Using a MOSFET gate driver as a high-voltage line driver for 50 ohm coax? - Page 1
Forum section: electronics / projects

author: tom66
author_rank: Super Contributor
timestamp: November 27, 2014, 09:26:35 pm
content: I've an application where I need to get about 5Vp-p swing at the end of a 50 ohm coax cable, this is logic so I do not need analog levels. I need high current drive and high voltage drive, as there will be 10V on the output, terminating through 100 ohms total (50 ohm output impedance, 50 ohm termination, over a short length of 50 ohm coax.) Data rate is about 2.5~3MHz.My logic input is standard CMOS 3.3V from an MCU.It's for driving some WS2812B LEDs. I encountered some issues with pure 3.3V and noise immunity, plus long lengths of cables caused reflection issues which would cause occasional glitches in the output.I was thinking of using an IC like MCP1416 as it has all that I need:- Strong capacitive drive supported, for long coax if necessary, although I expect to be using shorter cable- 18V rated (I'll be running at 10V max)- Can source 1.5A (I'll be using 0.1A maximum, or 0.2A if the cable is shorted, in a fault condition.)- 3.3V logic compatible- Can drive up to 8MHz assuming 60ns delay time is symmetric on both edgesRealistically since the data is a well-distributed mix, and occurs in bursts, the average current will not be more than 50mA.It's also only £0.45 from Farnell, so the price is very attractive compared to alternatives like a differential line driver with one output floating, or an op-amp solution.The only real concern I have is whether the device will be OK with a continuous conduction current of about ~0.1A, when it's designed for pulse outputs. But, the Rds of the internal power stage is 7.5 ohm max (high side), so even in the fault condition, this is 300mW. This is a lot for a SOT23-5, but with the 250C/W thermal resistance, it'll only 75C above ambient (~30C max) so I think it'll be ok. Typical load shouldn't put it above 18mW/4.5C rise.Can anyone see any obvious issues with this idea? Has anyone tried using a MOSFET driver as a high speed single ended line driver before?

author: nuno
author_rank: Frequent Contributor
timestamp: November 27, 2014, 09:54:06 pm
content: What kind of cable length?

author: langwadt
author_rank: Super Contributor
timestamp: November 27, 2014, 09:56:23 pm
content: Quote from: tom66 on November 27, 2014, 09:26:35 pmas long as you keep the dissipation in mind no problem, many drivers are little more than few transistorsas long as the far end is terminated you don't need to have source termination there should be no reflections, so a few standard gates on 5V in parallel could do just fine

author: tom66
author_rank: Super Contributor
timestamp: November 27, 2014, 09:56:47 pm
content: Quote from: nuno on November 27, 2014, 09:54:06 pmA few metres, maximum.

author: tom66
author_rank: Super Contributor
timestamp: November 27, 2014, 09:57:32 pm
content: Quote from: langwadt on November 27, 2014, 09:56:23 pmOk, that's interesting. I'm still new to transmission lines but why would I not need source termination? Wouldn't that cause a line mis-match, which would cause reflections?The WS2812B has pretty poor noise immunity and I found that even a 300mVp-p edge transient/ringing could cause the occasional bit error. So I do need to keep reflections to a low level.

author: pardo-bsso
author_rank: Frequent Contributor
timestamp: November 27, 2014, 10:41:39 pm
content: Quote from: tom66 on November 27, 2014, 09:57:32 pmIf the far end is properly terminated every pulse you send will not "bounce back" , so it doesn't really matter that the source is not matched.In this case the source mismatch can be neglected because there are no incident waves from load to source, so nothing happens.Do note that if you try to match it (say, by putting a series resistor) the pulse will be half the output voltage.

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: November 27, 2014, 10:44:09 pm
content: I'd use RS422 over twisted pair - should be good for many tens of metres.

author: tom66
author_rank: Super Contributor
timestamp: November 27, 2014, 10:59:52 pm
content: Quote from: pardo-bsso on November 27, 2014, 10:41:39 pmInteresting. Yes this is why I have a 10V supply for the driver, so I get about 1/2 on the other end, or 0V-5V swing. I've actually got a tracking supply - the LED voltage Vcc varies to optimise power saving (when only red is displayed, the Vcc can reduce, RGB LED needs lower voltage to operate at red colour), so the 2xVcc will also vary to keep a consistent Vcc on the other end.Mike, I did consider using other protocols, but I'm trying to keep it as simple as possible on the other end. This solution requires only a single termination resistor and should give me nice clean and fast logic. With the RS422, would I not require a receiver, with its own power supply (presumably the same as the LED supply), and an RS422 driver? What would be the advantage of doing this for a system with about 2m max of coax/cabling per LED strip?

author: moffy
author_rank: Super Contributor
timestamp: November 27, 2014, 11:01:43 pm
content: If I used say a number of 74HC14 devices in parallel, I would use a decoupling resistor, say 10 ohm to decouple the output stage from the load and prevent parasitic oscillations. 6 in parallel each with 10 ohm on each output is only 1.67 ohms output impedance.

author: tom66
author_rank: Super Contributor
timestamp: November 27, 2014, 11:04:49 pm
content: Quote from: moffy on November 27, 2014, 11:01:43 pmThat could work but the 74HC14 has a Vcc dependent logic input threshold so I'd have to level shift the input to get it to switch with 10V Vcc. I also don't know what its actual output impedance is - it's rated something like 1V sag at 20mA, so it seems reasonable to say ~50 ohm per driver in addition to the output resistance. How tolerant is it about all outputs being shorted to ground in a fault condition too?

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: November 27, 2014, 11:10:33 pm
content: Quote from: tom66 on November 27, 2014, 10:59:52 pmMany potential avantages, some or all of which may or may not matter in any given scenario.You don't need coax - cheap cat5 is fine.Yes you need power at the rx end but this can just be the LED power supply you have there anyway so no big deal.You also have a ton of margin so you know it will just work without any dicking about.Bear in mind the WS2812 is somewhat sensitive to pulse and gap widths, so you don't want to risk introducing any asymmetry.As you will need a local supply for the LED current, if your source has its own supply you may have ground loop or noise issues, and again, RS422 will just work.2m will _probably_ work single-ended.However instead of driving at 10V, I'd drive at 5V with HCMOS and series damping resistor, and use a schmitt trigger HCT buffer at the far end, which will switch at the 2.5v you'll get at the far end.

author: electronic_eel
author_rank: Regular Contributor
timestamp: November 27, 2014, 11:30:51 pm
content: Quote from: tom66 on November 27, 2014, 11:04:49 pmThe 74HC series is only rated up to 6V, so 10V would probably kill it. But if you only terminate the receiving end and not the source, you won't get the 50% voltage drop. So you won't need 10V in to get 5V out, 5V in would be enough.Nonetheless are the input thresholds of the 74HC14 dependent on the vcc voltage. So if you want to drive it with 3.3V logic, but power it with 5v, it won't work reliably. So better use a 74HCT14, that is designed for level-shifting from 3.3 to 5v.

author: tom66
author_rank: Super Contributor
timestamp: November 27, 2014, 11:50:26 pm
content: I am still not sure why I do not need the series resistance, given it is a transmission line. Is it just because the frequency is too low to notice transmission line effects? Video signals are only 4MHz and they use series resistance... So I would expect ~2.5MHz digital signal to have some considerations too?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 28, 2014, 02:45:02 am
content: If the channel is single point, you can get by with source termination.If the channel is multi-point (multiple receivers), you can use source termination, provided the receivers are connected with sufficiently short stubs, have low capacitance, and have some form of glitch prevention: hysteresis, input delay, synchronous input window, etc.The reason is, at the source end, the transmitter goes from 0 to VCC, and the line instantaneously goes to VCC/2 (the line looks like a resistor, locally). This VCC/2 wavefront propagates down the line, until it bounces off the unterminated end. As the reflection propagates back, it still carries an amplitude of VCC/2, but this now superimposes upon the VCC/2 already present, and the line goes up to VCC. After 2*t (t = electrical length of the line), the voltage is settled. So, the inputs need to tolerate an indeterminate logic state (i.e., VCC/2) for a short period (zero duration for the last load, where the voltage flips instantly from 0 to VCC as the wavefront reflects off that point, up to 2*t for a receiver directly at the transmitter).A synchronous input window (that is, anything where the signal goes into a clocked flip-flop, subject to setup and hold times: typical examples include synchronous serial, SPI, strobed parallel buses, etc.) is fine with poor signal quality, so long as the duration of the disturbance (of the order 2*t, but may be several times longer for a very badly terminated line that exhibits multiple reflections) does not violate the setup and hold times.An historical example: the PCI bus ran up to 66MHz, a frequency which was limited by the maximum length of the motherboard expansion bus, with all cards equipped, assuming all cards are using the maximum allowable stub lengths. PCI is source terminated, so cards in different physical slots experience different signal quality, and all must tolerate this effect. PCI is a strobed parallel bus architecture, so all the data and signal lines were fine, but the control and clock signals (which define timing) must be treated differently, using one of the other methods I listed above for deglitching.Load or source-load termination is generally used where very high bandwidth is required: you cannot get a stable logic signal before 2*t on a source terminated line, and that's that. Just as source-only termination does not depend upon a load terminator, load-only termination does not depend upon a source terminator. Indeed, for strong logic levels, you must drive the line with a source impedance much less than the line's. Which is what you've been intuitively following: a beefy driver, a big line, and a big old termination resistor.For sending SPI long distances, I would recommend:1. Don't. Use a robust long distance method, like asynchronous serial, or one of the -buses (CANbus, Profibus, ...), or you could do worse than whole friggin' Ethernet. These are designed to be, or can be made to be, very robust in the presence of signal degradation and environmental noise. Many will require isolated receivers (although Ethernet holds advantage here, as it's already transformer isolated).2. If you must, at least use a good signaling standard, like RS-422 (as Mike suggested).NOTE: you won't succeed sending this down CAT5 much distance, because the propagation delays between pairs are NOT matched; in fact, it's explicitly made different, to reduce crosstalk (namely, the twist rates are different -- which changes the effective velocity factor per pair). You will require matched  length cable, which usually means  shielded twisted pair: cable of this construction can have equal delays, and therefore can be used to transmit parallel synchronous data a moderate distance (you're still limited by the delay mismatch tolerance though).3. To be fair, coax isn't bad, and the fact that it is its own ground and shield has some advantage over the twisted pair plus implicit ground.Woe be unto those who dare attempt RS-422 without a common ground, and a good one at that -- long RS-422 cables either require contiguous shielding or individually isolated receivers!But coax is expensive, bulky and inconvenient to handle.There's also the option of ribbon cable, if you're feeling cheap and don't need it to be tough -- ground every other wire and it kinda looks like a balanced twisted pair, except untwisted, and more than a pair alone. Delay match should be good, but don't expect noise performance to be great.Tim

author: AG6QR
author_rank: Frequent Contributor
timestamp: November 28, 2014, 03:29:55 am
content: Quote from: tom66 on November 27, 2014, 11:50:26 pmWhether you notice transmission line effects has a lot to do with the length of the line relative to the wavelength of the signal. 3MHz is roughly 100 meter wavelength. 2m is a small enough fraction of 100m that the transmission line effects will be small.But even if you were wanting to send a signal down a coax multiple wavelengths long, what transmission line effects were you expecting to see and guard against? If the receiver end is properly terminated, there will be no reflections there, and if there are no reflections bouncing back to the source end, then an impedance discontinuity at the source won't cause reflections.Terminating at both ends can have an advantage over terminating at only one end if there are any impedance discontinuities in the middle of the line which might cause reflections. It can also be helpful if the termination at the far end is imperfect. But if the line is good, then doing a good job of terminating at only the far end is enough to eliminate reflections.

author: Zero999
author_rank: Super Contributor
timestamp: November 28, 2014, 01:25:07 pm
content: Quote from: moffy on November 27, 2014, 11:01:43 pmThat won't drive a transmission line properly. With such a low source impedance, there will still be ringing unless the the load is terminated properly. The correct way to avoid this is to connect the 74HC14s in parallel and connect it to the cable via a resistor with the same impedance as the cable.Quote from: AG6QR on November 28, 2014, 03:29:55 amThat is true if it's purely a 3MHz signal, but if it's a square wave with very fast rise/fall times then harmonics could be a problem. The sharp rise/fall times can excite resonances in the cable, resulting in overshoot and undershoot, causing the logic signal to bounce. The easiest way to mitigate this is to terminate the transmission line properly, either at the source or load.I've encountered this before when sending a 16kHz square wave signal over a 40m piece of co-axial cable, using a line driver with a very low impedance. In theory the wavelength was 21.4km, huge compared to the length of the cable but the very fast rise/fall times caused lots of ringing. This was easily mitigated by adding a 51R resistor in series with the output of the line driver.

author: tom66
author_rank: Super Contributor
timestamp: November 28, 2014, 01:52:37 pm
content: So, yes, the rise time will be quite fast, necessarily so. About 20~30ns. This requires a bandwidth of at least 30MHz. Not sure how it affects transmission line behaviour but better to be safer than sorry and do it properly, ~51 ohm on output, ~51 ohm termination over 50 ohm coax? (Can also use 75 ohm, but not sure whether there are any major advantages/differences between the two.)

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: November 28, 2014, 03:20:36 pm
content: Quote from: T3sl4co1l on November 28, 2014, 02:45:02 amA nice solution is NVE's passive input isolators, which don't need power on the input side

author: Zero999
author_rank: Super Contributor
timestamp: November 28, 2014, 03:21:16 pm
content: Quote from: tom66 on November 28, 2014, 01:52:37 pmYes, 30ns rise time is very likely to cause ringing. It only needs to be terminated at one end. I preferred to terminate at the source because I couldn't modify the load. I also think terminating at the source would suit your application because it will reduce power dissipation. Another method of terminating at the load is to AC couple the resistor via a capacitor, that way it only terminates the high frequencies where transmission line effects occur.In the case of termination at source, to get rid of the oscillation, the resistor just needs to be equal or greater than the impedance of the transmission line. If the resistor value is equal to the characteristic impedance, the rise/fall time will be optimal and it will be critically-damped. If the resistor is higher than the impedance of the transmission line, then it will be over-damped, so it will behave like an RC filter, with the higher frequencies being attenuated. A lower value resistor will result in some oscillation, due to the wave reflecting back and forth, in which case it's under -damped.In your case, 51R will be fine, since it will be only slightly over-damped, which should be no problem for your signal, even 75R shouldn't be a problem, the rise/fall times may be slowed slightly but not enough to be an issue.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 28, 2014, 03:52:07 pm
content: Quote from: Hero999 on November 28, 2014, 01:25:07 pmIt'll drive a line perfectly fine, or you can use a single device, since the 74HC output stage is approximately 50 ohms Rds(on) at 5V supply. They did that for a reason...QuoteWhose retarded theory says that?The wavelength of the harmonics was on the order of the cable length.Bandwidth is ALWAYS the highest harmonic(s) in your signal.Note that buying a line driver (let alone a freaking MOSFET driver) is pointless if you're source terminating it with a pissy 50 ohm resistor...Also, 74HC outputs transition in a few ns, much faster than the gate driver. The gate driver will probably not be very happy above 2MHz, even with essentially an unloaded output, and the risetime, delay and skew are all excessive.Tim

author: Marco
author_rank: Super Contributor
timestamp: November 28, 2014, 04:45:04 pm
content: Quote from: T3sl4co1l on November 28, 2014, 03:52:07 pmWon't meet the logic levels of the WS2812B then ...Any way, why not just use a 2n7002 and a couple 100 Ohm pull up resistor to drive a 2n2222 emitter follower to drive the transmission line? (Need parallel termination at the end to keep the load approximately resistive to allow the emitter follower to work.) A couple of cent worth of parts.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 28, 2014, 05:02:37 pm
content: Quote from: Marco on November 28, 2014, 04:45:04 pmWell, none of these source-load terminated options people keep bringing up will meet that. As I mentioned before, for low capacitance receivers, or a single point target, source termination is fine, in which case an 'HC output will more or less do the job. A low resistance driver is necessary for load termination, but this doesn't require a huge driver; a couple bus drivers in parallel will do the job, or you can build your own "CMOS" inverter with a 2N7002 + BSS84, or similar structures with BJTs.FWIW, 74HC14 outputs shouldn't be connected in parallel, because the input circuits will likely transition at different times. They can be paralleled with, say, 50-100 ohms in series with each output; which basically reduces each output to nearly CD4000 grade levels (not quite; those are more like 250 ohms, and that's at 15V), but using a whole chip (hex) in parallel improves things a bit. Ideal would be a 74HCU04, which is just the output transistors, no buffering.Tim

author: Marco
author_rank: Super Contributor
timestamp: November 28, 2014, 05:09:43 pm
content: Quote from: T3sl4co1l on November 28, 2014, 05:02:37 pmDamn it, you're too fast ... I recognized that a little late, was just about to remove my postQuoteWould the WS2812B really have the bandwidth necessary to recognize the messy edge? With 10 ns typical propagation delay I can't see there being more than a couple ns between them.

author: Zero999
author_rank: Super Contributor
timestamp: November 28, 2014, 06:15:23 pm
content: Quote from: T3sl4co1l on November 28, 2014, 03:52:07 pmAll right, so if six outputs are connected in parallel it would make the situation worse, not only because they may transition at slightly different times (whihc as you say will be mitigated if they're all on the same IC) but because the impedance will be lower, so one 74HC14 and no resistor is better than 6.Quote from: T3sl4co1l on November 28, 2014, 03:52:07 pmI agree.