Why doesn't this work?

[McSpunckle]

OMGWTFLED.png (2.84 KiB) Viewed 1722 times

Can't figure this out.

All LEDs are the same 3mm white LEDs. All resistors are 1%.

D1 and D2 work. With SW1 closed, the first 3 LEDs come on as expected, but D3 is dim. Closing SW2 turns off D3, and D4 is brighter than D3. The voltage across R3 is higher than R2.

I've tried adding diodes (silicon and LED) and resistors between D3 and D4. it seems like, at least with D4 off, D3 would be the same brightness-- but it isn't. It's quite dim. And I've tried different LEDs, resistors, resistor values, etc. Nothing fixes it. I'd blame the switch, but D1 and D2 are on the same switch as D3 and they work fine.

I know LEDs with different voltage drops don't play well together in parallel, but that doesn't seem to quite be the issue here... unless I'm missing something.

[McSpunckle]

Ok, so just disconnecting D1 doesn't work, but disconnecting both D1 and D2 does make D3 and D4 pay right together... feh

[McSpunckle]

Ok, 100 ohms between D1&2 and R2 fixes the problem. Thanks for the help!

[McSpunckle]

Wait shit no it doesn't. D3 is brighter now, but it still turns off when I close SW2. >.<

[Schlatte]

have you tried seeing it through the eyes of a loaded voltage divider? have you tried seeing it through the eyes of a voltage/current divider for parts with linear and diode characteristics? have you tried using Kirchhoff's laws (also keeping in mind that you have linear and diode characteristics)?
ohh god.. I hate such circuits... we did loads of them in 2nd grade...

[multi_s]

What exactly are you trying to achieve?

D1 and D2 work. With SW1 closed, the first 3 LEDs come on as expected, but D3 is dim.

Of course it will be. Consider D1 is in parallel with D2 so the voltage across both must be the same, and assuming identical LEDs, the current will be the same. But wait, now you have (R1 + D3 in series) in parallel with D2 so the voltage across R1 + D3 must be the same as the voltage across D2.

With out doing any real math, intuitively you know there MUST be a voltage drop across R1 IF there is any current going through that branch -> the voltage across D3 MUST be less then the voltage across D1 or D2 SINCE VR1 + VD3 = VD1 -> VD3 = VD1 - VR1 -> assuming all currents are in the same direction -> VD3 < VD1

Closing SW2 turns off D3, and D4 is brighter than D3. The voltage across R3 is higher than R2.

Is this a typo? if it TURNS OFF D3 then assuming D4 is on at all, it would be brighter then D3.

Or if you mean its brighter then D3 was when its on, the previous answer explains why. IE Vd4 is not clamped by D1 or D2 as was the case with D3 so it behaves more as you expected.

edit: its implied that less forward voltage across a diode is caused by less current. so either showing that there is less voltage drop OR that there is less current implies that the brightness will also be less.

[multi_s]

hi

sorry i missed your second post.

Wait shit no it doesn't. D3 is brighter now, but it still turns off when I close SW2. >.<

ok think of it like this. when you close SW2 MORE current is going to flow through R1 right? Because the resistive load it sees just decreased (so current increases).

Now go back to what i posted before about VR1+VD3 being clamped by VD1 and or VD2.

If the current in R1 increases, the voltage across it also INCREASES. Since VR1 + VD3 is FIXED by VD1, Obvisously an increase in VR1 MUST mean a DECREASE in VD3 -> as stated earlier less forward voltage means less brightness. If VD3 is less then some characteristic of the LED it will not emit photons at all.

what are you trying to setup? there is probably an easy way to do it.

[McSpunckle]

D3 was supposed to be dimmer than D1 and D2-- that's what that R1 is for. It just wasn't supposed to be dimmer than D4.

I got it working properly by disconnecting D1 and D2 from the cathode of D3 and giving them their own resistor to the switch. I set it back up the old way with some alligator clips just to take some measurements and see what was happening. Voltage across the LED when it's working right is 2.7 volts, but hook the other LEDs directly to its cathode and it drops down to 2.2V. Voltage across R1 goes from 5.7V to 5.2V. R2 goes from .5 volts to over 5 volts. So, yeah, the increased current through R2 (made worse by R1 with SW2 closed)) is creating too large of a voltage drop for D3 to operate.

w00h! Solved. Thanks.

[multi_s]

in this case you can also get rid of R1 and just increase the value of R2 and R3. then you still have the same amount of parts as your original, but working how you wanted it.

[McSpunckle]

Yup. That's exactly what I'll do if I'm doing this again. What I was doing is setting up the LEDs for a glow ring on a pedal to come on with the bypass LED, and a second LED to work separately. Alas, it was not having any of my shenanigans.

[multi_s]

sounds cool. good luck.