Re: does this circuit use conventional current?


Mike wrote:
KodKodKod Learning Consulting wrote:
Hi,

Sorry, I didn't read carefully! You used the right term as John Fields
said.

For a diode, the bar is the cathode and the edge of the triangle is the
anode. And current flows from anode to cathode.

LED is just a diode. It can have both hole or electron flows (depends
on how it is made). LED is forward biased when the anode has higher
potential then the cathode and current (conventional) flows from anode
to cathode. For electron, it flows in the opposite direction (from -ve
to +ve as it is attracted by the E-field) and for hoe, it flows from
+ve to -ve as repelled by the E-field. However, since electron has -ve
charge and hole has +ve charge, they contribue to the conventional
current in the same direction.

So when you deal with diode, to determine whether it is forward biased
or reverse biased, it is better to inspect the terminal voltage instead
of how it is connected. In this case, we are sure the anode always has
higher voltage than the cathode (which is connected to the ground). So
the LED is forward biased. (Of course, the anode voltage is changing
during operation and it can be too low (but still > or = ground) to
turn on the LED and it doesn't lid).

Regards,

KodKodKod Learning Consulting


Thanks for the explanation of hole/electron flow. So, the answer to my
question is: "Yes, this circuit uses conventional flow"?

--
Mike

Hi, Mike. I'm not sure whether you're confident you've gotten a
satisfactory answer yet. Here's a picture that might help (view in
fixed font or M$ Notepad):

|
| ~
| ___ ~ LED1
| .---|___|-->|-----.
| | R1 ~ |
| | ___ ~ LED2 | A |. | K
| o---|___|--|<-----o ------| \|-------
| | R2 | |/ |
| | |
| +| |
| --- |
| - |
| | |
| | |
| '-----------------'
|
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

Looking at the simple circuit diagram, there's a 3V battery and two
resistor-LED strings. In the picture, LED1 will be ON, and LED2 will
be off. If the anode (A) is more positive than the cathode (K), the
diode is said to be forward-biased, like LED1. If the cathode is more
positive than the anode, the diode is said to be reverse-biased. A
forward-biased diode permits current to flow, and a reverse-biased
diode prevents current flow.

I believe you're asking if the output of the last gate is sourcing
current or sinking current. Here are two logic gates, hooked up with
two resistor-LED strings:

|
|
| __ VCC
| -| \ +
| | )o---. |
| -|__/ | V ~
| .-. - ~
| R| | |
| | | .-.
| '-' | |
| | | |
| V ~ __ '-'
| - ~ -| \ |
| | | )o---'
| === -|__/
| GND
|
| Output A Output B
|
|
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

Output A is said to be current-sourcing. Of course a 4093 output can
either source or sink current. When the output is a logic 1 (output
voltage about equal to the battery +), current will flow out of the
gate (thinking conventional current flow), through the LED and
resistor, to COM, and turn on the LED. When the output is logic 0
(output voltage approx equal to the battery -), both sides of the LED
will be at basically the same voltage, so no current will flow.

Output B is said to be current sinking. When the gate output is 1,
both sides of the LED will be at the same voltage, so no current will
flow. But when the gate output becomes 0, current will flow from the
battery + thru the LED and resistor, and into the logic gate output.
The output acts like a sink for current (again, using standard
conventional current flow).

So, in your circuit, the LED will only be on when the NAND gate output
is high. That's a current sinking output.

By the way, don't connect the battery negative or anything else in this
circuit to ground. The symbol is a way of saying circuit common (COM),
or the negative terminal of the battery.

You sound like an intelligent guy, and just need some background in the
basics to be able to have a lot more understanding. I'd recommend
finding a copy of Don Lancaster's "CMOS Cookbook" for more
straightforward, non-technical background on CMOS circuits. Actually,
most of what's in your application circuit is covered in his book.
It's available at many libraries, Amazon, or Mr. Lancaster's website:

http://www.tinaja.com/

Good luck
Chris

.

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