Create a +5.00V source from a +5.15V power supply?

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po210

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Hello.
I´ve a 5.15V power supply (in fact,is my laptop USB port V+) and I´d like to create a +5.00V supply from it.I can´t use a LM317 regulator because it needs some volts of input/output difference.I´d like the source to provide 100mA maximum (or so).
Is there an easy way to get this? (I have no 5V Zener diodes)
 
Why?
5V Zener will not work!
The next USB computer you use will not be 5.00V. It might be 4.91 volts.
Use a longer cable and the voltage will drop with load.
If you really need 5.00 volts, (knowing the USB voltage could be 4.5 to 5.25 volts) then you are looking at a boost buck supply.
When you say 5.00V what do you really need? 5V+/- 1%, 2.5%, 0.1%????
 
Thanks for the answer.
In fact,I´m measuring temperature with a 12bit ADC,and I´m afraid my max sample value 4095,could depend on the V+ supply value.
 
What is the hardware device? I use devices like this one for ADC all the time running off USB on my laptop. Devices like this have internal references for the ADC. Just a matter of what the external hardware device is.

Ron
 
That linked unit is right along the lines of this type unit. It will work fine off any USB ports voltage of about 5 volts. Any USB or even a wall wart will power it. Measuring temperature is just a matter of signal conditioning your temperature sensor output and scaling it to 0 to 5 volts.

It will work just fine.

Ron
 
Yes I know it works that way.But I´m not sure if 4096 level is exactly +5.00V or the voltage I measure after connecting to my laptop,which is 4.77V in regulated output and 5.13 or so in unregulated output.
So,I think i can make a 5.00V from those 5.13v at my USB and feed with that both the uChamaleon and my circuit.
But how can i do 5.00V from 5.13V (as easy as possible)?
 
If you have the uChamaleon sitting there try it. Feed a known voltage into one of the analog channel inputs. Install the driver disk.

From the manual:


Use the test software. If you are really curious about the bit count apply a known 5 volts and write a script so you can read that channel and see what you get. The unit does not need to be powered by a precision 5.000 volts.

There is nothing (nothing simple) that will convert your 5.1 volt USB power to a perfect 5 volts. According to USB specifications Supplied voltage by a host or a powered hub ports is between 4.75 V and 5.25 V. Your device is designed to work from the source. If I supply the device with 4.8 volts or 5.2 volts and apply 5.00 volts to an analog input and read the bits for that channel I will see 4095. The USB input voltage has nothing to do with the Vref of the A/D as long as the input power is in normal USB specifications. Have you read the manual?


The unit is designed to work from a USB port and as long as that port meets USB specifications it will work just fine. You do not need a precise 5 volts for the power source.

Ron
 
the normal way of doing things would be to get a very good reference. measure the reference and calculate a supply gain error factor
 
Yes,i read the manual and that why i bought the thing.What it says is

<<or to get an accurate 5 Volts level for some analog applications, the μChameleon has an internal linear regulator>>

but if i measure 4,77 in the output marked as "regulated" i don't see a 5volt level.
If i connect an analog input to that line, I get values from 4085 to 4095.
So,I´m going to "sacrify" an input line to measure a 2.5V reference voltage derived from USB source,and then scale my measurement appropiately.
 
I admit I just don't get it. Devices like this are pretty common and I have never seen one behave that way. The attached is the open view of your unit and most units like this all look the same. It looks like USB power is directly fed to the main chip and the external DC is run through a common regulator chip. The main chip should take care of its ADC reference.

Seeing some bounce in the straight A/D count is not unusual. There will always be some noise so even if you apply a precise 5 volts you will not always see the exact expected count. Generally in devices like this I right justify the data a few bits to get reliable stable readings. Doing the math in my code.

Several years ago I did an experiment to measure several computer case temperatures using a real simple 4 channel A/D device. The results can be seen here. I actually calibrated the thing.

I even went to the forums for your device but did not see mention of anything of value. The USB voltage on the system I was messing with was about 4.8 volts and no problem with similar devices. Matter of fact, thinking about it the one in the link was real old as I used RS232 port for communication.

I just do not see, using that device, why it isn't working for you. Beats me.

Ron
 
<<or to get an accurate 5 Volts level for some analog applications, the μChameleon has an internal linear regulator>>

The Chameleon does not regulate the USB power. It says in the manual:

Simply connect a standard wall-mount transformer, with a typical output
voltage between 9Volts and 12Volts to the black connector beside the usb
connector. The external voltage will be internally regulated to a clean
5Volts, and the μChameleon will switch from USB power automatically.
 
Hi Mr T

He wants to run off a laptop. If the analog inputs need a precision 5 volts for the thing to work off a USB port that sort of sucks.

Ron
 
Thank you very much for your opinion and experience.
So, 4095 should be exactly 5.00V and i must not worry.
What confused me is that if I measure 4.77V in an output marked as "regulated" it could be,i though,the one used to reference the samples. This phrase "The external voltage will be internally regulated to a clean 5Volts" is not true, if that regulated voltage is the one i measured,but maybe that voltage has nothing to do with the one used to reference the ADC samples.thas was my reasoning line,but surely,the real reference is inside the chip and i must not worry.(although then, what are those 4.77V for? )
 
This is a guess but when they say "The external voltage will be internally regulated to a clean 5Volts" I believe they refer to if you apply an external DC voltage to the DC PWR connector. I am guessing IC2 in the picture to be a 5 volt regulator. IC2 only comes into actual play when an external DC voltage is applied through that DC PWR connector. Diodes D1 and D2 look to be polarity protection diodes. Again, I am guessing but it looks that way.

Next, I seriously doubt the actual uC is using that voltage as a Vref. For example one similar little device I have been playing with is USB powered and has an analog input range of -10 to +10 volts. The lower end units have a fixed analog input range but the slightly better units allow the use of code to change the input ranges.

My guess looking at the image is those 2 output connectors are used to provide 5 volt power to any other small devices you may wish to power. That was based on what I did read. However, the actual manual really doesn't make mention of them. I would venture if you for example externally powered the device and as a test just used a 9 volt battery you would see about 5 volts on those terminals. Simply because the 9 volts would exceed the dropout voltage of IC2 which I guess to be the regulator. The reason for the 4.77 when you USB power with just over 5 volts would be loss. Again, a best guess case.

Overall, I would not worry about it. I would feed the thing a few known voltages and see what I read. Also, remember what I mentioned earlier as to noise so do not expect to see a dead stable count if you look at the bits.

Ron
 
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