Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Regulating low current 900mv from a 1.8v source without an LDO

Status
Not open for further replies.

jstefanop

Member level 2
Joined
Nov 9, 2011
Messages
48
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,286
Activity points
1,789
So I have a single PLL pin that needs .9v and draws no more than 5ma of current. I don't want to waste a whole LDO just for such a small task, so i want to use the 1.8v output from an existing LDO and drop it down to .9v for this pin. What 2-3 low cost component count solution do I have for this problem?

My first thoughts was I would just use a zener diode, but after some search a .9v breakdown voltage zener does not seem to exist. A simple resistor voltage divider also won't work since I don't know the exact current draw of the pin(could be 1ma..could be 4.4ma).

What I'm thinking now is to just use a regular diode, that has a .9v forward voltage at the low current I need. Any thoughts?
 

What I'm thinking now is to just use a regular diode, that has a .9v forward voltage at the low current I need. Any thoughts?

What about a silicon and a germanium diode in series? Instead of the Ge diode you could use a (possibly high current, s. PDF below) with a forward voltage of about Vf=0.3V.
View attachment SS8P4C_Schottky-Diodes.pdf

As you are posting in this forum (and not in the Analog Integrated Circuit (IC) Design, Layout and Fabrication one), I hope you can use discrete devices!
 

What about a silicon and a germanium diode in series? Instead of the Ge diode you could use a (possibly high current, s. PDF below) with a forward voltage of about Vf=0.3V.
View attachment 123424

As you are posting in this forum (and not in the Analog Integrated Circuit (IC) Design, Layout and Fabrication one), I hope you can use discrete devices!

Wouldn't a single .9v Vf diode like this https://www.digikey.com/product-detail/en/SD101CW-TP/SD101CWTPMSCT-ND/717418

work?

Why use two diodes to obtain a series Vf of .9v when you can do it with one?
 

I think one possible reason is germanium and silicon compensate each other temperature-wise.
 

Vf=0.9V @ 15mA is a specified max. value for the SD101CW diode. Its typ. Vf may be much lower, perhaps half this voltage. Unfortunately the data sheet doesn't show min/typ/max Vf values or its temperature dependency characteristic, like the SS8P4C data sheet above.

- - - Updated - - -

I think one possible reason is germanium and silicon compensate each other temperature-wise.

Unfortunately this isn't true: they have nearly the same Vf temperature dependency.
 

Vf=0.9V @ 15mA is a specified max. value for the SD101CW diode. Its typ. Vf may be much lower, perhaps half this voltage. Unfortunately the data sheet doesn't show min/typ/max Vf values or its temperature dependency characteristic, like the SS8P4C data sheet above.

Yea I realized this after looking at a few datasheets. This one is almost there if I connect both of them in series: **broken link removed**

Gets me a Vf of .6-.9 between .5-5ma. Unfortunately even this range is too great. I guess regulating with diodes won't work after all...The only way it could work is if I can find a diode that has a very stable Vf over that current range.

I did come across this as well in my searches...not sure how this works, but it seems like that circuit keeps a stable voltage up to 20ma.

**broken link removed**
 

These simple methods produce semi-regulation.

Left: A single diode produces 0.7V. Add a potentiometer above it. Dial your desired output.

Right: Two series diodes. Add a potentiometer across the upper diode. Dial your desired output (between 0.7 and 1.4V).



The resistors result in 'loose' regulation. You might get within a few percent with loads from 1 to 5 mA. It may be good enough, depending on your requirements.
 

The question you should be asking is, what voltage tolerance is required?

Because the simple diode methods described above will have both current and temperature dependencies. I would be surprised if you get a combined tolerance tighter than about +/- 50 mV without trimming.
 

+/-100mv from 900mv is completely fine. Which is why Im even considering this. I guess using a pot to figure right resistance then just using that fixed resistance for final design could work.
 

The only way it could work is if I can find a diode that has a very stable Vf over that current range.
You wouldn't find any - it's physically impossible.

I did come across this as well in my searches...not sure how this works, but it seems like that circuit keeps a stable voltage up to 20ma.

**broken link removed**
In my eyes: can't be better than any of the previous 1 or 2 diodes solutions. A diode is a diode is a diode ;-) . Not yet talking of its not to be forgotten temperature dependency.

Brad's solution is definitely better, but should be simulated after estimating.

I know you don't want a further LDO, and actually there aren't so many for this low voltage. But seeing the rather large voltage variations from the previous solutions I guess it's not too bad a solution: I think the best solution is a dedicated voltage regulator which I found meanwhile: Page 12 of its data sheet (available via the link before) shows a 0.9V output circuit with the LT3022 from Linear Tech. A 4 devices solution - additionally needs 2 resistors and one (ceramic!) 10µF cap.
 

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top