Additional series resistance on buck converter feedback - Page 1
Forum section: electronics / projects

author: HwAoRrDk
author_rank: Super Contributor
timestamp: July 08, 2025, 12:33:37 pm
content: I'm looking at datasheets for DC-DC buck converters, in particular the AOZ6682CI. In theirdatasheet, they give the usual typical application schematic:They show a series resistor RTon the feedback node. But it's not mentioned anywhere else in the datasheet, and no discussion whatsoever of any need for such series resistance in the section on selecting resistor divider values.What does 'T' stand for? Can I omit this resistor? Or would the need for such series resistance depend on the R1/R2values used? I'm thinking maybe it's needed for low R1value. However, I am thinking of using something close to the recommended 52.3k/10k for 5V output: 43k/8.2k, which is not particularly low.

author: Peabody
author_rank: Super Contributor
timestamp: July 08, 2025, 02:15:16 pm
content: That is indeed puzzling. The resistor would have an effect on the voltage at FB only to the extent that current is flowing through it. But the datasheet says the maximum current into FB is 200nA.

author: HwAoRrDk
author_rank: Super Contributor
timestamp: July 08, 2025, 02:54:34 pm
content: I found that thedatasheetof sibling part AOZ6683CI does mention RT:QuoteSo I guess it's just named RTbecause it's a 'T'-type network. But they still don't mentionhowthe RTvalue affects the feedback loop.

author: David Hess
author_rank: Super Contributor
timestamp: July 08, 2025, 03:44:44 pm
content: Quote from: HwAoRrDk on July 08, 2025, 02:54:34 pmThe functional block diagram on page 4 shows what is going on. The FB pin leads to an inverting amplifier which has pole/zero compensation, so the DC input impedance is high with low input bias current, but the AC input impedance depends on the compensation capacitance (pole) and series compensation resistance (zero). So the resistance presented to the FB pin adjusts the frequency compensation, with higher resistance lowering the bandwidth of the control loop.I do not see anything about how to select the series resistance. A transient response test could be done while adjusting the series resistance to dial it in, then increase the series resistance by some margin.

author: HwAoRrDk
author_rank: Super Contributor
timestamp: July 08, 2025, 06:59:33 pm
content: Ah, I see. I've seen such series resistors described in datasheets of other buck converters as "make the loop more stable", so I guess a higher resistance on the FB pin makes the control loop react slower, meaning it won't over-react and introduce instability in response to very fast changes in the output load.Perhaps I shall just specify a zero-ohm resistor for the time being - because I need to put something; a series resistor is already in the circuit of an existing design where I am aiming to substitute the DC-DC converter for one that's been discontinued - then see if anything about the loads presented shows any instability.I just wasn't sure about whether RTwas absolutely required, because you never know.

author: HwAoRrDk
author_rank: Super Contributor
timestamp: July 08, 2025, 07:06:31 pm
content: By the way, on a related note, I am puzzled by another thing in the datasheet of another buck converter I was considering: the TITPS56339.It requires the use of a series resistor, of 30Ω, on the bootstrap input. Why? The only justification stated is "to make the device more robust". I've not come across any other parts that require this. So I had to reject the TPS56339 because there's no spot on the existing design for one.

author: David Hess
author_rank: Super Contributor
timestamp: July 08, 2025, 07:18:32 pm
content: Quote from: HwAoRrDk on July 08, 2025, 07:06:31 pmThe series resistance limits the current charging the capacitor to a safe value. Pin 2 is driven by the output switch which is fine, but pin 6 relies on a small diode with limited current handling capability.

author: Paradroid
author_rank: Contributor
timestamp: July 09, 2025, 05:56:23 am
content: Quote from: David Hess on July 08, 2025, 07:18:32 pmIs "charge" really the right word here? I thought these boot caps are for generating a small negative voltage for the internal FET drivers, and if that's true, the series resistor is for protecting the diode on pin#6.Jens

author: mtwieg
author_rank: Frequent Contributor
timestamp: July 09, 2025, 11:19:53 am
content: Quote from: HwAoRrDk on July 08, 2025, 06:59:33 pmRT is not absolutely required. Whatever behavior you can get with R1+R2+RT, you can also get with just R1+R2. But having RT gives more flexibility in choosing component values. Maybe convenient since the datasheet gives no guidance on stabilizing the feedback....Quote from: Paradroid on July 09, 2025, 05:56:23 amActually, when a resistor is added in this location, it's usually intended to slow down the turn-on of the high side FET.It's possible the intent here is to smooth out the spikes of current which would otherwise be drawn by the bootstrap cap (such spikes might be problematic if they contaminate other circuitry. Normally if that's the intent, the resistor would be put on the anode side of the bootstrap diode. However that's not possible with the TPS56339, so can't be sure what the exact intent is.

author: langwadt
author_rank: Super Contributor
timestamp: July 09, 2025, 11:34:14 am
content: Quote from: Paradroid on July 09, 2025, 05:56:23 amnot a negative voltage, just a voltage floating on to of the sw node for the upper gate driver

author: Paradroid
author_rank: Contributor
timestamp: July 10, 2025, 06:38:01 am
content: Yep, my mistake, it's surely to generate a voltage higher than Vin to make gate switching on the high side faster. Still, there must be a diode behind the "boot" pin that can be protected with a resistor.Quote from: mtwieg on July 09, 2025, 11:19:53 amWhy would you want to slow that down? The slower you make the gate charge change rate, the more heat you'll produce. The whole point of a switching regulator is to preserve power, not to waste it as heat.There might be some chip internals that I don't know about - like the gate drivers having limited current capability. If someone could shed some light on what role that might play on such a small device (that probably shouldn't dissipate more than 0.5W sustained), that would be great.Jens