# How termination resistors work; what happens if I use lower values?

I am going to try to interface low-speed 8bit DDR2 chip to FPGA, and I've got some questions crucial to make it work :-)

Is that correct that the idea of termination resistor is to sink most of the signal to GND, so that only small part of it reflected back? Have anyone tried to put let's say 2-3 resistors of smaller value so that multiple remining reflections will be out of phase & cause less interference?

• The general solution to the telegrapher's differential equations -- which are derived from the equivalent RLC circuit of a differential length of a two-conductor transmission line -- yields both forward and backward propagating waves. From the fact that the impedance of the load is ZL = VL/IL, you can derive that V- = [(ZL - Z0)/(ZL + Z0)]V+, where Z0 is the transmission line's characteristic impedance and V- and V+ are the voltage amplitudes of the reflected and incident waves respectively. Thus if ZL = Z0, the amplitude of the reflection is 0 and there are no standing waves. – Eryk Sun Mar 1 '11 at 18:28

A transmission line can be modelled as an infinite set of capacitors and inductors (lossless). You start to use this model as your electrical line becomes large enough that you cannot think of the line as an instant connection.

### General Idea

First, an LC circuit is going to have ring, and if it suddenly hits an "open" instead of another LC circuit it will bounce very high. If you were to make a model using 10 inductors and 10 capacitors this would easily happen. When you place termination on the end you are damping the signal. If you have a perfectly matched resistor at the end you will have 0 overshoot as the resistor will dissipate its power.

### Source Termination

If you instead place a resistor that matches the transmission line in series between the source and the transmission line you get one of the most effective termination techniques. In this case the line can only be driven to 1/2 of the target voltage, but the signal travels down the line and when it hits the open at the other end (most inputs are almost opens with very high impedances) it bounces, doubling, and giving you a full voltage at the receiver. The signal then travels backwards and, when it reaches the source, terminates on the resistor.

This may not be instantly clear, I would very much suggest "High Speed Digital Design: A Handbook of Black Magic", but this means your line does not drive nearly as high at one point, and noise is a function of dV/dt. This only terminates noise on the line at the source, which helps a large amount. I would heavily suggest you tear into my favorite handbook of black magic.

### Trace Impedance

Most people have heard of the simple equation forms of inductance and capacitance. Capacitance goes up with area and down with distance. Inductance goes up with the size of the loop.

If you think of of a trace above a ground plane, as your widen the trace, the area increases but distance does not. This means that your capacitance increases while your inductance stays the same. As your distance increases, your area must increase a lot to keep the same impedance.

There are many many different calculators out there. I found one instantly with a google search.

Just match your impedance, add some termination, and try to avoid bad practices like bridging across a break in a ground plane (No embedded traces around these signal lines). I hope this also makes the physical effects a little more clear.

### Too Small a Termination?

You will actually get reflections, but instead of bouncing up, it will bounce down. An open will double your voltage, it will all reflect backwards. A short does the opposite, giving you zero voltage. It also greatly increases your power absorption from your driver.

Imagine a transmission line as being a bunch of hanging weights connected by springs. If everything is uniform, and one gives a weight at the north end of the line a brief southward shove and return it to its original position, a very nice wave will propagate southward down the line; the energy that gets put into each weight from one side will be perfectly delivered to the other, so that once the wave has passed by a weight that weight will be motionless in its original position. All very good until the wave hits the end of the line.

At that point, one of three general things can happen:

1. If the last weight on the south side has can move freely with nothing connected on its south side, it will accept energy from the second-to-last wave, but won't have anything to push against. The northward push-back that it didn't receive from the south side won't cancel out the southward push it got from its north side. The uncanceled momentum of the weight will thus cause it to pull the weight north of it south, and start a wave that propagates to the north. Note that while the original north-south wave was a compression wave which resulted in waves traveling briefly south from their starting point, the reflected wave will be a tension wave with waves traveling south.
2. If the last weight on the south side has its south-side spring attached to an immovable wall, the wall will push back harder than would one of the normal weights. This harder push-back will cause the weight to send a wave back toward the starting point; this new wave will be a compression wave like the original, but will cause the weights to shift briefly north of their start point.
3. If the south spring of the southernmost weight is connected to something that offers just the right amount of resistance, all the energy of the wave will be dumped into that resistance, and there will be no reflection.

The scenario where the last weight has some resistance, but not the right amount, will behave as a combination of (1) and (3), or (2) and (3) above. The scenario to shoot for is #3.

• Analogies are always a good idea to understand, +1. A diagram would make it much more easy to understand though, especially with all those north and south... – Mister Mystère Oct 11 '14 at 11:38

They match the impedance to the trace impedance. That's why there isn't a reflection. The fact that they may sink current is just a side effect. Their values should be calculated based on the trace impedance and that of the receiver and driver. High-Speed Digital Design by Johnson & Graham is the book I recommend on this topic.

Multiple smaller value resistors will attenuate the signal too much. It also may be more current than the driver can handle.

The principle behind termination resistors is to match the impedance of your inputs to the impedance of your transmission line (PCB) traces) and your source. Typically, input pins have high input impedance, since they are CMOS. Adding a small value resistor in parallel with the high impedance input pin will effectively set the input impedance to the resistor you added. This is useful, because output impedance is usually fairly low, and it is easy to make a micro strip transmission line with low impedance.

The goal when using a termination resistor is to make it a close to the input pin as possible. Using multiple resistors would be less optimal since the resistor is less like a lumped element. The other thing is that you should know your target impedance. Having a resistance larger or smaller than your impedance will result in mismatch, which causes reflections.

I don't fully know the mechanics of it, but the purpose of the termination resistor is to make it appear as though the transmission path carries on forever. Any change in impedance will cause reflections, such as connectors, damage to the transmission path, or (obviously) transition to a path with different impedance.

Using a lower value resistor (I'm not sure what you mean by multiple smaller-value resistors—if you put them in whatever configuration, you'll just get some other effective resistance with worse HF performance as it's spread out) will cause your drivers to source and sink higher powers than normal, which may cause damage.

The reflection coefficient would be negative, so the reflected wave would have a 180° phase-shift as a result of transitioning to a lower impedance medium.