Altera_Forum
Honored Contributor
14 years agoTransceiver Pseudo-CML equivalent circuit?
I'm currently testing with the Stratix IV GX development kit and would like to know if anyone has details on the equivalent circuit for the transmitter.
A CML driver typically consists of a differential pair terminated in 50-ohms to a termination voltage, a current sink, and current steering logic, as shown on p9 of this data sheet http://www.analog.com/static/imported-files/data_sheets/adcmp572_573.pdf (nothing special about this part, it just had a CML driver circuit diagram). The CML transmitter outputs high and low voltages by steering current through one of the differential signals, i.e., in the data sheet above, one of the transistors is on, while the other is off. The Stratix IV handbook has the circuit diagram for a DC coupled link between a Stratix IV GX transmitter and receiver on p1-45 of Volume 2 (p489 of the PDF). http://www.altera.com/literature/hb/stratix-iv/stratix4_handbook.pdf However, this diagram does not show how the transmitter buffer is implemented. If I assume that the buffer sinks current according to the CML driver circuit in the Analog Devices data sheet above, then I can draw an equivalent circuit in LTSpice to calculate the DC currents and voltages (yeah, I can do the same with a piece of paper and a calculator too). For a transmit termination voltage of 0.65V and receiver termination voltage of 0.82V, the result is a logic high voltage of 735mV and a logic low of 435mV, assuming 50-ohm terminations and a transmitter current sink of 12mA (the transmitter Vod setting can be used to adjust this). The problem is that I observe a logic high of 910mV and a logic low of 540mV. The fact that the observed logic high is greater than 735mV, means that the receiver termination resistor is actually sinking ~2mA (to develop ~100mV across it), meaning the driver is sourcing ~2mA current. This is inconsistent with the CML driver circuit. However, perhaps that is the meaning of pseudo in pseudo-CML :) Anyone have any insight into this? Cheers, Dave