Wrong output caused by termination resistor?

[LoveLaika]

I built a circuit using an LVDS receiver, the DS90C032B, and I'm confused about something. My schematic is shown below. It simply converts an LVDS differential signal into a single-ended signal. The issue is when no differential signal is applied while the IC is powered on. My oscilloscope shows the output being in the LOW state (Vout = GND, or very close to 0 volts); when you look at the other unused receivers, their outputs are in the HIGH state (Vout = VCC at +5V). According to the datasheet regarding receiver failsafe, it says that the IC requires external failsafe biasing for terminated input failsafe in the case that the driver is unplugged from the bus. I assume that this is the case since there's no input being driven. Thus, I have the 100-ohm resistor across the inputs as shown in my schematic. It says that the  the resistor provides a small bias while line is un-driven which will cause output to be in the high state, but it's not. Why would this be the case? If I remove the resistor, it would be essentially the same as the other receivers, so I think I might be misunderstanding something.

[TimFox]

My experience with similar differential receivers are that they have an internal bias network so that if the inputs are unterminated (no resistor between + and -), the output goes to a defined state.
However, with the usual termination resistor (100 ohms in your case), that bias network supplies too small a current to the inputs to develop a valid input voltage.
Therefore, if you add the termination resistor, you need to add appropriate external resistors to induce a differential voltage above the minimum requirement when nothing drives the resistor.
Obviously, the external resistors must not supply enough current to overcome normal drive signals.

[LoveLaika]

Thanks for your reply. Regarding what you talk about with external resistors, would the scheme be something like they show in Figure 3 of this App Note? It is annoying that this wasn't expounded upon in the datasheet and was only briefly mentioned.

Yes, it appears that for unused inputs, internal resistors set the output to a defined HIGH state as per the datasheet.

It's kind of late to be asking, but is it really necessary to have the termination resistor? The receiver isn't being driven by another LVDS driver IC, but rather it receives an LVDS signal-pair (at least, that's what the device manual says; no indication of how this signal is being driven, no schematic. All it says is signals are LVDS type) and converts it to a single-ended signal. I was testing this scheme with another part, the AM26LS32, and it worked when there was no resistor at the inputs (yes, I know the part not meant for LVDS, but as a proof of concept, it achieves the goal).

[TimFox]

Yes, we used the circuit in Fig. 3.  That was for a 26LS32 receiver on RS-422 or RS-485 differential signals.
The termination resistor is only needed if the differential transmission line is long enough that you need to worry about reflections.
LVDS and RS-422 have different voltage specs, but the same function.
Read the datasheets to verify compatibility for your application.

[LoveLaika]

Thanks. I appreciate the feedback. I'll have to do some experiments with and without the resistor to see if there's any difference. With any luck, the signals are slow enough and the cable short enough that it won't make a difference.

[TimFox]

Usually, you want to compare the signal rise time to the electrical length (in seconds) of the transmission line to see if the termination is needed.
Typical cables are 1 nsec for 20 cm of cable (one-direction), depending on the cable dielectric (insulation).