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I'am developing a simple badge with a fading led. My idea was a LMC555 timer powered by a cr1025 battery, these batteries have a nominal voltage of 3v. I chose the LMC555 for its low operating voltage. https://www.ti.com/lit/ds/symlink/lmc555.pdf

I have ordered my parts and made the circuit following this instructable: https://www.instructables.com/id/LED-fading-with-a-555-timer/ and this schematic.

instructable

I used these parts:

  • LMC555 timer
  • 220uf electrolytic capacitor
  • 2N2222 npn transistor
  • two 1206 red LED's 1.7V forward voltage
  • 1.5k resistor instead of 20k for better timing

Powered by a my lab bench powersupply set to 3V. The circuit starts only working above 4.5V, I tried changing the current limiting resistor to a higher and lower value and changing the npn transistor for a ao3402 mosfet http://aosmd.com/res/data_sheets/ao3402.pdf , but that did not help. I even tried removing the current limiting resistor so that there is no voltage drop over the resistor.

Because the fading is to fast to be measured with my multimeter and my oscilloscope stopt working last month I tried to simulate the circuit in LTspice. ltspice Here you can see the LMC555 is giving the full 3V on the output, but on the end the led gets only 700mV. What did I do wrong, is it even possible? I hope somebody can help me solving this problem.

Lux
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  • You say you used 2 LEDs, but your circuit only shows one? If they are in series, then thats 2x 1.7V drop. So that is a drop of 3.2V. You also have a drop over the transistor which can be found in the datasheet. 3V will not be enough for that. Unless you had them in parallel instead? Please update the schematic to show your actual design, because the simulation and reference don't seem to match your parts list – MCG Aug 15 '19 at 10:13
  • Thanks for your reply, you are totally right I forgot to add the second LED. The two LEDs are placed in parallel, so I think this does not really matter to the voltage drop story. You say that I can find the voltage drop off the transistor in the datasheet, am I right that it is the Base Emitter Saturation Voltage? – Lux Aug 15 '19 at 10:39
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    Yeah, the Vbe will be the drop there. It says a minimum of 0.6, plus the 1.7V drop of the diode and you are left with 700mV. That is likely what you are seeing. I don't use LTSpice so not too sure of the exact point that is being measured as it doesn't show probes – MCG Aug 15 '19 at 11:07
  • That is sad to hear, Is there something I can do? Are there special mosfets or transistors with a low Vbe? On the picture you can see the text "LED" "Resistor" and "Out" these are the points where the probes are placed, you can see the names and the colours on the top row off the simulation screen. – Lux Aug 15 '19 at 12:30
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    I would just use a different battery. 3V clearly isn't enough. Taking the LED and transistor drop into account, you'll have 700mV left. You're using a 220 Ohm resistor, which gives you about 3mA, which is barely going to light that LED, so you'll never be able to see the fading effect. Practically, the CR1025 won't be able to power it for very long. You are best to change the battery to a 9V – MCG Aug 15 '19 at 12:40
  • Use an n FET and swap locations of it and the LED. – MadHatter Aug 15 '19 at 13:19
  • You also may be able to drive the LED directly from the 555 output. you can check the data sheet and see what the max current draw from it is. Then you don't need any transistor at all. – MadHatter Aug 15 '19 at 13:20
  • @MCG Thanks for the idea, but a bigger battery is not an option. Yeah I know it is stupid but I was on a tight schedule and already ordered the PCB and all of the components. I made a test setup with a normal 555 on 9v thinking that I can go to a 3V cell with a low power 555, not taking in to account the voltage drop. – Lux Aug 15 '19 at 14:52
  • @MadHatter I think that is not possible because there is a 1.5k resistor in the way for the timing which is limiting the current for de LEDs. – Lux Aug 15 '19 at 14:52
  • Unfortunately it isn't going to work properly on 3V. You could add another in series for 6V though. Why is a bigger battery not an option? @MadHatter yes, you can drive an LED direct from a 555 but it will remove the 'fading' effect that is wanted – MCG Aug 15 '19 at 15:24
  • @MCG The pcb is only 20x20mm a 9v battery would be too large. Because the PCB and all of the components are on there way I think I make a bodge on the PCB so that the LEDs and there resistors are directly coupled to the battery. This will be the best way to not let everything going to waste. – Lux Aug 15 '19 at 18:53

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