Electronic – Drop voltage a bit

diodesmicrocontrollerresistorsstep-downvoltage

I have this ATtiny13a mcu led setup that uses total of 33mA when both leds are on and it works perfectly when powered through usb output.

My problem is that now I want to power it using a 6V 500mA wall adapter, so I want to drop the 6V to below 5.5V since ATtiny13a mcu works at 1.8-5.5 Volts range.

I have come up with two different circuits to achieve this small voltage drop.
Please advice which one of these circuit is better. If both are not good then do tell me other options.

  1. Use a 1N4007 diode in series with Vcc.
    enter image description here

  2. Use a small 33 ohm resistor in series with Vcc.
    enter image description here

I checked the wall adapter with a multimeter and it outputs 5.89 volts when no load is connected.

Best Answer

Neither is good because the voltage drop is current dependent. If the LEDs are off (such as at power-up) the voltage drop will be minimal. In addition, your target voltage is the absolute max of the micro and this is not a good place to be!

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Loading the diode voltage dropper.

If you want to proceed with this approach I suggest that you put a minimum load on the diode dropper and get the voltage well into the safe operating voltage range of the micro. A fixed resistor pulling a few tens of mA should suffice and you could make use of it to light a "power-on" LED.

The power waste will be insignificant on a mains power supply.

Problems with the resistor option:

  • Let's say you're aiming for a 5 V supply and that the LED resistors drop 3 V. We can calculate the LED current then as \$ I = \frac {V}{R} = \frac {3}{220} = 13.5 \mathrm {mA} \$.
  • Let's also assume that the micro draws 0.5 mA. (Wild guess.)
  • There are three possible conditions: no LED, one LED and two LEDs. Let's look at the voltage drop through your 33 Ω resistor in each case.

Table 1. Resistor voltage drop calculations.

               I         V drop
No LED         0.5 mA    0.5 x 33  = 16.5 mV
 1 LED         14 mA     14 x 33   = 462 mV
 2 LEDs        27.5 mA   27.5 x 33 = 907 mV

This clearly shows that there is no regulation with the resistor (meaning that the voltage varies with load) and that on power-up or with all LEDs off that the voltage on the micro will be only 16.5 mV below the supply.

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