I am looking at using a CMOS totem pole to drive a digital signal. I am aiming to match the output impedance to 50 ohm with a series resistor. My question is how do I determine the impedance added by the CMOS component?

I from reading this link:
https://www.physicsforums.com/threads/how-to-determine-the-output-impedance-of-cmos-gates.433643/
I gather that (1) the output impedance differs between the N-channel and P-channel FET and (2) the impedance can be approximated by (Vcc - Voh)/Ioh. I do not know what Voh and Ioh are... are they values from the data sheet?

The data sheet for the part I am considering is: https://www.fairchildsemi.com/datasheets/FD/FDG6332C.pdf
A follow up question: Is Rds(on) equivalent to the output impedance?

 

Are you just driving a 50Ω load out of the CMOS pin and wonder what the load voltage will be?

Why do care what the effective source impedance of the CMOS driver? Is your 50Ω load at the end of a long piece of 50Ω coax cable?

 

Yes, assuming you are making a digital signal it is Rds on.

 

I think he wants to know what series resistance should be used at the driver output.

Try 33 ohms.

 

Thanks for the responses. I am trying to match a 50 ohm coax cable impedance with a series impedance at the driver. Therefore, the CMOS source impedance adds to the resistance of the resistor I believe.

The load at the other end is high impedance. I have follow up questions dealing with reflection etc, but will post that in a separate thread.

Thanks!

 

If you are using a very long 50-ohm coax cable with high frequency content then you need 50-ohm terminator at the far end.

I will quantify long and high a bit later.

Why not ask on this same thread.

 

Ah, I guess I can discuss it in this thread.

The clock driver shown below will be attached to a 50-ohm coax cable. The rise time is ~2ns, so I approximate the bandwidth as: 0.35/2ns = 175MHz. So transmitting through copper, things start getting ugly around 4 feet when transmission line characteristics start occurring, yes?

Understood that an easy solution may be to simply terminate with 50-ohms at the load. However, the load is on a vibrating surface and other engineers have voiced concerns with the terminator causing additional failure points. I am trying to wrap my head around what happens with the high impedance load.

I read the following app note: https://www.fairchildsemi.com/application-notes/AN/AN-610.pdf
So the high impedance load at the receiver causes the voltage there to double when it is reflected... Does that mean I just need to pump in a 2.5v signal level at the driver side?

Thanks!

 

Propagation speed down a coax cable is about 5ns/m. Yes, you are right there in the thick of things looking for trouble.

A high impedance load is going to leave the end of the transmission line un-terminated, causing reflections.
I know of no other solution than to install a 50Ω terminator at the receiving end.

And yes, installing a proper series resistance at the driver will reduce the return reflection.

 

The load impedance of a CMOS output is not constant. It is low impedance when the output is low or high. It is is high while the signal is in transition, but only for a short period of time.

 

Understood that the output impedance is not constant. I don't think there's anything to be done about that.

So simply using series termination will not adequately reduce the reflection? I don't really understand the explanation of that Fairchild app note.

 

The series termination at the driver will absorb the secondary reflections (2nd, 4th, 6th etc., i.e. the return reflections) while the parallel termination at the receiver will absorb the 1st, 3rd, 5th. etc.

 

I am not getting it - what is a secondary reflection?

If you have a parallel terminator at the receiver that is matched to the transmission line, isn't there no reflection at all?

 

Secondary reflection is the reflection coming back from the receiving end to the driver end of the transmission line. If not properly absorbed this will reflect back down the cable. This results in the ringing that is observed at the rising and falling transitions of the signal.

Transmission line termination is never 100% perfect. You will always get some reflection. The idea is to minimize as best as you can.

 

There seems to be some confusion here of what you really want. It sounds like your driving a high impedance at the far end of the line. If that is the case you want a 50 Ohms source resistance. The source resistance forms a voltage divider with the transmission line so you get a half height signal until it hits the high impedance receiver. At that time the reflected voltage adds to the signal to provide full signal level.
If you have a fast enough scope you can see this: the output of the driver will be a sharp rising edge, the input of the receiver will be a fast rising edge, the receiver side of the resistor will have a sharp rising edge to half height and then a few nano seconds later (the time for the signal to travel to the receiver and back) the final rise to full signal height.
The 33 Ohms mentioned by Mr. Chips is a really good value to start with. Look at your receiver input with a scope to make sure you've got a good looking signal. If you don't then you can play around with the resistor value until it looks better.
The 50 receiver termination is what is needed to prevent reflection, but for good signal integrity at a high impedance receiver you need that reflection to get your full signal height.
It sounds like your goal is a good signal at the receiver, so you should be able to measure when you've achieved that. If you're wanting the maximum power into the load, and no reflections, ignore everything I said.

 

Cowades has it correct. If you properly match the coax impedance at the source then a high impedance load signal should have only small perturbations.
Make sure that you use a 10:1 probe with a high frequency oscilloscope when viewing the signal at the load.

 

Excellent. Thank you all for the assistance.

I am in the process of designing the driver board that will interface with the high impedance receiver. I was really confused as to why this expensive piece of test equipment would have a high impedance receiver because it seemed to cause an insurmountable problem due to signal reflection. It seems like a terminated receiver would be better, but I guess that places current requirements on the driver?

What cowades explained is shown in the graph below, I just didn't understand it. It boggles my mind to think of the transmission line as a voltage divider.

 

crutschow makes a really good point about the probes. Lousy probes, or good probes with a long ground lead, will cause ringing that you can't tell from a real termination problem. If you can minimize the ground lead it will help even more. If you've got probes with the little spring tip grounds (pink banded probe) it will help minimize ringing when you're looking at the signals. If your lab has ever bought good probes, the ground springs are probably squirreled away in a cabinet somewhere. If not you can probably make something. Hopefully there is a bare ground nearby.



BTW, good picture from the Fairchild appnote. I've added that to my collection.

 

............
What cowades explained is shown in the graph below, I just didn't understand it. It boggles my mind to think of the transmission line as a voltage divider.

To a high frequency signal, a transmission looks like a resistance with a characteristic impedance due to the cable distributed inductance and capacitance. So when you apply a step voltage, energy equal to the voltage squared divided by the characteristic impedance is stored on this distributed capacitance and inductance as the step propagates down the cable.
When the pulse reaches a load at the end of the cable equal to the characteristic impedance the energy is dissipated and no reflection occurs. If the pulse reaches a high impedance load, the energy has nowhere to go so it is reflected back down the line.
The reflected energy adds to the energy already on the line so a reflected pulse travels back down the line on top of the initial incident pulse.
If the source impedance equals the cable impedance then the returned energy is absorbed and no further reflections occur.
Otherwise part of it is again reflected (depending upon the difference between the source impedance and the cable impedance) and travels back to the load. This back and forth reflection continues until the excess energy is dissipated.