Signal Generator output impedance question - Page 1
Forum section: electronics / beginners

author: Unknown
author_rank: Unknown
timestamp: September 17, 2013, 03:22:37 am
content: Please forgive my ignomorance and lack of edumucation here, but why do certain function generators require that you use an impedance load across the output?According to the manual, this HP 3312A requires that,"the output signal MUST be terminated into a 50 ohm load."They then recommend a 50 Ohm feedthrough terminal.I can't help but wonder why, if it is so important, was it not just installed in the factory across the output terminals?This is probably right in front of my eyes but please do enlighten me.Thanks.

author: Unknown
author_rank: Unknown
timestamp: September 17, 2013, 03:37:04 am
content: How would that work if you connect it to a 50 ohm load?

author: AG6QR
author_rank: Frequent Contributor
timestamp: September 17, 2013, 04:03:12 am
content: Quote from: Gazucha on September 17, 2013, 03:22:37 amLook up transmission line theory.The classic 50 ohm signal generator looks like an ideal AC voltage source followed by a 50ohm resistor (in practice, it's normally a less than ideal voltage source followed by a less than 50 ohm resistor).That feeds into a 50 ohm transmission line (a piece of coax, usually) which carries the signal to a load of some sort. In order to eliminate reflections within the transmission line, the impedance of the source, the transmission line, and the load must all match.If you put the load resistor on the output terminals of the signal generator, that would satisfy the generator, but then you wouldn't be able to attach a 50 ohm coax transmission line to carry the signal somewhere else. Furthermore, all of the signal generator's available output power would be consumed by that load resistor.You may want to use the signal generator to actually send power into some kind of load. Signal generators aren't only for driving oscilloscopes. If you're driving a load of impedance higher or lower than 50 ohms, you can put resistance in series or parallel to make the total impedance at the end of the transmission line look like 50 ohms. Or you can use a transformer to match impedance, so that the end of the transmission line "sees" 50 ohms.With most signal generators, terminating the transmission line with the perfect 50 ohm load is only important if you need precise voltages out of it, and if you want to avoid reflections in the transmission line. The generator itself will not be harmed by driving an open circuit.

author: Unknown
author_rank: Unknown
timestamp: September 17, 2013, 04:31:08 am
content: Quote from: AG6QR on September 17, 2013, 04:03:12 amWhat's wrong with source termination?

author: AG6QR
author_rank: Frequent Contributor
timestamp: September 17, 2013, 05:41:15 am
content: Quote from: alm on September 17, 2013, 04:31:08 amOK, I was exaggerating when I said you can't carry the signal to somewhere else.Source termination certainly be made to work for a single high impedance load. But source termination needs to match the transmission line to work right. A signal generator already has a 50 ohm impedance at its source. IOW, it's already source terminated. Soldering another 50 ohm resistor across the output terminals, as suggested, would give it a 25 ohm impedance, so it wouldn't match the transmission line, and would generate strong reflections at both ends of the line. And it would lower the power available to drive a load at the other end.Backing up to the case where you use only ideal 50 ohm source termination, it gives you have a high reflection coefficient at the unterminated load end of the transmission line. By itself, in the ideal case, this shouldn't matter, as the reflection will be absorbed at the source. But if there is any discontinuity in the line, it does start to matter, as described here.http://www.sigcon.com/Pubs/edn/bothendsterm.htm

author: Unknown
author_rank: Unknown
timestamp: September 17, 2013, 01:54:26 pm
content: Thanks for the replies.Perhaps I should have just considered the generator like an amplifier?It is 'standard' that amps have a particular load impedance range and the generator is actually amplifying and outputting waveforms, - very much like an amplifier.The main difference perhaps being that the generator is creating those waves, as opposed to an amp receiving them from some external source.Very interesting article regarding both-end termination too AG6QR.Well, that's that one cleared up then....Now just need to buy a 50 Ohm test lead or two.

author: AG6QR
author_rank: Frequent Contributor
timestamp: September 17, 2013, 02:02:17 pm
content: I slept on the thread, and another thought came to mind overnight.An unterminated transmission line, open at the receiver end, will send reflections back to the transmitter. Depending on the ratio of the length of the line to the wavelength of the signal, these reflections can interfere constructively or destructively, or at any arbitrary phase. So the transmitter will see some sort of complex reactive impedance.In the case where an open transmission line is exactly 1/4 wavelength long, or (2n+1)/4 wavelengths, the impedance seen at the signal source looks like a dead short.If the signal generator manual contains explicit warnings that it may only be used to drive a terminated line, the reason is likely that the driver isn't designed to safely handle the current required to safely drive a dead short, nor can it safely drive a transmission line that looks too much like a dead short, i.e. a 1/4 wavelength line with an open circuit at the end.This is a common situation with radio transmitters. The final output transistors often can't handle a poorly terminated line, and may overheat or otherwise destroy themselves.

author: robrenz
author_rank: Super Contributor
timestamp: September 17, 2013, 02:16:48 pm
content: It took me a while to just understand the simple voltage divider aspect of input output impedance. I posted my diagram in another thread on the same topic. Yeah.. there is more to it than just this but you need to get this part first.

author: Mechatrommer
author_rank: Super Contributor
timestamp: September 17, 2013, 03:14:22 pm
content: spot the difference?fwiw: LCs network is highly simplification of a coax cable.

author: vk6zgo
author_rank: Super Contributor
timestamp: September 17, 2013, 03:30:19 pm
content: Quote from: alm on September 17, 2013, 04:31:08 amAt one of my jobs a young Engineer was looking at a Pulse & Bar video Test signal.He thought that the Generator was faulty,until I pointed out that his termination was attached to a BNC "T" piece connected to the Generator output end of the cable."Oops!" said he, connected the cable the way he originally intended to,& all was well!

author: edavid
author_rank: Super Contributor
timestamp: September 17, 2013, 03:32:40 pm
content: Quote from: Gazucha on September 17, 2013, 03:22:37 amYou left out the rest of the sentence: "In order to meet the specifications listed in Table 1-1, the main signal output must be terminated into 50 ohms." It will work fine without the termination, it's just that some of the specifications depend on it. (I'm not sure which ones, and actually the specifications are fairly vague, but probably square wave rise and fall time.)

author: marshallh
author_rank: Unknown
timestamp: September 17, 2013, 04:06:52 pm
content: You will see severe overshoot (ringing) where the peak amplitude is around twice the stable level. Depends on the cable length.Once the signal is driven into the cable, it reaches the far end which is essentially open (I.e. you have a 1mohm scope input.).This presents an impedance discontinuity and 99% of the signal reflects back to the source.Once the reflected signal bounces to the FG output driver it will see another impedance discontinuity but not nearly as severe.This process repeats until the energy dissipates in the cable and i/o drivers on each end.In a digital system, as long as the setup time (somewhat related to the clock frequency) is long enough that the signal is stable at its sampling point there's no problem. (It's still considered tacky and unrefined to allow thisIf the signal is a digital clock, it needs to be monotonic, which a ringing square wave definitely is not. If it's bad enough, the second ringing excursion (the dip after the first major overshoot) will fall below the circuit's Vih and maybe Vil threshold and cause clock gitching.

author: Unknown
author_rank: Unknown
timestamp: September 18, 2013, 03:10:21 am
content: Quote from: marshallh on September 17, 2013, 04:06:52 pmThat is exactly what I found whilst testing today without a 50 Ohm cable.Looked very pretty on the scope but..... Not part of the plan.Thanks for all the input.