Filter Bandwidth Current Sense Amp - Page 1
Forum section: electronics / projects

author: WWW
author_rank: Newbie
timestamp: June 17, 2025, 08:43:01 pm
content: I posted this question on TI's E2E forum, and the Electrical Engineering discord, but no one will take a stab at it.I'm trying to read current of a DC motor driven with PWM using an H bridge (20kHz). I am using low side current sensing with a shunt and a TI INA180 amplifier. This will then feed the ADC of a microcontroller (STM32G4).The amplifier output would be roughly a 20kHz square wave.I found this reference design:https://www.ti.com/tool/TIDA-010056The reference input filter bandwidth is ~2.4MHz, the output filter bandwidth is ~1.6MHz. But the amplifier they use only has a bandwidth of 150kHz. I'm assuming TI knows a lot more about this than I do, but I was under the impression that the amplifier had to be faster than the filter or things would become unstable. Can someone help me understand this?Also, the INA180 has a max output of 8mA. How is it capable of driving a 1nF cap through a 100ohm resistor?

author: Benta
author_rank: Super Contributor
timestamp: June 17, 2025, 09:21:06 pm
content: The 150 kHz is closed-loop bandwidth (the INA180 has fixed gain). Do not confuse it GBW product.

author: WWW
author_rank: Newbie
timestamp: June 17, 2025, 09:28:21 pm
content: I figured I was missing something so obvious they didn't bother to put it in the spec sheet.Does that mean since the gain is 100, the actual bandwidth is 100 * 150kHz = 15MHz?Is this the same as small signal bandwidth vs large signal bandwidth?Sorry if this sounds really dumb. Analog electronics is outside my wheelhouse.

author: Benta
author_rank: Super Contributor
timestamp: June 17, 2025, 09:43:44 pm
content: Figure 7-16 in the TI datasheet gives the answer. And no, your 15 MHz makes no sense. Forget GBW like I said. You've no idea how this device is compensated internally. No one has.Just accept that it has a gain of 100 with a -3 dB point of 150 kHz.

author: mtwieg
author_rank: Frequent Contributor
timestamp: June 17, 2025, 09:55:05 pm
content: It does seem odd to have external filters much faster than the amplifier itself. But there are practical reason to do it this way.In a real current sense resistor, there is some series inductance which effectively creates a highpass filter effect. The input filter is likely there to counteract this (basically cancel a zero with a pole). If you don't, then current sense waveform will have "spikes" on each edge of the switched current. In some cases, these can be so large that they actually exceed the input range of the amplifier, causing all sorts of strange distortion.The design guide for the TIDA-010056 says:Quote

author: WWW
author_rank: Newbie
timestamp: June 17, 2025, 10:40:52 pm
content: I do recall reading somewhere that resistors smaller than .001 ohms would cause inductance issues.If we only know the closed loop bandwidth how do we design the filter and determine stability? These current sense amplifiers seem to be black boxes with limited application information.Also, I don't understand this statement:INA180A3 offers a bandwidth of 150kHz at a gain of 100V/V, along with a slew rate of 2V/us ensurethat for a full swing step change in current through the sense resistor (during FET switching), the transientsettling time of the current sense amplifier is less than 5us.Is that determined experimentally?I'm either way over-thinking this, or I'm completely out of my depth.

author: Benta
author_rank: Super Contributor
timestamp: June 17, 2025, 10:58:35 pm
content: I don't know how to explain this better.You keep on about opamps and filters and stuff, but it's irrelevant.The INA180A3 is a function block that will amplify the voltage over its inputs 100x. Period. It will do this linearly on inputs up to 150 kHz.Whatever you want to filter or waveshape or clip or phase-shift or... you do AFTER the ouput of the INA180.

author: WWW
author_rank: Newbie
timestamp: June 17, 2025, 11:16:51 pm
content: It's still an amplifier. Driving a capacitive load, i.e. an RC filter can cause oscillations. So, how do you size the RC filter to do its job without causing oscillation given the information in the INA180 datasheet?There are numerous methods for designing charge bucket and anti-aliasing filters, but they are all for regular op amps, not these black box current sense amps.For example:https://www.analog.com/media/en/analog-dialogue/volume-46/number-4/articles/front-end-amp-and-rc-filter-design.pdf

author: Whales
author_rank: Super Contributor
timestamp: June 17, 2025, 11:20:18 pm
content: QuoteThese devices are designed to be connected to a shunt resistor, a VCC-GND cap and nothing more. You must not have any high-frequency signal path between the IC's output and its input pins. ie you do NOT put feedback resistors (or caps) on these chips.In this intended configuration they are guaranteed stable, you do not have to do any analysis.Technically an outer control loop (ie the rest of your circuit) can ruin this because it probably changes what happens to the current shunt's voltage depending on the output of this amp, but that's a whole other can of worms you tend to have to do approximations for and test anyway. Yeah it's annoying not to have all the details about chips.QuoteNope. This is wrong.Generally you want the amp to have more bandwidth than the signal going into it, but this doesn't cause instability. It's for other reasons (signal distortion, bad CMRR, RF pickup). The input filter on this device is probably "good enough".Quote from: WWW on June 17, 2025, 10:40:52 pmIf that's true then it doesn't matter what amplifier you use.QuoteYes it would have been determined by them testing actual samples in a real circuit. There is only really one config for this chip so it's very easy for them to measure and spec.The "gain of 100V/V" is important to understand. That means for every 1V of input this circuit will try to output 100V. Eg if there is 0.036V across your current shunt then this IC will try and output 3.6V.QuoteIt can do it for small output voltages AND slow output voltage swing. Indeed it will hit limits under some situations. The original circuit might not hit these conditions, it might not matter or it might just be a limitation of the whole design. No idea if intentional or not.

author: Whales
author_rank: Super Contributor
timestamp: June 17, 2025, 11:23:07 pm
content: Quote from: WWW on June 17, 2025, 11:16:51 pmAh OK.The series 100ohm resistor is likely enough to solve this. The internal opamp (inside this IC) is sampling the voltage on the non-capacitor side of this resistor. It's not the same as having the cap right on the output.Technically speaking, as the resistor value decreases and/or as you saturate the output current limit of this chip then this behaviour might change. It's likely that the chip is stable as long as you stay below that output current limit. If you're worried then yes you have to test.

author: mtwieg
author_rank: Frequent Contributor
timestamp: June 18, 2025, 12:20:44 pm
content: Quote from: Benta on June 17, 2025, 10:58:35 pmIf a spike on the input causes the output to saturate, then all these assumptions go out the window.I have seen multiple cases where a large enough input spike causes output distortion persists long after the input spike is over, and adding the input filter corrected the issue. This was with TI's INAxxx devices as well.Quote from: WWW on June 17, 2025, 11:16:51 pmRegarding stability, I don't think there's any cause for concern here (especially if their reference designs and datasheet mention the use of such filters). Like maybe it you put a large ceramic cap directly on the output you could get it to oscillate, but for any reasonable RC filter (i.e. with resistance greater than the closed loop output impedance) it should be fine.However there are different types of current sense amplifiers which would not play well with this type of input filter. For example the "current steering" type that linear tech likes to put inside their PWM controller chips. Reading the datasheet thoroughly will usually keep you safe.

author: Benta
author_rank: Super Contributor
timestamp: June 18, 2025, 07:47:20 pm
content: "If a spike on the input causes the output to saturate, then all these assumptions go out the window."Agreed. The RC filter on the input will somewhat alleviate that, but high crest factor is always a problem.One of the reasons why I love current transformers for this kind of application.

author: WWW
author_rank: Newbie
timestamp: June 18, 2025, 09:07:55 pm
content: Re-reading the datasheet, it says the INA180 can drive a 1nF load with no oscillation, so the reference design is fine.There is a graph of the output impedance.This is where a 1nF capacitor (blue line) would cross over.I don't know what that really means, if anything.Is it safe to assume that since the max output impedance of the A3 version is around 400 ohms, a filter capacitor larger than 1nF could be used as long as there is a series resistor of at least 400 ohms?

author: Benta
author_rank: Super Contributor
timestamp: June 19, 2025, 12:02:23 am
content: You're reading the datasheet as the devil reads the bible.I can't help anymore, I'm out. Good Luck.

author: WWW
author_rank: Newbie
timestamp: June 19, 2025, 02:49:45 am
content: No loss. You haven't said a single helpful thing.https://www.ti.com/lit/an/sboa443/sboa443.pdfThis document seems to show how to design an output filter for a current sense amplifier. But, if you go through the calculations using the reference design from the original post, it's not even in the ball park (R23 should be 15k, C23 should be 100pF).I just want to know if the amplifier will be stable with a given output filter.It's either impossible to determine mathematically, it doesn't matter, or it's so easy everyone is just laughing at me. Someone tell me which it is.