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I am trying to build a guitar pedal but am unfortunately failing. I have about a high school physics understanding of electronic circuits, but the readings I am getting seem impossible to me. I might obviously be measuring wrong, or I might have a flaw in my understanding somewhere, so I turn to you for help.

First, here is the circuit diagram:

circuit diagram

My issue is this: If I connect my multimeter to the 9V input and the ground, and measure the resistance, I measure about 500 ohms. This seems about what I would expect, as most guitar pedals consume somewhere in the range of 10 to 200 mA, and so this circuit would to my understanding require 18 mA given a 9V power source.

measuring resistance

However, when I connect my power source and measure the voltage from those same two points, I read about 0.8V. The adapter I'm using can supply only 200mA of power, and so by Ohm's law there should only be a 4 ohm resistance. I own another power supply, which is supposed to supply 1700 mA. Plugging this in leads to a voltage measurement that fluctuates between 0.5 and 0.7 volts.

enter image description here

Now, if I were to have a short circuit somewhere, I would still expect to read 9V over the whole circuit. If I did not have a full connection, I would expect to read infinite ohm resistance, and no voltage either. My only explanation is that there is a complete circuit, but the real resistance is much higher than I am measuring. How could this be?

EDIT: Here are two pictures of the top and bottom of the PCB:

pcb top

pcb bottom

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    \$\begingroup\$ If I connect my multimeter to the 9v input and the ground, and measure the resistance, I measure about 500 ohms. Then you're "lucky" but realize that this test means nothing. You cannot expect every circuit which consumes 20 mA at 9 V to measure 9 V/20 mA = 450 Ohms. Your multimeter will not apply 9 V to the circuit. Many circuits behave in a non linear way, consuming much less current at lower voltages. \$\endgroup\$ Oct 5, 2021 at 8:35
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    \$\begingroup\$ I see you're applying Ohm's law everywhere. Don't! It doesn't work like that since you have non linear devices in your circuit: JFETs, diodes. What you do now will confuse you as you simply cannot use Ohm's law everywhere like that. Ohm's law mainly applies to resistors only. To debug your circuit, apply 9 V then measure the voltages in the circuit. \$\endgroup\$ Oct 5, 2021 at 8:35
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    \$\begingroup\$ The adapter I'm using can supply only 200mA of power, and so by Ohm's law there should only be a 4 ohm resistance NO. Stop applying Ohm's law everywhere. There is no 4 Ohm in a 200 mA adapter. It means that the adapter is designed to supply a constant voltage (like 9 V) up to a current of 200 mA. What is true is that the smallest load such a supply can power is a resistor of 9V / 200 mA = 45 Ohm. \$\endgroup\$ Oct 5, 2021 at 8:39
  • \$\begingroup\$ Okay! Thanks for that. \$\endgroup\$
    – Linus
    Oct 5, 2021 at 8:41
  • \$\begingroup\$ A short in series with a diode, or a diode junction of a transistor, would give you a reading in the 0.7 V ballpark. Leave the supply connected for a while, and feel all the components to see which one is getting warmer than the rest. Many circuits contain an input power supply diode to make sure it's powered the right way round, so you could just be seeing the voltage on that diode - should be pretty easy to spot if you trace the tracks form the supply wires - meaning a dead short somewhere on the actual circuit, look for solder splashes! \$\endgroup\$
    – Neil_UK
    Oct 5, 2021 at 8:41

2 Answers 2

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You have diode D2 installed backwards:

enter image description here

The blue/green arrow points to the banded end of the diode as marked on the PCB. The red arrow points to the banded end of the diode as you have installed it.

The 1N4001 can handle a good bit of current, so it probably wasn't damaged.

D2 is there as a reverse polarity protection. The idea is to short circuit the power source if it is connected backwards.

With D2 installed backwards, it shorts the power supply when you connect it correctly.

Remove D2, and install it as marked on the PCB.


The 0.8V you measured was the give away. That's about the forward voltage of a diode conducting a large current. I know that guitar pedals often have a reverse polarity protection made using a diode across the power supply inputs, so I looked for one in your schematic and found D2. I then checked your photos to find D2, and found it there next to R13. The photos are fuzzy, but clear enough to see the cathode marking on the PCB for D2 and the cathode marking on the diode itself.

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  • \$\begingroup\$ Thank you so much!! \$\endgroup\$
    – Linus
    Oct 5, 2021 at 10:48
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    \$\begingroup\$ @Linus while you're at it, check to make sure you didn't reverse polarity of the large capacitors either. \$\endgroup\$
    – LShaver
    Oct 5, 2021 at 17:19
  • \$\begingroup\$ Just for my understanding, how come I still read that 500 ohm resistance from input to ground? I would expect the improperly installed diode to give no resistance at all? \$\endgroup\$
    – Linus
    Oct 6, 2021 at 7:04
  • \$\begingroup\$ A diode isn't a short circuit when forward biased (conducting.) You can see that on the 0.8V that you measured when the power supply was connected. Your ohm meter is interpreting the forward voltage of the diode as a resistance. \$\endgroup\$
    – JRE
    Oct 6, 2021 at 7:10
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Further to JRE's answer, something else looked odd.

Having found a picture of the bare PCB:

PCB Silk Image Source

It is clear that all of your variable resistors are inserted the wrong way also. The two metal pins visible from the top of the trimmers are each end of the resistor (pins 1 and 3). The middle pin (not visible from the top) is the wiper (pin 2):

Wiper pinout from PT-10 datasheet Image Source

You have connected your trimmers 120 degrees out from the correct rotation - notice on the PCB there are numbers marking 1, 2 and 3. You've connected pin 1 to pin 2, pin 2 to pin 3, and pin 3 to pin 1.

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  • \$\begingroup\$ Ohhh, stupid. I'd googled whether orientation mattered but was apparently misinformed. Thanks for the help!! \$\endgroup\$
    – Linus
    Oct 5, 2021 at 10:48

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