USB2.0 current drivers?

[Bambur]

Could someone explain the theory behind the current drivers used in USB2.0 signalling? Why are current drivers used instead of conventional push-pool totem-pole outputs? What benefit does it provide? I study the specification and have got very curious about these drivers

[amspire]

I cannot actually speak on why certain decisions were made, but it would look like it is a way for the high speed drivers to provide high output while maintaining the 90 ohm differential impedance.

The low speed drivers are the usual voltage drivers with series  resistors in each leg, and this works fine as they only have to supply up to about 0.3 volts.

The high speed drivers have to be able to provide up to 3.6V output per driver and to do this with a voltage driver plus series resistor, it would need a supply of over 7.2V. A problem since they only have a 5V supply.

So instead they have current drivers which have a very high impedance feeding across two 45 ohm terminating resistors which results in the required 90 ohm final impedance. As long as the current driver can source their current up to an output voltage of 3.6v, then this arrangement solves the problem of running of a 5V rail.

So basically it is a solution that maximizes the output voltage from the drivers while maintaining the requirement of 90 ohms total termination at both ends. A voltage driver + series resistor would also work, but would probably need a 10V supply instead of a 5V supply.

Richard.

[Bambur]

Richard, thanks for the reply! According to the spec LS/FS single-line signaling is done in the range 0-3.3V, and HS 0-0.4 V (with current drivers) without taking into account possible under- and overshoots. Why current drivers are needed is a question to me I do not know answer to at the moment.

[amspire]

I can only speculate, but as I said above to output 3.3v from a voltage driver through a 90 ohm resistor to give a 90 ohm source, you would need over 6.6 volts for the supply. 10v practically.  Using a constant current driver allows 3.3V to be sent from a 5V supply. 90 ohms in parallel will give a 90 ohm source.  So it is a way of getting the most out of a 5V supply.

[Bambur]

Richard, you might be very right in fact. In order to reduce the signal swing in each wire (EMI and etc.), they decided to use current drivers, thus limiting the current and voltage to some acceptable levels. Otherwise, the voltage swing would be 3.3V / 2 = 1.65V instead of 0.4V as per the spec. Now it looks to me like a very wise decision after all. I wonder why I didn't think about it at first place. Thanks a lot!!

[ejeffrey]

One nice thing about current steering logic is that the transmitter is a constant current load on the chip.  This means that the power supply pins and decoupling caps (and their inductance) is not in the signal path.  This makes it easier to go fast, reduces the generated noise on the supply bus that other parts of the chip have to deal with, while at the same time being less vulnerable to the same types of interference.

[Bambur]

ejeffrey, good point! Thanks! Indeed, such a driver would produce much less common-mode noise due to parasitic inductance of package pins and etc. It looks like there is a current source which output is directed to either D+, or D- lines when signaling, and to the ground when idle. It all makes a lot of sense to me now. What a brilliant design after all!