50 ohm line driver  - Page 1
Forum section: electronics / projects

author: LukeW
author_rank: Frequent Contributor
timestamp: December 22, 2019, 08:51:17 am
content: I want a small, cheap, common IC that can buffer a single 5V TTL signal and drive it down a couple of metres of RG58 into a device with a 50-ohm input.Surprisingly it seems uncommon to find such 100mA-capable logic buffers.I think the MCP1402 might be a good choice.http://ww1.microchip.com/downloads/en/devicedoc/20002052d.pdfAny thoughts on this?

author: Yansi
author_rank: Super Contributor
timestamp: December 22, 2019, 09:06:20 am
content: Why 5V amplitude? There is a shitton of cable drivers, but they usually use different (lower) voltage levels.If you that much need 5V amplitude, parallel a couple of CMOS gates together.

author: David Hess
author_rank: Super Contributor
timestamp: December 22, 2019, 01:10:08 pm
content: 50 and 75 ohm TTL drivers are intended to operate with a parallel termination to about +2.5 volts so they only have to sink 50 milliamps. If you intend to drive a 50 ohm transmission line which has the typical 50 ohm termination to ground, then other arrangement will need to be made like using a specialized 50 ohm line driver. These exist but not formally as part of the TTL family of logic chips. More commonly higher impedance transmission lines were used to reduce power consumption.The easiest way is to use a couple of discrete bipolar transistors driven by a logic gate.

author: Yansi
author_rank: Super Contributor
timestamp: December 22, 2019, 01:13:53 pm
content: ...or just a couple paralleled CMOS gates, with a series - parallel termination scheme

author: Marco
author_rank: Super Contributor
timestamp: December 22, 2019, 01:44:43 pm
content: Quote from: Yansi on December 22, 2019, 01:13:53 pmWith series termination on the driver he'd need to drive at 10V and be burning 1 Watt driving high ... oops, not in the IC of course. Still a lot of power to burn just to drive a signal.

author: Yansi
author_rank: Super Contributor
timestamp: December 22, 2019, 03:28:49 pm
content: 10V not necessary. You can modify the receiving side, don't you? 2.5Vpp is more then enough drive even for quite an amount of cable length.Not sure what the OP wants this for, nor am I sure if by "TTL" he indeed meant real TTL.And regarding your 1W, how much do you think yall waste when driving 25ohm (double parallel termination) from 5V? Isn't that the same 1W?There's a reason why most coax cable digital standards use smaller amplitude levels.

author: David Hess
author_rank: Super Contributor
timestamp: December 22, 2019, 05:20:43 pm
content: Quote from: Marco on December 22, 2019, 01:44:43 pmIt is not clear what levels are required at the receiver. LukeW only mentioned starting with a TTL signal.If the receiver is expecting TTL, then 5 volts into a double terminated transmission like will deliver 2.5 volts which is plenty but will require 50 milliamps. A CMOS output with multiple gates in parallel would be required to achieve this.

author: LukeW
author_rank: Frequent Contributor
timestamp: December 22, 2019, 11:57:38 pm
content: I can’t void the warranty on a $200k machine.The input is 50-ohm parallel terminated and can’t be removed.The minimum threshold voltage given in the manual is 3V, and this seems true from my tests.It needs to be operated by scientists who aren’t EE specialists - “plugging a T-piece and 50 ohm terminator onto anything is always good for signal integrity, right?” and they may sometimes patch it into devices that can’t drive a 50-ohm load or 3.3V devices or devices with series termination at the source, it could vary.Recently a controller device was causing problems even though it was asserted to be 5 volts - yes, checked and found to be 5V, and seemingly capable of strong current drive, but it turns out the driver device was 50series terminated so the receiver device was only getting 2.5V and not triggering.It seems like a flexible, general-purpose buffer is what’s needed.Sure, a totem pole push-pull sort of thing makes sense - and I think a MCP1402 delivers that with minimum BOM count and cost.

author: David Hess
author_rank: Super Contributor
timestamp: December 23, 2019, 12:20:34 am
content: So the receiver has a 50 ohm termination to ground and 3 volt transition. That makes it seem like it uses a 5 volt CMOS input; TTL has a lower threshold.The MCP1402 will certainly work as long as series termination is not used. So would a bunch of CMOS gates in parallel. I would use a PNP transistor or maybe a PNP current mirror for short circuit protection but there are lots of options.

author: magic
author_rank: Super Contributor
timestamp: December 23, 2019, 08:34:11 am
content: Not sure if this gate driver is short circuit protected.Notice that anything that will drive a 50Ω terminated 50Ω line will also drive an unterminated 50Ω line if you parallel two.BTW, what are your rise time and delay requirements? Notice that the part you found has 35ns delay, for example.

author: daqq
author_rank: Super Contributor
timestamp: December 23, 2019, 11:35:46 am
content: Maybe you could abuse some gate drivers?http://www.ti.com/lit/ds/symlink/ucc27517.pdfThat said, putting several gates in parallel should work just as well. Also, beware of short circuits.Or, use an analog switch?

author: duak
author_rank: Super Contributor
timestamp: December 23, 2019, 05:45:27 pm
content: If the source has to have a 50 R Zout then using a gate driver operated from 10 to 12 V could be a good choice. Slower, wastes power and could frap an unterminated 5 V input.I'm impressed at what the MCP1402 can do - back in the day it would have taken a bit of board area to do the job.

author: Marco
author_rank: Super Contributor
timestamp: December 23, 2019, 05:56:20 pm
content: Double sided termination might be the ideal, but with these voltages is simply omitting source termination really a problem?

author: OldEE
author_rank: Contributor
timestamp: December 23, 2019, 06:34:22 pm
content: Depending on your data rate can you use RS422/485 drivers over twisted pair?

author: Yansi
author_rank: Super Contributor
timestamp: December 23, 2019, 08:11:53 pm
content: Quote from: Marco on December 23, 2019, 05:56:20 pmIt's not about voltage, but about the cable length and signal transition times.If the cable length becomes significant portion of a wavelength of the travelling signal, you get reflections.So for a sharp square wave, you ge reflections even with short cable.

author: langwadt
author_rank: Super Contributor
timestamp: December 23, 2019, 08:37:21 pm
content: Quote from: Yansi on December 23, 2019, 08:11:53 pmyou don't get a reflection from a correct termination

author: Yansi
author_rank: Super Contributor
timestamp: December 23, 2019, 08:40:26 pm
content: Quote from: langwadt on December 23, 2019, 08:37:21 pm

author: langwadt
author_rank: Super Contributor
timestamp: December 23, 2019, 09:09:27 pm
content: Quote from: Yansi on December 23, 2019, 08:40:26 pmbut the receiver is terminated?

author: Yansi
author_rank: Super Contributor
timestamp: December 23, 2019, 09:17:17 pm
content: Both ends of a transmission line must be terminated.

author: langwadt
author_rank: Super Contributor
timestamp: December 23, 2019, 09:48:00 pm
content: Quote from: Yansi on December 23, 2019, 09:17:17 pmnope

author: Marco
author_rank: Super Contributor
timestamp: December 23, 2019, 10:00:16 pm
content: Quote from: Yansi on December 23, 2019, 08:11:53 pmOnly interference and the slight reflection from termination mismatch at the device travels to the source, those getting bounced to the device isn't really a big deal.

author: Yansi
author_rank: Super Contributor
timestamp: December 23, 2019, 11:15:19 pm
content: I disagree.I'd gladly wait to explain your theories about "just one end properly terminated is enough" to the engineers, that send 3gbps digital signals over coaxial cables.

author: jonpaul
author_rank: Super Contributor
timestamp: December 24, 2019, 12:10:41 am
content: See my SMPTE and AES papers re digital transmission, 2014, all these issues are addressed along with reference designs and cable comparisonContact me off list with PM for links.Jon Paul

author: langwadt
author_rank: Super Contributor
timestamp: December 24, 2019, 12:19:02 am
content: https://books.google.dk/books?id=oCaUw-LSt7EC&lpg=PA155&ots=DPVQTGzrLF&dq=near%20end%20far%20end%20termination%20double%20termination&hl=da&pg=PA153#v=onepage&q=near%20end%20far%20end%20termination%20double%20termination&f=false

author: Yansi
author_rank: Super Contributor
timestamp: December 24, 2019, 12:34:48 am
content: Very good explanation in that book. In fact, indirect confirmation that a good system should have the transmission line terminated at both ends such way, that provides good match for the frequency content of interest.(Which, for SMPTE above, is return loss better than 15dB at up to the bit rate frequency)

author: langwadt
author_rank: Super Contributor
timestamp: December 24, 2019, 12:54:16 am
content: Quote from: Yansi on December 24, 2019, 12:34:48 amsure but you don't always have to wear both belt and suspenders, considering the OP mentioned TTL and a 10s of nanoseconds risetime/delay buffer I'd assume we are not taking gigabit speeds

author: tggzzz
author_rank: Super Contributor
timestamp: December 24, 2019, 08:39:57 am
content: Quote from: Yansi on December 23, 2019, 09:17:17 pmBoth ends ofeverytransmission line are terminated. If they weren't they wouldn'tbethe ends.

author: magic
author_rank: Super Contributor
timestamp: December 24, 2019, 08:51:09 am
content: Quote from: langwadt on December 24, 2019, 12:54:16 amNever allow OP to get in the way of overengineering

author: David Hess
author_rank: Super Contributor
timestamp: December 25, 2019, 01:19:40 am
content: Only one end needs to be terminated which is easy enough to see in single ended party line applications where the drivers are open collector and a pull-up is used at either end. When an output is pulled low, it is driving two separate transmission lines but there are no reflections so receivers can be located anywhere. Party line applications require incident wave switching.ECL works the same way with a parallel termination only at the receiving end and the driven end being an emitter follower and ECL is fast enough that even short traces act as transmission lines.If you look through old oscilloscope schematics, you will often find the analog transmission line between sections being driven with a current source and only parallel termination used at the receiver. Double termination could be used, and sometimes is, but would halve the voltage gain wasting gain-bandwidth product.This application note from TI covers it:http://www.ti.com/lit/an/snla027b/snla027b.pdf

author: bachvaroff
author_rank: Newbie
timestamp: July 11, 2023, 03:59:52 am
content: My 2¢:https://www.ncbi.nlm.nih.gov/pmc/articles/PMC8342968/