Diode AND Logic; Pull Up Resistor

[digitaleyes]

I would like to build an AND module for my Eurorack, but I'm having trouble understanding how this circuit works.

On Doepfer's DIY page, they say the following can be modified to their OR diagram to realize an AND circuit:

As a modification of this circuit even "and-wired" sockets can be realized. But this type of combination is not used very often: all inputs have to be "high" to turn the output to "high". For all other conditions the output remains "low". On this the diodes have to be flipped (i.e. the cathodes at connected to the input sockets) and the resistor has to be connected between the output and +12V (so-called pull-up resistor).

I drew the following, but I have no idea what I'm doing.


 
If this is correct, I'm completely perplexed. I copied the idea from what I gathered on diode logic on Wikipedia. Any help, thoughts, or suggestions would be great.

TIA
Mike

[sleemanj]

If Input A is low, then current flows through diode A, pulling that node (and therefore output) to Vfwd of the diode (say 0.7v, 0.3v schottky).

Similarly if Input B is low, same thing.

If both input A and B are low, same thing.

So that leaves where input A and B are both high, and you can see then that the diodes can not conduct (reverse biased)  meaning that the output is high.


NB: you have an extra ground/reference, center bottom of diagram, you're just pulling the output low permamantly with that, eliminate it.

[digitaleyes]

Like this?

I guess that's one thing that concerns me when the voltage is low. Will all 12V pass through the diodes when either input is low? In my basic understanding, I would think this would cause 12V+ to come out of my jack and potentially damage the signal source.


Thanks again

[Andy Chee]

I guess that's one thing that concerns me when the voltage is low. Will all 12V pass through the diodes when either input is low? In my basic understanding, I would think this would cause 12V+ to come out of my jack and potentially damage the signal source.

Assuming it's correctly sized, the pull-up resistor will limit the current that is delivered through the diodes when the input is low.

[digitaleyes]

Thanks for the feedback. I guess I'm going to dive in. I just bought a DC power supply and will soon get an oscilloscope. I want to learn all I can about electronics.

So current = voltage/resistance. (let's assume I go off of the 5V power supply in my case with a 10K pullup resistor). This would then equal .0005 amps of current? And this must be a minuscule amount of current?

I'm only asking because I don't know. I just looked up and saw that .1 - .2 amps can kill a human. Granted, that's 200- 400x the amount of current of .0005 amps. I don't want to kill myself once my power supply arrives.

Assuming it's correctly sized, the pull-up resistor will limit the current that is delivered through the diodes when the input is low.

[Andy Chee]

I'm only asking because I don't know. I just looked up and saw that .1 - .2 amps can kill a human. Granted, that's 200- 400x the amount of current of .0005 amps. I don't want to kill myself once my power supply arrives.

Theoretically speaking, 30mA (0.03A) AC current or 300mA (0.3A) DC current across the heart can kill a human. 

But human flesh usually has high enough resistance to prevent electrocution death from anything less than 50V, so I'm sure you'll be fine