TTL signals: output impedance, overshoot, and other characteristics - Page 1
Forum section: electronics / projects

author: ampetrosillo
author_rank: Newbie
timestamp: December 06, 2014, 07:03:24 pm
content: Hi, I'm new to the forum: I'm an Italian Electronics and Telecommunications Engineering student (in my third year), and, well, I'm here for some helpI have an assignment for my Electronic Measurements and Instrumentation course, specifically me and my group were asked to perform time and voltage measurements using reflectometry on coaxial cables. We used an Agilent 33120A waveform generator and a HP 54600B oscilloscope.One specific task was to measure the propagation delay over an adapted 5m RG58 coaxial cable: having plugged T connectors to both the scope's channels, we connected the waveform generator to the first channel, plugged a 5 m coaxial cable to both T connectors, and we adapted the line connecting a 50 ohm plug to the second T connector. This way, we could see approximately the same signal at the start and at the end of the cable, and measure the delay between the two signals. We executed this task using both a 1kHz square wave and the sync signal from the waveform generator, which apparently is a TTL output. From the manual: "For sine, square, triangle, and ramp waveforms, the sync signal is a TTL “high” when the waveform’s output is positive, relative to zero volts (or the dc offset value). The signal is a TTL “low” when the output is negative, relative to zero volts (or the dc offset value)".I'm not familiar with TTL modules (I know they are a BJT technology used in the manufacture of logic gates, but that's it), a bit of searching revealed to me that TTL signals are very badly suited to transmission lines because of their varying output impedance on their "high" and "low" levels. In fact, the displayed sync wave (connecting the sync output directly to one of the channels of the scope, reading voltage over an open circuit basically) is an approx. 4V square wave with a 400mV overshoot (approximately 100us long), and with an approx. 200mV DC offset, which is consistent with my findings (TTL low levels are 0-0,4V, while TTL high levels are 2,2V-VCC). If I read the voltage differential on the coax with the 50 ohm terminator plugged in, the overshoot completely disappears (and of course voltage approximately halves). Why is this? Is this related to that variable output impedance?(I have attached a captured screenshot from the scope, comparing the 1kHz square wave generated by the 33120A to its sync signal. Both signals were sent into separate oscilloscope channel inputs, basically constituting an nearly open circuit).Another question is: time measurements are best taken using the ordinary square wave or the sync wave? The sync wave has a much faster rise time, so I'd point to the latter.

author: w2aew
author_rank: Super Contributor
timestamp: December 07, 2014, 03:47:52 am
content: These videos of mine might help.

author: hlavac
author_rank: Frequent Contributor
timestamp: December 13, 2014, 08:56:35 pm
content: TTL is totally unsuitable for driving 50ohm transmission line.The TTL signaling works by pulling a small current from inputs to ground to get Low, hence the asymmetric driving capability (much weaker High than Low).TTL "totem-pole" output stages usually drive "high" state thru a 130 ohm-ish resistor, thats pretty high compared to 50 ohm line impedance you are driving.Also the standard TTL outputs are rated only for something like 1.6mA load.It would be best to use some sort of buffer to get sharp edges suitable for the measurement onto your transmission line...

author: TerraHertz
author_rank: Super Contributor
timestamp: December 14, 2014, 07:41:33 am
content: Since your objective is only to measure the cable delay, it doesn't matter that TTL outputs are not well suited to drive 50 ohm coax. So long as you have one good fast edge, you can do the measurement.TTL outputs can't source much current in the high state, so your waveform into a 50 ohm load doesn't meet the TTL voltage 'H' spec. But for this use, doesn't matter. Also the rising edge will be slow, again due to TTL's poor pullup current source ability.But the falling edge should be fine. It will go to 0V, since the 50 ohm terminator at the cable end isn't supplying any current, but sinks it.The overshoot you saw with the terminator removed is exactly what you'd expect from an unterminated coax, seeing an edge. Keep the terminator attached.A TTL standard input's source current to a L output is 1.6mA, and a TTL output has fanout of 10, ie it can sink 16mA. Enough to get a good enough edge to measure the delay of your longish coax. Offhand I recall coax propagation is something like 1nS per foot, so 5m gives about 15nS. You'll definitely be able to see it.Good idea to first check for timing skew between your scope channels by feeding them the SAME signal.

author: zapta
author_rank: Super Contributor
timestamp: December 14, 2014, 02:52:50 pm
content: The signal generator has TTL voltage levels but it doesn't necessarily have the same output impedance of a TTL gate. These are two separate characteristics.