Trying to understand constant off time circuit state when shorted - Page 1
Forum section: electronics / beginners

author: amaschas
author_rank: Regular Contributor
timestamp: March 23, 2022, 06:59:37 am
content: Hey all, I've been working on designing a circuit that uses the shunt-FET approach documented in the LM3409 data sheet, wherein a MOSFET is place parallel to the LED load, and is driven by a PWM signal to control the brightness of the LEDs, rather than driving the PWM pin of the driver. The driver is a constant current buck converter that utilizes a constant off time to set the frequency of the regulating switch, and from what I can understand, that off time is determined by a resistor and capacitor. The time it takes for the resistor to charge the capacitor to a certain voltage (1.24V I believe), is what determined the off time of the cycle.The datasheet includes a node about the possibility of the maximum off time occurring when the output is shorted when the shunt get is turned on by the PWM signal, and recommends using a second voltage and second resistor to prevent this from happening. What I am confused about is what exactly this second resistor is doing. Is it simply providing a minimal amount of current to ensure that the capacitor is charged in a minimum time? The formula the data sheet provides seems to indicate that the choice of resistor is related to the current output of the converter, so that is my best guess. I've included a screenshot of the relevant portion of the data sheet, and the entire thing can be accessed here:https://www.ti.com/lit/ds/symlink/lm3409.pdf.I specifically interested in what ROFF2 is doing in the circuit below, and what is being represented by the formulaROFF2 = ROFF1 X VDD / ILED X RDSON

author: xavier60
author_rank: Super Contributor
timestamp: March 23, 2022, 08:11:42 am
content: Yes, the current through Roff2 is to prevent the OFF time becoming infinite, causing the inductor current to drop to zero. As well as there being no current to immediately flow though the LEDs when the DIM MOSFET is turned off, the whole thing may not even start for a long time.I would set the value of Roff2 by testing various values. It wouldn't matter much if the OFF time is a bit on the long side causing higher than normal current ripple. But it's possible for the frequency to drop low enough to be audible.The value of Roff2 will interact with Roff1 but not not visa versa.EDIT: I cant see the need for the diode in series with Roff2.

author: amaschas
author_rank: Regular Contributor
timestamp: March 23, 2022, 03:30:07 pm
content: Quote from: xavier60 on March 23, 2022, 08:11:42 amThanks for explaining, that does make sense. The value of resistance I get from the formula is pretty high, on the order of 40MΩ, so I guess the idea is a little trickle of current would always come through from VDD, insuring that the capacitor charges even if Vout is too low to effectively provide enough power. In essence I am setting a lower bound of the frequency that the controller will run at.I guess I'm also wondering what the formula they provide is trying to represent. Like I can see that the bottom of the equation is I X R, which would give us the value of Vout when the output is shorted, but I'm struggling to understand what exactly they are doing to calculate a usable value for ROFF2. So someone pointed out to me that the formula provides a value for ROFF2 relative to ROFF1 that is the same as the ratio between the shorted Vout and VDD.Just so I understand this correctly, this is my sense of what is going on:ROFF1 and COFF create an RC network, and the time constant of this network is given by T = RC. However, in this case the internal logic of the converter doesn't wait for the capacitor to charge up to the Vout voltage, but rather discharges the capacitor when it reaches 1.24V. Thus the formula provided in the datasheet:TOFF = -ROFF X (COFF plus (parasitic capacitance)) X ln(1 - 1.24V/Vout)I'm guessing this formula tells us how long the capacitor takes to charge to 1.24V, rather than the full Vout. This works fine in the normal case, where Vout will be > 1.24V, but when the output is shorted, the voltage falls below 1.24V because of the low RDS(on) of the shunt mosfet. In this case we use VDD as a fallback, and use a resistor with a value that provides an RC timing constant that is the same as what the ROFF1 resistor would generate if it was being fed with a voltage high enough to trigger the 1.24V cutoff in the controller.In terms of what's going on in the controller, I want the frequency to be as low as possible without exceeding the maximum off time, which would shut the controller off. It would also be nice to not have the controller make noise, so maybe set a minimum of 20khz or something, in which case I could use the formulas in the datasheet to calculate a resistor value that would result in the minimum frequency I want the controller to operate at in the shunted state.Is there any downside to having the controller operate at a higher frequency than is strictly necessary in the shunted state?

author: xavier60
author_rank: Super Contributor
timestamp: March 23, 2022, 09:14:20 pm
content: A higher frequency than is strictly necessary in the shunted state increases switching losses and also the controller might have trouble properly controlling the minimum ON time.I don't understand the formulas at all.I don't have to deal with the complication of shunt PWM in my LED products. For new designs, I temporarily insert a low value current sensing resistor into the inductor's current path so that I can view the current ripple on an oscilloscope while setting the timing resistor to get a reasonable compromise between frequency and current ripple, neither of witch are too critical anyway assuming that the inductor value is large enough.Also be aware that ripple current can cause core losses in powdered iron type inductors.

author: Giaime
author_rank: Regular Contributor
timestamp: March 24, 2022, 01:05:16 pm
content: Hi, I've used those converters a lot in the past.First your 40MOhm Roff2 is way too large, I usually get 100-500kOhm, probably you've made a wrong assumption in the calculations or you have choosen a very high Roff1/very small Coff, or a strange value for Vdd.Other than what xavier60 has already said, I can tell you that usually is not possible to keep the switching frequency above udible range while having the output shorted. In my experience that's not a problem, SMT inductors do not really "sing" at those frequencies (if you keep the current ripple small, let's say under 50%). The only noise source in a circuit like this is when you place capacitors in parallel to LEDs to try to solve electromagnetic noise, try to avoid that or place a 10-50R resistor in series with the capacitor. Sometimes the LEDs themselves will "sing" a bit but that's unavoidable. Usually the noise made by PWM dimming frequency is much much worse than the noise by the converter itself operating at an audible switching frequency, mostly because the converter is operating in continuous current mode (CCM, where the inductor current does not fall to zero within the off time). This latter noise source is curable by increasing the PWM dimming frequency above human hearing range.The diode in series with Roff2 is just needed to balance the effect of the diode in series with Roff1, to make calculations simpler and to avoid that Toff while shorted is dependent on ambient temperature (that affects diode forward voltage). It's not strictly needed but nice to have. The diode in series with Roff1 is mandatory because otherwise the PWM signal shorting the output (that can be asynchronous to the ton/toff operation of the converter) will mess up with the charge/discharge of Coff.

author: amaschas
author_rank: Regular Contributor
timestamp: March 24, 2022, 09:05:00 pm
content: Quote from: Giaime on March 24, 2022, 01:05:16 pmSo just validating my math here, for VDD I'm using my input voltage of 28V, ROFF1 is 13.5k, ILED is 360mA, and the RDS(on) of my MOSFET is 28mΩ.So my calculation is (13500Ω x 28V) / 0.36A x 0.028Ω) = 378000 / 0.01008 = 37.5MΩ. This would result in a switching frequency of ~3khz. I did a bit of calculating to see what the value of the resistor would be if I wanted to run the regulator at 20khz shorted out of curiosity, and got 2.25MΩ, which is still large but more reasonable. For comparison, the reference design I'm basing my implementation on (https://www.ti.com/tool/TIDA-01415), uses a 3.16MΩ resistor for ROFF2, and uses that same resistor value for different values of ROFF1, so it seems like there is a bit of wiggle room in terms of setting the frequency for the shorted state.I'm struggling to find it now, but when I delved into the user submitted questions for the LM3409 there was a reply that indicated that setting the frequency higher than necessary would essentially result in a higher peak to peak ripple current. I think I might just have to try a few different values and watch what happens on the scope (I have a current probe).

author: xavier60
author_rank: Super Contributor
timestamp: March 25, 2022, 02:01:49 am
content: The "ILED × RDS (on )" part of the formula gives the voltage drop of the parallel MOSFET. I cant see how this is relevant to the value of Roff2 when the forward drop of the freewheel diode is being ignored although it having a significant affect on the inductor's discharge rate.

author: Giaime
author_rank: Regular Contributor
timestamp: March 25, 2022, 09:12:33 am
content: Quote from: amaschas on March 24, 2022, 09:05:00 pmAh, I see what's the issue. I usually use a logic power supply of 3.3V or 5V rail for Vdd (different than my LED power supply, I mean) and my LED current is much higher, that's why I was suggesting a lower value for Roff2.Keep in mind that, in shorted condition, since the inductor current slope in off-time is very small (due to small Vout), one could go unreasonably small for the inductor peak to peak ripple. But keep in mind equation 11 in the datasheet, that poses a minimum limit for the peak to peak ripple, it is valid also when shorted.

author: Giaime
author_rank: Regular Contributor
timestamp: March 25, 2022, 09:15:44 am
content: Quote from: xavier60 on March 25, 2022, 02:01:49 amThe calculations in the datasheet are very simplified, providing "guideline" numbers. Variations due to LED temperature (thus forward voltage), inductor value tolerance, exact output voltage when shorted (dependent also on wiring, PCB traces, temperature...) are all factors that impact the switching frequency, peak-to-peak ripple etc... and most of the time ignored in those simplified calculations.

author: xavier60
author_rank: Super Contributor
timestamp: March 25, 2022, 09:37:39 am
content: Speaking of freewheel diodes, I have found that the Vishay branded ES1D-E3/61T to have much faster reverse recovery than some other brands that I tested.

author: amaschas
author_rank: Regular Contributor
timestamp: March 25, 2022, 04:39:52 pm
content: Quote from: Giaime on March 25, 2022, 09:12:33 amYeah in this case I am driving 8 LEDs in series, so 17-25V at 360mA. That's an interesting note about the ripple current, I'll have to do some testing but it's possible that increasing the frequency, and therefore the ripple current, might be advantageous?

author: xavier60
author_rank: Super Contributor
timestamp: March 25, 2022, 10:02:46 pm
content: Quote from: amaschas on March 24, 2022, 09:05:00 pmThat could be explained by the ICs feature of swapping its current comparator inputs every cycle to cancel out offset error.It's explained in page 15 of the data PDF.

author: amaschas
author_rank: Regular Contributor
timestamp: March 26, 2022, 05:07:58 am
content: Quote from: xavier60 on March 25, 2022, 10:02:46 pmInteresting, I don't quite follow. How would swapping the comparator inputs lead to high ripple current at higher frequencies?

author: xavier60
author_rank: Super Contributor
timestamp: March 26, 2022, 05:32:47 am
content: Quote from: amaschas on March 26, 2022, 05:07:58 amI don't think I can explain it much better than TI have on page 15. The input polarity swapping causes a half frequency component in the current waveform. This is what actually causes the increase in ripple. The amount of comparator offset error might vary a lot from one part to another, affecting the amount of ripple caused.And after all that, it's likely to not even matter. The best way to optimize the thing is by experimentation.Lets us know what frequencies and ripple percentages that you eventually settle on.BTW, all of my products use the HV9910 which has the option for fixed COT which only results in good line regulation and not so good load regulation but because the LEDs clamp at a predictable volatage, it's all ok.

author: xavier60
author_rank: Super Contributor
timestamp: March 26, 2022, 05:45:16 am
content: How time sensitive is the application? Would a few microseconds of current ramp up be a problem?Also a few added microseconds of ramp down when the IC is disabled.

author: amaschas
author_rank: Regular Contributor
timestamp: March 27, 2022, 08:29:25 am
content: Quote from: xavier60 on March 26, 2022, 05:32:47 amOoooh, that's really interesting, specifically because that part is actually in stock in a bunch of places, whereas the LM3409 and TPS92515, which were the two parts I originally started designing with back before covid, have leads times of over a year. I have enough LM3409 to make a few more prototype but that's it, and I've been looking for a replacement with availability.Quote from: xavier60 on March 26, 2022, 05:45:16 amYou mean when I initially power on the device? That shouldn't be a problem. I'd also be using the shunt FET to drive the high PWM frequency I need in this case, so the current ramp up time shouldn't affect the timing-sensitive parts of my implementation.

author: xavier60
author_rank: Super Contributor
timestamp: March 27, 2022, 09:14:58 am
content: Quote from: amaschas on March 27, 2022, 08:29:25 amI was wondering why you were using the shunt method of PWM. High PWM frequency is a good reason.This is a rough example of how I use the HV9910 to high-side drive an n-ch MOSFET.It's actually the HV9910B that I have settled on, the "C" version has a belly plate which is unnecessary to me.It has to be Microchip branded for COT to work properly.Because the whole IC is live with switching signal, I take the precaution of minimizing noise picked up by the RT pin like putting the resistor close to the pin.Some resistive load might be needed in some applications to help it start.Added: Because suppliy of the HV9910B is drying up also, I tried some that I purchased on Aliexpress. They work fine.