Using diode breakdown voltage/similar to determine if voltage is over a given value

[Jamdan]

Is there a way to run voltage through something (zener?) with a breakdown/bypass/etc., at 5.25 volts, and go from there to a resistor/diode that would use .25 volts, leaving 5 volts remaining? (without changing the amps much).

Basically, is there a simple method of determining if the voltage is over 5.25, and if so, subtracting .25?

(I asked this in the middle of another thread, but it kind of got buried and unnoticed (or it's just a dumb thing to ask!))

 

[duffy]

I can't quite understand what you're asking. Sounds like you might just want a 5V zener and a resistor, but you could need a "low droput" regulator like an LP2954. Need more details.

 

[ronsimpson]

If I remember right: Jamdan has power that is about 5 volts. (4.75 to 5.25) and he wants 5.0 volts. At 1 amp.

The LP2954 is only good for 250mA with a drop out of 0.47 volts.

 

[Jamdan]

Hi Duffy,

Thanks for responding.

So I was right that I could have a 5.25v zener followed by the proper ohm-value resistor to dump .25 volts and get 5v?

If the incoming voltage was less than 5.25 volts, that would be usable for my application, and I could run that power straight to it. But if the incoming voltage got over 5.25 volts I would want it to go through the zener/resistor instead. How would I get it to do that?

 

[Jamdan]

Hey Ron,

Yes, I have anywhere from about 4.5 to 5.5 volts or so, that I would like to get to 5 volts. For this thread I would like to ignore any voltage below 5 volts (and slightly more complex voltage regulators, boost/buckers, etc.), and explore very simple ways of knocking the higher voltage down to 5 volts.

It seems the zener/resistor method is feasible, but I don't know how to arrange it. It seems like there is the potential of making several parallel paths of:

If the incoming voltage is: 5.50v to 5.74v --> 5.50v zener - .25v dumping resistor = 5.25 to 5.49v, which would then go to...
If the incoming voltage is: 5.25v to 5.49v --> 5.25v zener - .25v dumping resistor = 5v to 5.24v, which is good enough.
And...
If the incoming voltage is: 5.00v to 5.24v --> no zener/resistor = 5v to 5.24v

Famous last words - It seems like it should work...

 

[Dean Huster]

Putting the supply in series with one germanium diode (normally very, very low current applications) or with three hot-carrier (Schottky) diodes would reduce it by about 0.3v. If you start with a regulated supply, the result should be pretty good. If not a regulated supply to start, the LDO regulator is the best idea.

 

[Jamdan]

Hi Dean, thanks for the reply. The current would be up to about 1.1 amps.

 

[duffy]

So I was right that I could have a 5.25v zener followed by the proper ohm-value resistor to dump .25 volts and get 5v?

No, you would use a 5V zener like I said, +V would be taken off the cathode of the zener (remember they are hooked up "backwards") and the resistor would be in series with the +5.25 and the cathode of the zener.

This is called a "shunt regulator". Here's a diagram - http://www.keterex.com/appsspot.php

 

[Mikebits]

Is .25v really a problem? Most circuits can tolerate that deviation.

 

[colin55]

Next, you have to work out the wattage of the zener.

 

[Jamdan]

Thanks for the input guys.

And thanks for the link, Duffy, that explains a lot.

 

[crutschow]

For 1.1A I would use a low-dropout regulator set to 5V. It would follow the input voltage with little drop up to 5V and then regulate the voltage to 5V.

With a shunt regulator, the resistor is always in series with the load.

 

[duffy]

For 1.1A I would use a low-dropout regulator set to 5V.

Find one that has a Vin-Vout differential of less than .25V @ 1.1A.

 

[Jamdan]

What do you guys think of this boost/buck switching power supply that Ron found?

https://www.ti.com/product/tps63002?CMP=AFC-conv_SF_SEP&partnerName=

There is no dropout all, am I correct on that? It only accepts up to 5.5v input, but that is not too bad.

 

[Jamdan]

With a shunt regulator, the resistor is always in series with the load.

Duffy had an interesting link that addressed that earlier: http://www.keterex.com/appsspot.php

The Series Regulator on the bottom of that page is especially interesting to me.

Does this form of regulation have its own/any kind of dropout? I keep noticing "0.6v" in the formulae... (this question is for anyone, but mainly Duffy since it was his link (see what ya get for helping?! ))

 

[crutschow]

What do you guys think of this boost/buck switching power supply that Ron found?

https://www.ti.com/product/tps63002?CMP=AFC-conv_SF_SEP&partnerName=

There is no dropout all, am I correct on that? It only accepts up to 5.5v input, but that is not too bad.

Yes, that would give a constant 5V output for your varying input voltage. It would likely be the most efficient solution to your problem.

 

[Jamdan]

Thanks for confirming that for me, Crutschow.

That buck/boost is probably what I will use. Unfortunately, it only accepts up to 5.5v incoming voltage. Can you think of any way to deal with voltages from 5.5v up to about 6v (which would likely be the most I would ever encounter)?

I didn't see in the specs what this boost/buck does with incoming voltages that are over its 5.5v limit. Any chance it would still reduce 5.5v down to 5v, and then start adding any additional voltage over the 5.5v limit? In other words: 5.75v would become (5v + .25v =) 5.25v, 6.0v would become (5v + .5v =) 5.5v, etc. That is probably just wishful thinking on my part.

If so, is there any way to peel off any incoming voltage over 5.5v before the boost/buck?

 

[Reloadron]

This thread originally began here. with a simple USB question. Your original question was to the effect of:

I have a circuit I would like to power by plugging it into any USB port I happen to come across. The USB standard specifies output as 5 vdc (+/-.25 vdc) at 1 amp, which is perfect. But many ports are significantly (+/-.5 to .75 vdc) above or below this. I would like to take whatever output I am given from any USB port and then alter that output to be as close to spec as possible.

So, in short, how would I take the power from any USB port and then groom it to 5 vdc (+/-.25 vdc) at 1 amp?

Every USB port I have had experience with always met the 5 Volt ATX form factor specification of 4.75 volts to 5.25 volts. You were specific in that thread which you left hanging with above and below voltage specification. Now suddenly you no longer care about low voltage? Why would you say:

That buck/boost is probably what I will use. Unfortunately, it only accepts up to 5.5v incoming voltage. Can you think of any way to deal with voltages from 5.5v up to about 6v (which would likely be the most I would ever encounter)?

Here is a simple solution. The ZUS60505 will give you an amp with an input between 4.5 and 9 volts and there are many, many similar DC / DC converters out there that will do the same including higher currents out. The problem is you do not seem to know what you want based on two threads and nobody here can accurately guess when the game plan keeps changing. You want a solution? How about what I suggested?

They cost about $30 USD. You want more current? The solution cost more and if you want less current the solution cost less, so what do you want?

Ron

 

[Jamdan]

Hi Ron,

I started this thread to pursue a different direction/solution than the other thread, which I stated in the first post. This particular thread was originally began to pursue a simpler, non-IC solution to a simpler problem (ignoring low voltage). Through the posts of others and myself, it has slowly morphed to include elements of the other thread and possible solutions for both (similar) problems. This is mainly because most of the people posting on this thread also commented on the other thread as well, so they are, naturally, referencing it.

I do know what the problem is, and I do know what I want. But I am also willing to consider multiple avenues of getting there, or getting "close enough to there" if the trade-offs are acceptable.

I admit this has turned out a bit sloppy for forum etiquette, but we have gotten here honestly. I personally consider both threads as still open, and I continue to check both and comment when someone else leaves a comment on either, but if the moderators would like to merge the two threads I would understand and not object.

Thanks for your input. Your solution fits the voltage range well, but the price is 20 times that of the TPS63002 (which can handle up to 1.2A, which is plenty).

 

[OlPhart]

an unconventional aspect

Howdy, I thought this was a good idea, I use it and... it Does work (and well).
Use a negative voltage regulator as precision subtraction,
I've described it before: Most positive & Most negative terminals change places:
Result: regulated voltage is from rated span to input, Not to reference

To your need: use a -3.3V reg: it'll give you a 1.7V offset to whatever the input is, from 5V.

.... Use it, But Please attribute it. I'd never heard, but thought of it, and still think it's special... <<<)))