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So I'm trying to hook up a simple circuit like this...

http://www.engineersgarage.com/sites/default/files/imagecache/Original/wysiwyg_imageupload/1/Steering%20Diode_0.jpg

V1 is a 3.3V signal that can randomly turn on.

V2 is also the same.

Sometimes one, sometimes both, sometimes none.

I thought I'll just get a couple schottky diodes and everything will be fine.

But when I put 3.3 volts at V1 to test. I read 3.0v at v2(with no power supply hooked up yet at v2). I understand the voltage loss for diodes, but I don't understand why I'm getting any voltage at v2.

I'm sure I'm missing something simple, but I don't want anything to go haywire...

I've tested the diodes, especially on v2 and it says its good...

What am I being dumb about?

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  • \$\begingroup\$ I wonder if I'm the only one who misread the title. \$\endgroup\$ – Harry Svensson Feb 16 '16 at 21:31
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I'm not 100% sure this is your only problem, but when you view a diode as a mechanical 'one-way-valve' you have to consider the 'reverse leakage' as a similar leak in a mechanical valve. As such, there is a possibility of the reverse leakage of your diode charging a voltage into your V2 node where you're getting the 3.0V reading.

To ensure this is not causing upset to your circuit, please add 'pull-down' resistors (10K should work well) from your inputs to GND (before the diodes).

enter image description here

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  • \$\begingroup\$ Quick question.... why 2 before diodes instead of 1 after? \$\endgroup\$ – Bowhuntr11 Feb 16 '16 at 15:37
  • \$\begingroup\$ @Bowhuntr11 (1 of 2) In short, you want to place the pull-down resistors on the inputs. If you placed a pull-down resistor between the diodes & 'load' in your orig. circuit, it would be 'pulling down' power supplied to the load (by allowing some current to bypass 'load' through resistor to gnd), but wouldn't have a notable effect on voltage buildup at V1/V2 due to diode reverse leakage (pulldown resistor in this case would be 'before' the reverse leakage current). \$\endgroup\$ – Robherc KV5ROB Feb 16 '16 at 15:59
  • \$\begingroup\$ Thank you! So this looks good to you? circuitlab.com/circuit/55h4dj/circuit-1 That would stop the 3v3 from getting to V3, correct? \$\endgroup\$ – Bowhuntr11 Feb 16 '16 at 16:04
  • \$\begingroup\$ (2 of 2) However, having a pulldown resistor between each V terminal & its corresponding diode (as in the answer schematic), allows the resistor on the terminal to which no 'forward'/'source' potential is being supplied to provide a path-to-ground which only the diode's reverse leakage current 'sees'. So the voltage at the 'non energized' terminal should show only the voltage drop of the diode's reverse-leakage current across the pulldown resistor. \$\endgroup\$ – Robherc KV5ROB Feb 16 '16 at 16:15
  • \$\begingroup\$ Thank you so much! Last question, I promise! How do you figure out what the wattage/Ohms are suppose to be for a resistor like this. Seems that it would be hard to measure this. My load takes 3.3V and 200mA. Would that really mean I would need a resistor that can handle .66W or higher? So a 1W 10k Resistor? That seems wrong..... \$\endgroup\$ – Bowhuntr11 Feb 16 '16 at 16:33
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Schottky diodes are quite leaky (worse when they are hot), and you have no resistor to ground. It's reading the leakage current through the diode and the input resistance your meter (probably 10M or something like that). Low breakdown voltage Schottkys tend to be worse.

Consider the common garden-variety 1N5819:

enter image description here

Edit: As you can see from the above specs there is no guarantee that 10K will be low enough, but I've found 'typical' curves .. which you can use if you feel lucky.

enter image description here

3V reverse voltage at 25°C represents about 1uA typically, so only 10mV across the resistor- probably no big deal. At 75°C it's more like 0.03mA so you;ll have about 0.3V across the resistor 'typically', getting significant.

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  • \$\begingroup\$ That's funny. I have the 1N5817 diode, so I should be fine with it. Correct? :) I'm guessing my problem is the pull down resistor like you and Rob have mentioned. I will look into that more. Thank you! \$\endgroup\$ – Bowhuntr11 Feb 16 '16 at 12:25
  • \$\begingroup\$ If you have lower currents the BAT54 is worth considering. \$\endgroup\$ – Spehro Pefhany Feb 16 '16 at 13:06
  • \$\begingroup\$ Dumb question. Why would this work? But not the OP circuit? Isn't it exactly the same (inside the BAT54C)? \$\endgroup\$ – Bowhuntr11 Feb 16 '16 at 13:24
  • \$\begingroup\$ The pull-down resistor (still necessary) can be much higher value because the BAT54 is a smaller diode so it has much less leakage (and somewhat higher voltage drop for a given current). \$\endgroup\$ – Spehro Pefhany Feb 16 '16 at 13:44
  • \$\begingroup\$ Thank you so much for your time. So if I didn't want to wait for the Bat54, you would do what schematic one is saying? Would #2 do the same thing? These would keep v3 from ever seeing any voltage correct? And When v3 has voltage, it would keep v2 from seeing voltage from v3? 1) circuitlab.com/circuit/55h4dj/circuit-1 2) circuitlab.com/circuit/df9796/unnamed-circuit-2 \$\endgroup\$ – Bowhuntr11 Feb 16 '16 at 14:33
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You are measuring voltage there because you are using a high-impedance voltmeter. Will the diode leakage cause any actual problems to the un-powered circuitry on the other side of the leaking diode? If not, you can ignore the leakage and there's no need for the resistors.

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  • \$\begingroup\$ V1 are AA batteries. And V2 is 3V3 from a usb hub. I don't think it would...but I guess id rather be safe then sorry. \$\endgroup\$ – Bowhuntr11 Feb 16 '16 at 23:49

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