# Why can't LED be used as diode?

Discussion in 'Homework Help' started by faeem.ali, Apr 2, 2014.

1. ### faeem.ali Thread Starter New Member

Apr 2, 2014
2
0
Hey All ...

I'm new to electronics and need help understanding something. I have a circuit (attached to this post) with a SPDT switch and 2 LEDs.

When the switch is in position S1, only LED L1 must be on. When the switch is in position S2, both L1 and L2 must be on.

Now, I realise a diode must be used but I didn't have one available. So instead, I tried using another LED (D1) that's exactly the same as L1 and L2. However, when I toggled the switch, either L1 or L2 lit up, but never both simultaneously. I later replaced D1 with a proper diode and everything works as expected.

So, why didn't the LED work as a normal diode? Isn't a LED just a shiny diode? I read that the voltage drop on a LED is greater than on a normal diode, but a LED and a diode are the same otherwise. I also read that LEDs don't have "resistance", but "voltage drop". I don't understand this either.

Any clarification will be appreciated. Thanks

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2. ### MrChips Moderator

Oct 2, 2009
14,289
4,195
When the switch is in position S2, the voltage drop is determined by L2, which is too low for both L1 and D1 together to turn on.

Put another diode the same as D1 in series with L2.

3. ### crutschow Expert

Mar 14, 2008
16,215
4,334
Aside from the larger forward voltage drop, LEDs also generally have a low breakdown reverse voltage rating.

The problem with your circuit is that when the switch is in the up position as shown, then there is one-diode drop for the top LED but two diode drops for the bottom LED. If both diodes are LEDs then the double forward drop is too high to conduct any significant current. If the cross diode is standard then the combined drop is low enough so that both LEDs can light (but the one in series will have a significantly lower current.

To make the circuit operate properly use a separate resistor in series with each LED from the battery. You will need to reduce the value of the resistor in series with the two diodes to maintain the same current as the other LED.

LEDs and standard diodes do have a resistance but it is very non-linear (logarithmic) thus a very small voltage change generates a very large change in current.

See below for a simulation of a typical LED showing current on logarithmic scale versus voltage on a linear scale. A straight line indicates a logarithmic relation. The slight curvature at the top is a result of a small amount of ohmic (normal) resistance in the diode. Notice that for desired currents in the several mA and up range where the diode light is visible, the voltage changes only a few tenths of a volt. This makes the LED appear as if it has a somewhat constant voltage drop when on. That's why you need to add a series resistor to stabilize the LED current and make it less sensitive to small changes in the supply voltage.

Last edited: Apr 3, 2014
4. ### faeem.ali Thread Starter New Member

Apr 2, 2014
2
0
Thanks guys. I'll be doing lots more reading, especially about voltage drop. Crutschow, for some reason I'm getting an access denied error when trying to view your gif.

5. ### MrChips Moderator

Oct 2, 2009
14,289
4,195
As crutschow suggests, put the resistor in both arms of the LEDs.

6. ### crutschow Expert

Mar 14, 2008
16,215
4,334
It works okay on my computer when I click on the thumbnail. Is that what you are trying to do?

Edit: Are you a signed-in member of the forum?

Last edited: Apr 3, 2014