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I ran across a open-drain design and am baffled by the resistor between the nFET source and ground. There are only two purposes that I see it could serve.

1) It's series source termination, matching the impedance of a transmission line attached to the drain. However if this is the case, R would be better placed on the drain terminal.

2) Protection against a ground short in the event of an oxide breakdown shorting gate to source. Again, even in this case R would be better placed at the gate of the nFET.

Am I missing anything? The drain is indeed connected to a transmission line, and is pulled up by a 16V rail at the end of the cable run.

EDIT: I'll add that on the other end it's a high-Z digital input along with that 30k pullup to 16V. Also the buffer's supply is only 3v3, driving a low Vgs nFET.

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ Maybe intended to be protection against the output being shorted to a positive voltage. \$\endgroup\$ – Spehro Pefhany May 19 '15 at 21:46
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I suspect it's designed to be a current source, not a hard pull-down. M1 acts as a source follower with R1 as the load. The IRF530 has a VT of 2-4V, so if the buffer is producing 16V, the output current would be around 150-170mA regardless of the pull-up voltage. The other end of the line could use current sensing to determine the state of the output.

That's as much as I can guess without knowing more about the rest of the system. What's connected on the other end of the line? What's the pull-up resistance? Is this a single-ended data line, or part of a differential pair?

EDIT: A low-VT transistor (which is more reasonable) being driven by a 3.3V buffer suggests a current of around 37mA. The weak pull-up would source 0.53mA at most. Since the input at the other end is high-impedance, that suggests to me that the goal of this circuit is to limit the slew rate. Instead of producing a very fast (and thus noisy) pull-down, the current sink gives a gradual rise whose rate depends on the line capacitance. This reduces the bandwidth of the signal.

For example, if the line capacitance is 100 pF:

$$\frac {dV} {dt} = \frac {i_{out}} {C_{line}} = \frac {37\ \mathrm{mA}}{100\ \mathrm{pF}} = 370\ \mathrm{\frac {V} {\mu s}}$$

$$t_{rise} \approx \frac {16\ \mathrm V} {370\ \mathrm {V/\mu s}} = 43.2\ \mathrm{ns}$$

$$BW \approx \frac {0.34} {t_{rise}} \approx \frac {0.34} {43.2\ \mathrm{ns}} \approx 7.87\ \mathrm {MHz}$$

The bandwidth is proportional to the current, so if your transistor can sink 500 mA, that would give you a bandwidth of over 100 MHz. It's not hard to radiate at that high a frequency! The formula I used for bandwidth is a rough approximation, so don't take it too seriously. The important thing is that a 10x difference in pull-down current can give you a 10x difference in bandwidth.

The weak pull-up is also interesting. It seems like the intent is for the line to stay low for a couple microseconds after it's pulled down. Based on the asymmetry, I suspect this is a reset or some kind of system-wide status signal, not a normal communications line.

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  • \$\begingroup\$ Ah good point. I haven't worked with discrete transistors in a while. It's a single-ended discrete, pulled up to 16V by a 30k on the other end. \$\endgroup\$ – Andrew Martin May 19 '15 at 21:53
  • \$\begingroup\$ Looks like with the gate driven to only 3v3, and with an ultralow Vgs, we're looking at current limiting around 20mA. Thanks for pointing me in the right direction Adam! \$\endgroup\$ – Andrew Martin May 19 '15 at 22:13
  • \$\begingroup\$ I've updated my answer with some more speculation. \$\endgroup\$ – Adam Haun May 19 '15 at 22:44
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    \$\begingroup\$ it's probably only an IRF530 because that's the default CircuitLab device for a N-channel MOSFET \$\endgroup\$ – KyranF May 20 '15 at 0:02
  • \$\begingroup\$ Good reading for anyone else looking to brush-up on the topic. Thanks again for the insight Adam. edn.com/electronics-blogs/all-aboard-/4424573/… \$\endgroup\$ – Andrew Martin May 20 '15 at 16:19

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