Anything wrong with using two diodes to get about 3.3v

[Brainless]

Hi,

In a small setup to test an 'pro micro' (mini leonardo clone) and a RF24L01 module i used two 1N4007 to lower the voltage from a measured 4.45v to 3.23v.

The RF24L01 needs a 2.8 - 3.3v power supply, the SPI bus has 5v tolerant pins.

Is it acceptable to use two diodes to get that 3.3v?

[mariush]

No, not really (nothing wrong with using 2 diodes).
Just look in the datasheet for those diodes to see how the voltage drop varies with current and heat. For example, you may have 0.7v voltage drop at 100mA but you have 0.9-1v at 1A.. fortunately that rf24l01 seems to use only around 30-50mA so it's small enough not to change that forward voltage much.
 

[hlavac]

You may get too little volage drop when the load is very small.
Like in some very deep sleep modes.
You may need some minimal load.
Check the voltage drop/forward current gaph in diode datasheet

[Brainless]

Wow, that is fast.

Thanks for the replies. I was unaware that the voltage drop changes with load.
The load can change quit a lot in this case. The current the RF24L01 module draws in sleep or transmitting mode is significantly different. The supply voltage can be between 2.8v and 3.3v so i have to check if it stays in that range.

Now i know what to look for in the datasheet.

[sacherjj]

I would just get a handful of LDO 3.3V Linear Regulators.  You will find many uses for them. 

[rexxar]

I would just get a handful of LDO 3.3V Linear Regulators.  You will find many uses for them.

Agreed. If you don't care to salvage one, they're in basically everything. If you have any old PCI cards, they're almost guaranteed to have one.

[Brainless]

I know that regulator is the easiest thing to do, it will then also make sure that the output stays 3.3v even when the input voltage goes up to 5v instead of the 4.5 i have on my particular USB port and cable length.

I was just wondering if there where any detrimental effects when only using a diode to drop the voltage to a desired level.

[SeanB]

If it works over the range and does the job no problem using it. You often see it used especially as the diodes can cost nothing if you are already using them in the design. I have seen this on copier logic boards to supply some 3v3 5V tolerant logic without having an extra regulator. Also used to provide separate analogue and digital supplies using another diode to up the monolithic 7805 regulator ground to compensate. 3 diodes cheaper than 1 extra regulator and the extra 2 decoupling ceramic disc capacitors.

[Feanor]

Series diodes (and transistors) is a hack I use as well. (During prototyping, not really as a design choice)

If you are worried about low load and power consumption is not an issue you can put a permanent load resistor on the 3.3V side.

[SArepairman]

I would not do it unless it was a mass market product that I needed to sell a billion of...

I don't think using cheap tricks like this is a good idea for the hobbyist, there are too many traps. 

Though I must say single transistor switchmode power supplies and other Chinese mass market electronics "hacks" are very interesting despite being completely bootleg. And I suppose you can impress your boss if you design a product and save him 50 cents... so long it does not turn out the diodes are shady and you get a bunch of recalls. A 3.3 LDO can be gotten for like 1$. Not to mention it will block out noise and handle power supply problems better.

Don't be afraid to throw some ICs at your projects... I know people on this forum are vicious when it comes to saving money (I bet you can find a 30 page argument somewhere about if a 3cent carbon resistor is necessary in some one off circuit) but you don't need to think like a Chinese counterfitter engineer. 

I would giggle right now if someone can list some examples of how circuit modifications were performed (during manufacture) to make broken/out of spec parts work, especially if it was out of the box thinking.

[N.M]

you can use a hexbuffer/converter, i m using the m74hc4050 for my raspberry pi. the output voltage would be the supply voltage +- 0.5V. in your example it will be the 3.3V Pin from your arduino as the supply voltage. the advantage  you can protect your input pins from accidentally shortcuts and limit the current drawn from arduino.