·3, I suppose.Then you could pick up their recommended Rf and come up with some Rg and Rm.You are right, it's not that simple and the noise gain in fig. 9 is 2.7 due to matching networks.It looks like there is a system of equations to solve and optimize for bandwidth (not sure how), or just go with trial and error, knowing that noise gain is slightly less than 3 per input anyway. (It would be N·3 exactly if Rsource||Rm were zero).

author: LoveLaika
author_rank: Frequent Contributor
timestamp: October 26, 2021, 09:10:02 pm
content: Thanks for replying. Trial and error does seem to be the fastest. As long as the bandwidth is constant up until a certain frequency, I'm satisfied.Yeah, trying to solve for a system of equations for optimization may not be the best approach. Is it fine to assume matching networks and work around that, increasing/decreasing the feedback resistor as necessary around that value? While I picked 200 MHz as a bandwidth, I chose it based on what TI accomplished in their example summing op-amp on the first page. I figured if they could achieve it with 5 inputs, it would be a good thing to try and achieve. Though, they didn't have RMat their inputs. Plus, the addition of the 30-ohm resistor helps with optimization by placing that feedback/NG sum closer to their optimal value as shown in my LTSPICE test run. The frequency response corner gets flatter as the series resistance increases. (Strangely enough, the bandwidth of that one isn't 200 MHz like advertised. I'll have to re-examine that.)..Speaking of which...what's the proper term for that frequency response corner? And what can be done to make it flatter?

author: magic
author_rank: Super Contributor
timestamp: October 26, 2021, 09:31:33 pm
content: The maths could be solved accurately, it's just a PITA.NG = 1 + N·Rf/(Rg+Rm||50Ω))Rg = Rf/2Rm can be calculated knowing Rg→ You get an equation for NG as function of Rf. Find the intersection(s) of this curve with fig. 9.I think a reasonable simplification is to take Rm||50Ω as 30Ω, because that's ±3Ω accurate over the recommended range of Rf and later this number is added to Rg which is another order of magnitude larger still.It looks like roughlyNG = 1 + N·1.666as an average over the range of recommended Rf.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: October 27, 2021, 09:46:10 pm
content: Thanks for the help. It gave me a good value to start at; as I ran sims, the feedback resistor value (and subsequently the gain resistors) was too low, causing instability. Increasing it helped, especially with the addition of a tiny series resistor like how they had their example on the first page. I know it's all simulation, but I hope results also showcase themselves in real testing.