Figuring switching characteristics of a low-side driver IC - Page 1
Forum section: electronics / beginners

author: HwAoRrDk
author_rank: Super Contributor
timestamp: March 10, 2020, 04:11:27 pm
content: I'm designing a circuit that makes use of a couple of protected low-side drivers. I want to able to handle loads of 1.5-2A. I was originally thinking of using an STVNLD5300-E, but when I did some calculations for the thermal performance, I realised it was going to be very marginal at 2A:Rds(on)= 0.3; P @ 2A = 1.2WRth= 90°C/W (with 1cm2PCB copper area); Tj@ 40°C ambient = 148°C (max. is 150°C!)Luckily, there are several models in that range of low-side drivers, so picking the next one up, theVNLD5160-Egives me thermals of:Rds(on)= 0.16; P @ 2A = 0.64WRth= 90°C/W (with 1cm2PCB copper area); Tj@ 40°C ambient = 97°CThat's better. But, because I plan to have these used to potentially PWM the load, I thought it would be prudent that I also run the numbers on the switching characteristics.Going by these for the '5300, I make it to adequately support a switching frequency of about 43.5 kHz, which is fine for my needs. However, looking at the same for the '5160:I only make that out to support 21 kHz! That's not great, as I want to at least be able to handle 25 kHz PWM. Why are the numbers so much worse? What is it that makes a driver with better load capacity have worse switching characteristics?

author: jfiresto
author_rank: Frequent Contributor
timestamp: March 10, 2020, 04:36:42 pm
content: Would an ST VNB14NV04 / VND14NV04 / VND14NV04-1 / VNS14NV04 be fast enough?

author: mikerj
author_rank: Super Contributor
timestamp: March 10, 2020, 04:50:36 pm
content: Quote from: HwAoRrDk on March 10, 2020, 04:11:27 pmIt won't be anything like that high. 43.5kHz is the maximum rate that you can fully turn on and off within a single cycle, but won't allow you to exercise any PWM control; as soon as you move away from 50% you are eating into the rise or fall time so the switch won't turn either fully on (<50%) or fully off (>50%). At a duty cycle of less than ~35% the device will never switch on, and above 87% it will never switch off due to the fixed delay times. The 5160 is obviously worse still.

author: HwAoRrDk
author_rank: Super Contributor
timestamp: March 10, 2020, 05:37:48 pm
content: Ah, I get you. I didn't work out the switching speed taking into account the PWM duty cycle.So, for the '5300, if the minimum switching cycle is 23 us, then if I take a required minimum of, say, 10% for duty cycle, then I make that to be 4.3 kHz maximum frequency. And for the '5160, with 47.7 us min. switching cycle, it's 2 kHz. Boo.I presume that violating the switching cycle timing will lead to it only partially switching on or off, which will increase the power dissipation by a lot.Quote from: jfiresto on March 10, 2020, 04:36:42 pmI don't know, that's only single-channel. I want something that's dual channel. However, I found an ST brochure PDF that says that the similarVNS3NV04DP-Eisdual channel. The switching characteristics of this thing are much better. If I am interpreting the datasheet correctly, total switching time of 3.15 us with a 220R series gate resistor. By my calculations, again assuming minimum 10% duty cycle, that makes a maximum frequency of 31.7 kHz, which is pretty good. Looks to have pretty good Rds(on) and Rth too.The only downside is that it doesn't have discrete status output pins, which I would really like to have. There is some kind of status feedback via the input pin, but from their description I'm having a hard time envisaging how it works.QuoteSo... in the case of a fault, it actively pulls down the input pin? Would I need to do something like additionally route each control line back to an MCU input pin?

author: jfiresto
author_rank: Frequent Contributor
timestamp: March 10, 2020, 06:30:52 pm
content: Quote from: HwAoRrDk on March 10, 2020, 05:37:48 pmIt has been a few years since I used the part, but if IIRC, the idea is to have an MCU output pin drive the VNx14 through a series resistor then return the voltage at the VNx14 input to a second MCU input pin. If you are going to have a PWM output, you could return the voltage to a edge/change-detecting input and detect shorts by the input not being or never pulsing high.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 10, 2020, 07:31:03 pm
content: Heh, adding up all the times isn't exactly the maximum frequency.More like, minimum and maximum pulse width is the sum of rise and fall, and some propagation delay give or take, but you'd really have to test the device to see how much of each really matters, because real devices are not ideal transmission lines: the propagation delay depends on pulse width.Which means for a 10-90% PWM range, you need to be more like1/10th the frequencyor a few kHz.But more importantly, pay attention to those switching energy figures: multiply by Fsw to get power dissipation. The, uh, 5160 probably has a typo, they changed the units to uJ but didn't update the column. I hope... Total 68uJ at 43kHz would be 2.9W, way more than you can afford; but at 4.3kHz that's only 0.29W, not terrible. Probably 1-2kHz would be what you'd want.Such low PWM frequencies are fine for lighting applications, and may be adequate for solenoids or motors, but you'll want to check that they have enough inductance for this to be reasonable. It's not reasonable for general power control purposes, for which you should consider a proper switching regulator instead.Tim

author: HwAoRrDk
author_rank: Super Contributor
timestamp: March 10, 2020, 08:04:06 pm
content: Quote from: jfiresto on March 10, 2020, 06:30:52 pmHmm... I think I would want to XOR the two. When the drive output is high and the status input is also high, the result is low; when the drive output is high, but the status input is low, then the result is a high and we have a fault state. But I don't think that's doable in software without constant interrupts, and nor do I think I'll have room on the board for a quad-XOR IC.Quote from: T3sl4co1l on March 10, 2020, 07:31:03 pmOh yeah, good spot. Yeah, Ireallyhope that's a typo. It's got to be uJ, gotta be.Quote from: T3sl4co1l on March 10, 2020, 07:31:03 pmThis is going to be multi-purpose load switching - so as you said, relays, solenoids, motors, lighting (incandescent/LED). The former is mostly why I was hoping to be able to do 20+ kHz PWM so that reduced-power hold state could be implemented. But that may have to be left out if I can't get a suitable driver.

author: jfiresto
author_rank: Frequent Contributor
timestamp: March 11, 2020, 09:49:50 am
content: Quote from: HwAoRrDk on March 10, 2020, 08:04:06 pmIf you have an MCU in the mix, at least the ones I have worked with, and available I/O pins, you can set up the pin change interrupts of the status (low side driver input) pins, not enable the interrupts, and then periodically poll the corresponding MCU input change and input state registers, and identify faults. The low side driver will shut down quickly enough to protect itself and then slowly recover, which often gives the MCU firmware plenty of time to detect and respond to a fault. (I don't know your application.)BTW, some automotive driver chips can rapidly and repeatedly restart, automatically – and then fairly quickly burn themselves up from the severe, thermal cycling. With these chips, it is prudent to have the MCU guard against cycling, and stop driving the driver before it thermal cycles itself to death.

author: HwAoRrDk
author_rank: Super Contributor
timestamp: March 11, 2020, 09:55:30 pm
content: Quote from: jfiresto on March 11, 2020, 09:49:50 amDoesn't look like this will be possible for me. The MCU I am using only allows to configure a trigger condition (rising/falling edge, etc) for an entire port, and then to enable interrupts for individual pins in a port - i.e. you can't set up a pin change interrupt but not have the interrupt occur. There are no flag registers for pin changes either.Quote from: jfiresto on March 11, 2020, 09:49:50 amMy strategy will be to shut everything down and abort all operations as soon as any output driver fault occurs, and stay that way. The only way the user will be able to recover will be to power cycle. So no worries about those kind of concerns.

author: jfiresto
author_rank: Frequent Contributor
timestamp: March 12, 2020, 07:20:51 am
content: What MCU are you using?

author: HwAoRrDk
author_rank: Super Contributor
timestamp: March 12, 2020, 12:48:57 pm
content: STM8

author: jfiresto
author_rank: Frequent Contributor
timestamp: March 12, 2020, 07:12:59 pm
content: A bit of skimming suggests the interrupts are fairly integrated and not very lightweight.You have my sympathy if you have to trick the part to do your bidding.

author: HwAoRrDk
author_rank: Super Contributor
timestamp: March 13, 2020, 05:27:02 pm
content: I'm trying to get to grips with the status feedback of these VNS3NV04DP-E parts, particularly for a circuit that implements my earlier idea of XOR-ing the output from the controlling MCU pin and the actual voltage at the part's input pin.Thedatasheetsays that upon fault condition, it actively pulls down the input pin. I've been trying to simulate the thing, but I'm not sure whether what I'm doing is an accurate representation of the part's behaviour.The block diagram in the datasheet shows the pull-down:That's the symbol for a current mirror, right? I believe current mirrors are commonly used in IC design rather than embedded resistors for die space reasons. So, it's a current sink. And the side arrow seems to imply that it's adjustable, driven by the over-temperature detection circuitry.So how should I simulate this? I cooked upthis circuitin the Falstad simulator.According to the datasheet, typical fault condition sink current on the input pin is 15mA, so I used a 330R pull-down resistor to approximate that. However, the actual amount of current this can sink is limited by the gate series resistor, and this has a knock-on effect on the input to the XOR gate. If I use anything less than a 1K gate resistor, then the feedback voltage to the upper XOR input is too close for comfort to the HC-series 'low' input voltage threshold (1.3V, I believe). But, I want to keep the gate series resistor fairly low (< 1K) so as not to impede switching speed. If I put an actual current source component in the simulation, it turns the voltage negative when switched on, but I assume that only happens because it is an 'ideal' current source. If the circuitry in the real part is indeed a current sink, then can I assume it is actually only capable of pulling the voltage to zero, and I don't need to worry about the effect of the value of the gate resistor?One additional question: would it be safe to assume that this internal pull-down operates all the time, albeit at reduced current, regardless of fault status, such that I don't need a discrete pull-down resistor? I can't find anything in the datasheet to indicate this. But, all the test/application circuit diagrams do not show one, so...

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 14, 2020, 06:27:36 am
content: Why use XOR when two NAND will do? Well okay that's not a very convincing component-level alternative, but how about one BJT?Connect a PNP like so: emitter to MCU pin, voltage divider from MCU pin to base to gate (say 470 each), collector to pulldown (say 1-10k) to GND. Collector is fault. Transistor only turns on when MCU pin is high AND gate draws current.Tim

author: jfiresto
author_rank: Frequent Contributor
timestamp: March 14, 2020, 12:22:01 pm
content: Quote from: HwAoRrDk on March 13, 2020, 05:27:02 pmThe above block diagram shows a gated current sink (the linked circles and pokey arrow) to the right of the input pin. From the datasheet, I tend to the notion that the sink only sinks current during a fault, and at least 10 mA to ground. As long as the input resistor sources less than that into the input, the sink should hold the input at close to ground, when there is a fault. The resistor could be as little as 360 ohms if the MCU holds the other end high, at 3.6V.Driving the input (through the resistor) with PWM pulses complicates matters as the input will be low during the PWM off-time and indistinguishable from a fault. You may need a latch to confirm the PWM is driving the input high and the driver is passing that on, if the MCU can not frequently sample the input to catch a narrow pulse. The latch may be no big issue or even an advantage, but it is hard to say without knowing more about the MCU hardware and the pulses it is outputting.You may find it easier to use a low side driver with a separate, dedicated status pin. Infineon is the other big supplier after they acquired International Rectifier. Page 35 of their automotive selection guide (15 Mbyte PDF) lists a few possibilities.As I am sure you know, the "best" solution will depend on the details of your hardware, processing and signals. The MCU makes possible some tricks that may lead to a more agreeable design.

author: HwAoRrDk
author_rank: Super Contributor
timestamp: March 14, 2020, 07:51:55 pm
content: Quote from: T3sl4co1l on March 14, 2020, 06:27:36 amThat's quite neat, thanks for the suggestion.Does it confer any advantages over using an XOR gate? I guess it's cheaper, but we're probably talking pennies difference here. I don't think it would save any layout space (quad XOR 74HC86 TSSOP-14 versus 4x SOT-23, plus resistor arrays for both).Quote from: jfiresto on March 14, 2020, 12:22:01 pmThat explains it so much better than the datasheet.So as my MCU is working off 5V, I shouldn't use a resistor of less than 470R (would allow > 10mA); but, nor do I want to use more than 1K, so as not to adversely affect switching speed.Quote from: jfiresto on March 14, 2020, 12:22:01 pmWell, that's what the XOR circuit I previously proposed is for. If the FET gate differs from the MCU output, there is a fault signal; if the same (normal operation, or both off), no fault. Obviously, with PWM, the fault signal will be pulsing high, but I only have to catch the first low-high transition.Quote from: jfiresto on March 14, 2020, 12:22:01 pmYes, I've already looked at some of the Infineon devices. The BTS3410G doesn't actually have any status output, despite what that guide categorises it as having, and also has very poor switching speed; the BTS3408G is only rated for 1A max, and only has a single fault output for both channels; and the AUIP* parts are obsolete. That leaves single-channel devices, but I'm needing something compact, so they don't really suit my needs.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 14, 2020, 10:31:02 pm
content: Ah, actually looking at the datasheet, it claims it pulls to zero. I wonder how... it can't do that forever if it's storing charge on an internal node. Maybe it's drawing extra leakage from the drain? Which, at 30uA+ Idss, that's... probably what it's doing, huh? Pretty leaky for a switch, something to keep in mind for light loads (including literal LED lights, which will glow dimly).If that's the case, a weak pullup (less than the minimum fault current, <10mA; a 1k resistor is good) should give a logic level signal, and the XOR (or NOT + AND, or..) will do. Or just read it from another MCU pin straightaway.Tim