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Back-Powering and Phantom Power From JTAG

Throughout Yellowstone’s development, I’ve struggled on and off with mysterious problems related to the JTAG programmer. When the programmer is connected to the Yellowstone card, and I turn the computer off and on again, strange things sometimes happen. The Apple IIGS will boot to a screen of green and pink checkered squares, or will beep continuously, or simply freeze. This condition will often persist across multiple on/off/on cycles of the computer’s power switch. Unplugging the JTAG programmer finally returns everything to normal.

I assumed the JTAG programmer was somehow back-powering the Yellowstone card, which in turn might even have been back-powering the computer. When the computer was off, but the JTAG programmer was on (it’s a USB device connected to my PC), it looked like a small amount of power was being provided by the JTAG programmer to the Yellowstone card. In the past I made some half-hearted attempts to troubleshoot the issue, but never really got anywhere, and stopped worrying about it. Yesterday I took another look at the problem.

The JTAG programmer connection has six pins: VCC, GND, TCK, TMS, TDI, and TDO. The VCC pin is used by the programmer to sense the target board’s voltage, so it can set the level of the other data signals properly. Under normal conditions, that means a small amount of current is flowing from the Yellowstone card’s 3.3V supply rail through the VCC pin to the JTAG programmer.


How does this work? I disassembled the JTAG programmer to see. Its output stage consists of a 74HC244 and a 74HC07, both of which are powered directly from the VCC pin of the target board. That ensures the output signals will be at the right voltage level. So far, so good. But I also saw several resistors (pull-ups?) between the VCC trace and other signals on the JTAG programmer PCB. Some of these go back to the programmer’s own 3.3V regulator, or to I/O signals from the programmer’s built-in microcontroller. With a multimeter, I measured about 1.4K ohms between the programmer’s 3.3V supply and the VCC trace for the target board. So when the target is off, it will be back-powered by 3.3V through 1.4K ohm from the programmer. That’s not great, but the true situation is likely worse, because there’s probably also additional unwanted current coming from the microcontroller I/O signals depending on what state they’re in, and the effective resistance may therefore be lower.

I put everything back together, reconnected the JTAG programmer, but did not turn on the Apple IIgs. With the computer still off, I measured the voltage on Yellowstone’s 3.3V supply rail, and saw 1.3V. Then I measured the voltage on the 5V supply rail and saw 0.5V. Doh! I’m not certain how the FPGA and the computer will behave with those low voltages, as they’re certainly too low to run anything normally, but they may be enough to keep some components half-alive in a strange zombie state. Then when the power is turned on, those components don’t initialize normally, and weird things happen. I’m not sure exactly what the mechanism is, but holding the supply at 1.3V instead of zero when the device is off certainly seems like it could cause strange behavior.

The solution seemed obvious to me: put a diode between the 3.3V supply and the VCC pin on the JTAG header. That would allow current to flow out to the JTAG programmer during normal use, but prevent current from flowing in on the VCC pin when the computer is off. The JTAG programmer would see a voltage that’s a diode drop below 3.3V, so the JTAG signals would also be the same amount lower, but the signals should still be plenty high enough to work. To my poor Yellowstone board that’s already suffered so many cuts and patches, I made one more cut and patched in a 1N914 diode.

With the diode installed, JTAG programming still worked normally. The mysterious green/pink checkerboards seemed to have disappeared. With the computer off, I measured 0 volts on Yellowstone’s 3.3V supply rail – hooray! But unfortunately I’d celebrated too soon. While the diode seemed to fix the problem when the JTAG programmer was connected in its default power-up state, it was a different story after exercising the JTAG programmer a bit. I found that if I turned on the computer, used the JTAG programmer to update the FPGA, and then turned off the computer again, Yellowstone’s 3.3V supply rail showed 1.25V and the IIGS would freeze when turned on again. What??

Phantom Power

My experiments ended there, but I suspect the difference is due to the state of the TCK, TMS, and TDI pins. If these are low, then my diode on VCC fixes the problem. But if any of these are high, as they may be after concluding a JTAG operation, I suspect that Yellowstone is being powered through the TCK, TMS, and TDI pins.

How? Like many ICs, Yellowstone’s FPGA has clamp diodes on its input pins. Under normal use, the diode is meant to protect the input buffer if the voltage ever rises too high. It’s just a built-in diode from the input pin to the VCC supply. But when the FPGA is off, this means the chip can actually be powered by supplying a voltage at any input pin, which will raise the VCC supply rail to a voltage that’s a diode drop below the input voltage. This is called “phantom power”, and there’s plenty of discussion about it on the web. It’s not something you want to do intentionally, because the clamp diodes aren’t designed for this purpose, and can’t handle very much current. But in my case, unwanted phantom power seems to be the cause of the problem.

But wait: if the TCK, TMS, and TDI signals are coming from that 74HC244 chip I mentioned, and that chip is powered from Yellowstone’s VCC (which is at 0V thanks to the diode I added), then how can those three signals ever be high? My theory is that the 74HC244 is itself being phantom powered by the JTAG programmer’s microcontroller.

Is there a way to fix this? Additional diodes probably won’t help here, because the direction of current flow is the same for the normal case and the problem case. The only simple fix I can think of is to add a dummy resistor between Yellowstone’s 3.3V supply and GND, guaranteeing a minimum load of at least a few milliamps regardless of what the other chips on the board are doing. That way even if some phantom power is applied, the load will be enough to prevent the 3.3V supply rail from rising high enough to cause trouble, because the phantom power can’t (presumably) supply more than a couple of milliamps at most.

But the devil is in the details. Under normal conditions a 1K resistor would create a continuous 3.3 mA load on the 3.3V supply, but with phantom power the situation is different. There’s already several kiloohms of resistance due to who-knows-what between the phantom power source and the Yellowstone supply rail, so adding what’s effectively a 1K pulldown resistor doesn’t change things very much. I calculated that I’d need a dummy load resistor of just a few hundred ohms to move the needle enough, and reduce the VCC voltage due to phantom power to a point where I could be confident it wouldn’t cause trouble. But a resistor value that low would put an unacceptably large load on the 3.3V supply during normal conditions.

There’s probably some clever solution here, but I’m not going to worry about it. This problem only affects me and my development work with the JTAG programmer, so I’ve decided to ignore it. It’s not an important enough issue to be worth additional time and effort, I think. But it’s an interesting little detour into unexpected failure modes of electronic devices.

Read 4 comments and join the conversation 

4 Comments so far

  1. Dru Nelson - July 23rd, 2021 2:13 pm

    The whole topic of systems that have different power sources intrigues me.
    But that is ok since I’m not an EE.
    It is just something I never thought about much until now.

    There were quite a few devices back in the old days that had their own power supply.
    How did old parallel port printers solve this?
    Ethernet solves this (surprisingly) by isolating with inductors.
    I believe that many electrical test devices had an ‘isolation’ transformer.

    It doesn’t seem like there is a standard/one-size-fits-all solution.

  2. Merlin - July 23rd, 2021 6:00 pm

    Steve, not sure about the phantom power, but regarding the 1N914 diode, you might want to replace that with a Schottky diode. The 1N914 is a standard silicon diode with a forward voltage drop of 0.6-0.7 VDC. Schottky diodes typically have a forward voltage drop of about 0.3 VDC. Not too much of a difference, but when your entire power supply is 3.3 VDC, every 0.1 V counts.

  3. Steve - July 23rd, 2021 7:56 pm

    Yes, a Schottky would have been better. Unfortunately I only had regular silicon diodes handy for testing.

  4. Edward Eastman - August 9th, 2021 12:40 pm

    Have you considered a tri-state buffer chip and a set of pull down resistors to ensure isolation of signals and kill phantom power? They can be high impedance resistors to minimize current drain. You can hook the VCC of the Jtag to the enable line of the buffer also with a pull down resistor of high impedance. I know this adds complexity, so maybe just add a dip switch to kill power from the main board and call it ‘programming mode’.

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