1
\$\begingroup\$

Suppose we have an SPI output, it is sourcing a standard logic level (say 3.3 V). Suppose the output is connected to a trace on a PCB with impedance 50 Ohms, and the signal is going to a capacitive load (another integrated circuit). Suppose we know the current associated with that logic level.

If I were to instead make my trace 40 Ohms and change nothing else, will the output current in that logic level change? Ohm's law says yes, it would increase.

However, because push-pull buffers have their own source impedance that depends on their driving signal, I do not know if this change in output current would actually happen. If the current stayed constant when the line impedance were dropped to 40 Ohms, that would mean the buffer's output impedance had to increase by 10 Ohms.

So what would happen in this hypothetical situation?

EDIT: I am not referring to the DC state when the signal has settled at its voltage level. To charge up the CMOS input, there has to be some current flow on the line until that charge up is completed. I'm trying to understand what is this current.

EDIT2: Let's just ignore things like reflections etc., just want to know how the driving circuit forms its output voltage and current, and how do those quantities propagate along the trace until they eventually reach the load.

\$\endgroup\$

1 Answer 1

3
\$\begingroup\$

Assuming the input is CMOS type input, no current flows into it in DC state.

Since current going into CMOS input is 0, it will be 0 no matter what impedance transmission line you have between logic input and output.

Even if it was not a CMOS input, and has some DC load, the transmission line impedance still has no effect to DC current.

For AC currents, if the transmission line is very short, it can be ignored, and only the driver output impedance defines the current the input capacitance charges.

If the transmission line is infinitely long, a high output will just forever see the characteristic impedance as load, 50 or 40 ohms, and that with the output impedance determines the output current to the transmissiom line.

For anything in between, the transmission line length determines how long the current is defined by the characteristic impedance and driver output impedance.

But it gets complex quite quickly, as there will be reflections if neither end of the transmission line is terminated.

For some frequencies, the transmission line may look like a short to the driver, and for some frequencies, the transmission line may look like open circuit.

So yes, in a general case, the characteristic impedance of a PCB trace defines how much current initially flows to it.

\$\endgroup\$
5
  • 1
    \$\begingroup\$ I'm not talking about the DC state. There has to be some current that propagates into the transmission line even during the signal transition, that's why buffers use any power at all. Look at any spec for any digital component and they give a current value. \$\endgroup\$
    – user224284
    Mar 18, 2023 at 20:09
  • 1
    \$\begingroup\$ Good lord.... I would have thought that "logic" would imply switching signal, but okay it's AC. \$\endgroup\$
    – user224284
    Mar 18, 2023 at 20:12
  • 1
    \$\begingroup\$ @sparaps It depends how the signal switching frequency, velocity of propagation in the transmission line, and transmission line length happen to relate in the system. \$\endgroup\$
    – Justme
    Mar 18, 2023 at 20:22
  • 1
    \$\begingroup\$ Now we're getting somewhere. So let's just say the line is infinitely long, the maximum output current during the signal swing would be I(max) = V(max)/(Z(out) + Z(line)), correct? That current has to go somewhere, so according to Kirchoff that current continue to persist on the line until eventually it reaches a load? \$\endgroup\$
    – user224284
    Mar 18, 2023 at 20:48
  • 2
    \$\begingroup\$ Yes, if you have 5V supply, 10 ohm output driver impedance, 40 ohm transmission line of some length, initially there would be 4V at 100mA going into the transmission line, and it would stabilize to 5V and 0mA after the reflections have been bouncing for long enough time that they have settled. \$\endgroup\$
    – Justme
    Mar 18, 2023 at 21:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.