Voltage divider for microcontroller? - Page 1
Forum section: electronics / beginners

author: mayor
author_rank: Regular Contributor
timestamp: September 29, 2014, 11:32:00 pm
content: I need to power a circuit with 5v (one ic requires it). I'll use a 5v regulator, but wondered if using a simple divider were ok to power a very low power 3.3v microcontroller (it _might_ draw 1 mA, maybe 2). I'd really like to avoid having to add a second regulator. Is this viable? TIA

author: Stonent
author_rank: Super Contributor
timestamp: September 29, 2014, 11:36:10 pm
content: Well if you know your 5V is stable it might be ok as long as your resistors have the wattage rating you need. A 1.8V zener diode may work as well if you have one.* - I'm fully aware that someone else may be coming along to tell me I'm a complete idiot.

author: mayor
author_rank: Regular Contributor
timestamp: September 29, 2014, 11:38:41 pm
content: Ah yes! A zener diode sounds like a good approach. In cases like these, is that the generallyaccepted approach?

author: Stonent
author_rank: Super Contributor
timestamp: September 30, 2014, 01:34:27 am
content: I found that in a Microchip tips and tricks PDF on Newark.http://www.newark.com/pdfs/techarticles/microchip/3_3vto5vAnalogTipsnTricksBrchr.pdf

author: ludzinc
author_rank: Unknown
timestamp: September 30, 2014, 03:05:50 am
content: A Zener is an 'acceptable' approach, but practically you only tend to use then when chasing lowest dollar options.Issues with a Zener are:1. You need to run a reasonable current through them to get them to 'zener' correctly. At low currents the 3V3 zener might only be acting as a 2V5 source. 5mA is not an atypical 'test' current, and that can be a lot of wasted power (especially if battery powered).2. You waste a lot of power in the dropper resistor in front of the zener - both the zener and load currents flow through that.You can pick up 3V3 regs for around $0.50 so my preference is just to use that.(FYI - my home products typically run from a 12V supply, with a switcher to provide a 5V rail, and a linear to give a 3V3 rail).

author: tszaboo
author_rank: Super Contributor
timestamp: September 30, 2014, 06:44:08 am
content: Why? A TLV7xx33 is a sot23 package, smaller than nothing, or you can buy a x1117 for 3 cents. There is no real reason for this is there?

author: con-f-use
author_rank: Unknown
timestamp: September 30, 2014, 08:58:51 am
content: For 3 cents? Where?

author: tszaboo
author_rank: Super Contributor
timestamp: September 30, 2014, 10:14:26 am
content: Quote from: con-f-use on September 30, 2014, 08:58:51 amThe big east country with the red flag. Aliexpress ships 100 for you for 4 dollars including delivery. Farnell has it for .25 if you need it tomorrow, and that is a TI part.

author: mikerj
author_rank: Super Contributor
timestamp: September 30, 2014, 12:00:51 pm
content: Three silicon diodes in series will get you from 5v to 3.3-ish with no extra quiescent current. Regulation won't be great, though better than a potential divider unless you can afford to throw away power on small resistor values.

author: katzohki
author_rank: Frequent Contributor
timestamp: September 30, 2014, 02:03:56 pm
content: Treat your micro nice and use a regulator, or Zener at least. If it were something else like a LED I might not bother, but for a micro I would.

author: SeanB
author_rank: Super Contributor
timestamp: September 30, 2014, 05:21:02 pm
content: If the current is low, and constant use a red LED, it will drop 1V6 or so, and will do double duty as a power on indicator as well. 10uF ceramic capacitor as decoupling across the second micro, and a 10R say chip resistor to keep the capacitor inrush current from blowing up the led long term.

author: David Hess
author_rank: Super Contributor
timestamp: September 30, 2014, 09:20:31 pm
content: A voltage divider producing about 3.9 volts may be followed by an NPN emitter follower resulting in a low impedance 3.3 volt output which tracks the 5 volt supply. An easy way to get a more accurate output is to use a 3.3 volt divider followed by a PNP emitter follower driving an NPN emitter follower. Then the PNP Vbe roughly compensates the NPN Vbe.

author: netdudeuk
author_rank: Frequent Contributor
timestamp: September 30, 2014, 11:00:11 pm
content: An LM2937 LDO 3.3 variety would run ok from the 5 volt rail.

author: tylernt
author_rank: Newbie
timestamp: October 10, 2014, 08:24:01 pm
content: Quote from: mikerj on September 30, 2014, 12:00:51 pmI believe the vF on a diode varies with current. As long as your load current doesn't vary wildly, this is probably "okay" though.