Debouncer with LED inside

Question

How It's possible to add a LED to a line that it's debounced (switch) with an RC circuit. If I place 2 LED, one on ground and one on VCC to know button position. Will they remove my charge/discharge rate of Capacitor since they have a low R load?

/* EDIT */

Of course with schematic is easier.

^{simulate this circuit – Schematic created using CircuitLab}

Switch closed, D2 ON and D1 OFF Switch open, D2 OFF and D1 ON

With debouncing on Input

This schematics is wrong, how can I correctly set it up?

/* EDIT 2 */

It's a reference for a Micro input.

Answer 1

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. D1 is normally on. When SW1 is pressed D1 turns off and D2 turns on.

Note that the input will fall to 0.7 V minimum. Check that this is OK. I've changed the R values so it will affect your debounce timing. Re-calculate those too.

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I've split R1 in two: R1 + R4. This raises the INPUT voltage by \$ \frac {V_{cc}-V_{D1}}{2} \$ to address Dave Tweed's comment.

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^{simulate this circuit}

Figure 2. A better design?

How it works:

• With SW1 open 'B' is pulled high via R2 and D2. Meanwhile D1 is lit via R1. V_A will be whatever V_f of the LED is so D3 will be reverse biased.
• When S1 is pressed 'B' will be pulled to 0 V. D3 will now be in parallel with D1 but it's 0.7 V forward voltage will steal the current from D1 so it will turn off.