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Offset in the differential output signals in an LVDS transmitter?

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gilbertomaldito

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Is it possible for an offset in the differential output signals to occur in an LVDS transmitter?
 

Re: LVDS transmitter

I think that the answer is yes, but you should elaborate your question a bit more so I really know what you are saying.
 

LVDS transmitter

Yes, but there should be balance related specs that constrain
such offsets to a fraction of what the receiver can tolerate.
 

Re: LVDS transmitter

Hi,

The attached file is the typical LVDS transmitter circuit.
"A" is the ideal LVDS signal at the other end of the transmission line.
I just want to ask if is it possible that there will be an offset like what happened to "B" and "C"?
I have doubts if offset will happen because the differential signal depends on the Resistor. Not unless the resistance varies in time, the voltage drop across the resistor still remains the same. But im not sure, perhaps there are other factors that I did not see. What do you think guys? What do you think are the factors that can make the offset possible?

--andrew
 

Re: LVDS transmitter

If A delivers a ''01010101" pattern, I think there can't be such big offset as B or C.
If A delivers prbs pattern, maybe there is offset like B or C.
There are some relationship with bit rate, loading, and corss point of A and Ab.

andrew_matiga said:
Hi,

The attached file is the typical LVDS transmitter circuit.
"A" is the ideal LVDS signal at the other end of the transmission line.
I just want to ask if is it possible that there will be an offset like what happened to "B" and "C"?
I have doubts if offset will happen because the differential signal depends on the Resistor. Not unless the resistance varies in time, the voltage drop across the resistor still remains the same. But im not sure, perhaps there are other factors that I did not see. What do you think guys? What do you think are the factors that can make the offset possible?

--andrew
 

LVDS transmitter

The received voltage difference is generated by switching the driver current between both polarities. So how do you think an offset can be generated without an additional current source? It can't.
 

LVDS transmitter

Not all LVDS drivers are built with that specific topology. Then
there is always the possibility of some defect that shunts
current asymmetrically or impairs the switch devices' headroom
required, etc. (pushing back onto the CMRR of the tail current
sources).

Imbalance may not be the "normal" case but real product
always carries a tail of the abnormal, which you have to
crop.
 

Re: LVDS transmitter

Yes. The question can be either answered for the given simplified circuit (obviously there can't be an offset) or a real circuit. The standard LVDS specification doesn't require actually switched current sources, so it can be nearly anything.
 

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