Voltage mode vs current mode signalling

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promach

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How to PROVE (mathematically if possible) that current mode signalling is always faster than voltage mode signalling ?

 

Prerequisites are unclear.

What's the circuit model? Schematic looks like distributed RC line, no L, no transmission line effects. If so, what's the motivation for this arbitrary choice? Which real world circuit problem is modelled here?
 

It's not true in general.

Current and voltage are different sides of the same coin. Voltage creates current and current creates voltage.

The system sees the voltage and current, it can't know which one the driver was 'thinking about' when it created the signal.
 

There should be a difference in impedance levels I think. Like with the CML (Current Mode Logic) circuits, where the logic gates consume DC current to push down output resistance. Parasitic capacitances are the same compared to normal static logic gates, but output resistance is lower, so the time constants are smaller, bandwidth and speed are higher.
If it is coming from a book and isn't explained better I would put down that book.
 

Voltage and current mode drivers are not one faster than the other inherently. The advantage of voltage mode is that it to produce the same voltage swing it requires lesser current. See for reference: https://www.ece.tamu.edu/~spalermo/ecen689/lecture11_ee689_tx_circuits.pdf
Current mode circuits are faster than voltage mode in general (think of a 1:2 current amplifier) as opposed to a voltage mode amplifier. Your overall speed also depends on the pre-driver stages as well which is often faster in current mode logic.
 
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Someone told me the following :

 

For current-mode drivers, why is it easier to control output impedance ?

For voltage-mode drivers, why "0.25 to 0.5" of the current needed for a given output voltage swing ?

 

1. For current-mode drivers, why is it easier to control output impedance ?

2. For voltage-mode drivers, why "0.25 to 0.5" of the current needed for a given output voltage swing ?

3. What is the purpose of the two opamps in the voltage-mode driver ?

4. How do the two opamps contribute to current amount reduction ?
 

Perhaps the thread title should be changed to "voltage mode vs current mode vs matched impedance". Adding termination to the driver cancels the difference between voltage and current mode. You can't see from the outside how the signal is generated, it can be either a voltage source (thevenin equivalent) or current source (Norton equivalent).
 

In designing a voltage mode, series terminated driver
you have to include the driver's internal resistance
as part of the "Zo" total goal. Since that internal
resistance varies a lot, it must be small, leading to
oversized driver output devices. Which are then a
short circuit protection challenge. One of my first
patents was a short circuit protection scheme for
RS-422 drivers.

Current mode, if controlled by a good reference, is
likely to use smaller drivers and have a high impedance
which in parallel with 50 ohms, you can forget about.

A lot of the speed difference you see, is from transition
from "5V" logic to lower, use of short channel CMOS is
going to bump up speed a lot (and LVDS allows down to
1.8V technology at least, although much remains at 2.5V
and above. Old 5V LVDS was much lower bandwidth (like
150MBPS) than newer parts (400MBPS and higher) with
the main difference being the process node.
 

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