Author Topic: Help understanding the reference voltage on a LM3914  (Read 4086 times)

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Offline biohazrdTopic starter

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Help understanding the reference voltage on a LM3914
« on: August 03, 2015, 05:51:42 am »
So I just watched #204 and saw Dave using the built in 1.25V reference to run the Vhigh and Vlow for the expanded range configuration on a LM3914.

I'm a little confused as to how he is getting 6.4V and 3V out of the reference, however.

As per the data sheet the Iadj sources ~75uA (I think, I don't have the video up right this second), we want the resistor to drop 3V (one side is on ground) so we can solve for R using Ohm's law and set our Vlow to 3V. But what about the high?

Am I even right about this part?
 

Offline nugglix

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Re: Help understanding the reference voltage on a LM3914
« Reply #1 on: August 03, 2015, 06:05:07 am »
Hi!

I found this picture very helpful.

In short:
There is 1.2V between REFOUT and REFADJ.
REFADJ doesn't accept current (it sources some), so Ohms Law rulez! (again). ;)


 +--REFOUT (2.65V) (Supply at least 1.5V higher)
 |
1k2        (1.2V difference)
 |
 +--REFADJ
 |
 +--REFHI (1.45V)
 |
450R
 |
 +--REFLO (1V)
 |
1k
 |
GND


Hope it helps!

* Stolen from:
https://www.mikrocontroller.net/topic/270457#2829728
Translation by me.

Cheers
 

Offline biohazrdTopic starter

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Re: Help understanding the reference voltage on a LM3914
« Reply #2 on: August 03, 2015, 11:45:21 pm »
So because the REFadj has a constant current, the voltage can be set to whatever by using a certain resistance between it and ground. Where do you get the value for that 1k2 resistor between REFout and REFadj?
 

Offline T3sl4co1l

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Re: Help understanding the reference voltage on a LM3914
« Reply #3 on: August 03, 2015, 11:58:01 pm »
The ref works exactly like an LM317.  The resistance ratio of the divider is determined by the desired output voltage, while the value of the resistors (how much resistance to begin with) is determined by acceptable error due to bias current (maximum resistance), and maximum load on the VREF supply (minimum resistance).

As I recall, the bias current is fairly high, relative to the capability of the output, so it is desirable to add a trimpot for maximum accuracy.  Or dump it entirely and use a good one (e.g., TL431, LM4041, etc.).

Tim
Seven Transistor Labs, LLC
Electronic design, from concept to prototype.
Bringing a project to life?  Send me a message!
 

Offline biohazrdTopic starter

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Re: Help understanding the reference voltage on a LM3914
« Reply #4 on: August 04, 2015, 12:25:14 am »
Sorry, I really don't understand what you mean. How is using a voltage divider going to increase the voltage? If the output is 1.25v in relation to the adjustment pin, how do I get it up to a higher voltage?

I was thinking that since the output is 1.25v above the adjustment (which is floating and not grounded), and the adjustment always sources a constant current, I could use a resistor to ground to "set" the adjustment pin to a particular voltage using Ohms law. Since the current is constant, and I want, for example, 5v, I could choose a resistor that would satisfy 5v = 75 uA * R, thus setting the reference adjustment pin at 5v, and the Reference output at 6.25v (all in relation to the common ground).

Then I could use those voltages with a divider to set the high and lower bounds for the LEDs
« Last Edit: August 04, 2015, 12:29:57 am by biohazrd »
 

Offline rs20

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Re: Help understanding the reference voltage on a LM3914
« Reply #5 on: August 04, 2015, 12:42:41 am »
REFADJ has a small (notionally zero) current. The regulator will maintain REFOUT at 1.2 volts above REFADJ, just like an LM317.  So, the 1k2 resistor will have 1.2 volts across it. Ohms law says that this means that constant 1mA current will be flowing through the 1k2 resistor. Since no current goes into the REFADJ pin, it must go through the other resistors below. So, in that way, the REFOUT, REFADJ pins and the 1k2 resistor combine to give you a 1mA constant current source. You can then make any voltages you want by connecting resistors to that current source.
 

Offline biohazrdTopic starter

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Re: Help understanding the reference voltage on a LM3914
« Reply #6 on: August 04, 2015, 01:04:37 am »
Okay, I think I get it.

So rather than use any resistors below, I could do this

 +--REFOUT
 |
1k2       
 |
 +--REFADJ
 |
GND

Which would put REFadj at 0v and REFout at 1.2

but by putting some kind of resistor in there

 +--REFOUT
 |
1k2       
 |
 +--REFADJ
 |
 |
10k
 |
 |
GND

and since I know the current (1mA) and know the resistance (10k), I can calculate that REFadj is at 10v, and REFout at 11.2v. Assuming of course, that the power supply is at least 11.2v (or higher since this isn't rail to rail).
 


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