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I recently programmed my FPGA with a logic analyzer core for the purpose of reverse engineering a crappy modem I have. I found on it what I think is either JTAG or TTL-serial pads. There are 4 blank holes and from what I can tell, 2 pins go to the processor and the other two pins go to power(3.3V) and ground.

The ground pin I'm for sure is ground however. It has near 0 resistance with one of the pins from the power supply. (I assume it's connected through a ground-plane). All of my measurements with a multi-meter seemed to say nothing exceeds 3.3V off of these pins

Now, my problem: I want to hook an FPGA up to this to figure out what this port actually does. The FPGA is hooked up to my computer and gets it's power over USB. I tried measuring with a multimeter what the ground difference is(GND from FPGA to GND of modem) and I got back a surprising result. There appears to be a difference of about 20-23V.. I sure don't want to hook my 3.3V FPGA up to this with that kind of voltage.

So what exactly should I do? I assume the difference is because the power supply of the modem is not grounded(no third prong)

How do I safely hook my FPGA up to this externally supplied modem without frying the modem or my FPGA?

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  • \$\begingroup\$ There are lots of mechanisms to do this, and I assume they'll be well-covered in the answers, but as a quick response you can run your FPGA logic analyzer on a battery-powered laptop with the grounds tied together. \$\endgroup\$ – Kevin Vermeer May 31 '12 at 0:58
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The 20v voltage which you see across 2 chassis is high impedance sourced from mains. The schematics is a diamond shaped bridge fromed by 4 capacitors, real and parasitic. Top and bottom of bridge is mains. Horisontal of bridge is your voltmeter. Shoulders of bridge is this small ceramic UL certified capacitors, capacitive coupling of power cable to chassis, intercoil capacitance of power transformers. There is also an inductive background. The non-zero figures on voltmeter are caused by asymmetry.

If your equipment (power supplies) is generally trusty, you can tie grounds safely, so the few microamperes AC will be shorted, and it will resolve the problem.

But in your case you use something home brewed in combination with insulated device: modem is insulated by design with output transformer. So, no easy solution.

First class solution is creating lab grade ground using real earth and adding isolating transformer. Check the local electrical code as well. Only after this you can tie grounds.

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  • \$\begingroup\$ My FPGA is straight hooked to my computer by USB for power. Would you still not consider that trusty enough? \$\endgroup\$ – Earlz May 31 '12 at 6:10
  • \$\begingroup\$ The modem runs off of 12V, 1.5A. Would it be easier to just hack the modem up to run off of my computer's power supply? \$\endgroup\$ – Earlz May 31 '12 at 7:40
  • \$\begingroup\$ Your computer has no 3rd pin on power cord ? What about using single power strip for computer and modem. Then connect grounds of USB/FPGS to ground near com port of modem. It should be allowed by design of modem power supply to connect modem to computer grounds, even if there is this AC problem. \$\endgroup\$ – user924 May 31 '12 at 11:24
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    \$\begingroup\$ My computer power supply has a 3rd pin, but not the modem's power supply. \$\endgroup\$ – Earlz May 31 '12 at 18:08
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If you measure the voltage between non-connected devices the voltage can be anything. If either of the devices runs off a battery it will probably be 0V. But for two mains powered devices there's a parasitic path via the power supplies and the mains.

That's static voltage, it can't supply any power. If you don't trust it, connect the two grounds via a 1k\$\Omega\$ resistor, and measure the voltage across it. It should be (near) zero volt.

Ground is the reference level of your circuits, and connecting them is the only way that a 3.3V for one device is also 3.3V for the other. Except for a few rare cases, if there's one connection to be made between the circuits, it's ground. Everything else will be properly referenced on both sides.

To be clear: don't leave that 1k\$\Omega\$ resistor there. Grounds should be connected with an impedance as low as possible. Even lower! :-)

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There are other ways to approach the problem, such as getting your two devices to use a common ground, but the most robust is to use some sort of isolators made for this kind of purpose on the signals between the your devices. For example, here are some lists of digitial isolators and optical isolators at digikey.

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