Drop voltage a bit

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.

2. Use a small 33 ohm resistor in series with Vcc.

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

• Is there a reason why you don't want to use a regulator? – Ignacio Vazquez-Abrams Oct 14 '17 at 9:19
• Yes, because I don't have any at hand. – Kokachi Oct 14 '17 at 9:21
• I suppose it's a bit late for this project, but I recommend you keep some of both TO-220 and TO-92 versions of 5V and 3.3V regulators on hand. – Ignacio Vazquez-Abrams Oct 14 '17 at 9:23
• Point taken, I'll buy a few AMS1117 to use on future projects. – Kokachi Oct 14 '17 at 9:42
• Note that you can embed the CircuitLab schematics directly into your posts. You don't need to take a screen-grab. It also means that you can edit, copy, paste, etc. between questions and answers. – Transistor Oct 14 '17 at 10:32

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!

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.

• I like the "power-on" status led idea very much when using diode dropper setup. If possible can you please tell me about the resistor dropper circuit's drawback. – Kokachi Oct 14 '17 at 9:45
• See the update. – Transistor Oct 14 '17 at 10:24
• I like the little t! I use to draw an outline but that small symbol is less jarring. – pipe Oct 14 '17 at 10:28
• I've been using it to get CircuitLab to size my schematics appropriately for a couple of years. It avoids the "circuits for the deaf" appearance when there are only a couple of components. You are the first to comment! Oddly enough it also helps me quickly recognise my own schematics when I revisit an old post. – Transistor Oct 14 '17 at 10:31
• The 7805 needs a couple of volts higher on the input - typically about 7 V in for 5 V out. It's not suitable for your application. It's worth a look at the datasheet to see how to figure this out from the specifications. – Transistor Oct 14 '17 at 10:55