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[SOLVED] 7 Segment Led Equation oddity.

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ShinedBrass

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Hello all,

I am working on a project involving 7 segment LEDs but have a little problem with a calculation. Normally when dealing with one LED I would use.
Input Voltage Less Led voltage * forward current and it works great but with this particular beasty I end up with this:

Iv = 3.3v
Lv = 3.3v
fc = 10ma

(Iv - Lv) * fc would be (3.3 - 3.3) = 0 * 10ma = 0

So my question is do i use a resistor with this Led or just jam in a bit of wire, or is it the case that when the voltages negate you would add the current value as ohms ?
 

If you're trying to calculate a resistor value for a 10mA current, you're multiplying where you should be dividing. Try R=(Vs-Vf)/I where Vs is the supply voltage, Vf is the nominal forward voltage of the LED and I is the design current. Even if your supply voltage is barely enough to drive the LED, you should have at least a few ohms of resistance to prevent overcurrent since the LED will draw more current as it heats up. You need more voltage (headroom) if you intend to control the current in the LED to some precise level.
 
i can' t understand your question..... when you use the 7 segment Led display better you connect resister...

---------- Post added at 11:55 ---------- Previous post was at 11:51 ----------

if you use 3.3v input voltage you can use a 330ohm resistor. if you change the resistance value the output brightness of the led will be change
 

ShinedBrass said:
Iv = 3.3v
Lv = 3.3v
fc = 10ma
Click to expand...

Are you saying that your power supply is 3.3V and the nominal forward voltage of the LED is also 3.3V? Please post the display's datasheet. Normally a 330Ohm resistor (but for 5V supply) is a reasonable value as Dinuwilson suggested.
 
Last edited:

KJ6EAD said:
If you're trying to calculate a resistor value for a 10mA current, you're multiplying where you should be dividing. Try R=(Vs-Vf)/I where Vs is the supply voltage, Vf is the nominal forward voltage of the LED and I is the design current. Even if your supply voltage is barely enough to drive the LED, you should have at least a few ohms of resistance to prevent overcurrent since the LED will draw more current as it heats up. You need more voltage (headroom) if you intend to control the current in the LED to some precise level.
Click to expand...

Now I'm even more confused R=(Vs-Vf)/I

Vs=3.3v
Vf=3.3v
I=10ma

the value for R would always be Error

(3.3-3.3)/10 does not work as you can not divide 0 by 10.

:???:
 

what output you want?
 

ShinedBrass said:
(3.3-3.3)/10 does not work as you can not divide 0 by 10.
Click to expand...

As already requested, please post the datasheet of the display. I find it a bit difficult to have LED's forward voltage = 3.3V.
 

ShinedBrass said:
Now I'm even more confused R=(Vs-Vf)/I

Vs=3.3v
Vf=3.3v
I=10ma

the value for R would always be Error

(3.3-3.3)/10 does not work as you can not divide 0 by 10.

:???:
Click to expand...

10 means 10A, for 10mA you should divide with 0.01 but your voltages are strange, do you have a 3.3v power supply and 3.3v led?

Alex

P.S. As an additional comment you can divide 0 with any number and the result will be 0.
 
Last edited:

alexan_e said:
10 means 10A, for 10mA you should divide with 0.01 but your voltages are strange, do you have a 3.3v power supply and 3.3v led?

Alex

P.S. As an additional comment you can divide 0 with any number and the result will be 0.
Click to expand...

O.K So my math was a bit off and I cant find the datasheet atm accept for the mechanical specifications ; however the question still stands.

If my input voltage is 3.3v and my led forward voltage is 3.3v and forward current is 10ma what resistor do I need, I'm hoping this time I can get an answer and not a circle.

If the answer is 0 then how do I have a 0 ohm resistor other than chucking in a hunk of wire or is it a case that I do something like add 10ohms just to protect things.

Normaly I have no problems doing this stuff setting up simple led circuits but in a case where I get 0 ohms as my result what do i do?
 

ShinedBrass said:
If my input voltage is 3.3v and my led forward voltage is 3.3v and forward current is 10ma what resistor do I need, I'm hoping this time I can get an answer and not a circle.
Click to expand...

Get another LED!:grin:
 
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ShinedBrass said:
If my input voltage is 3.3V and my led forward voltage is 3.3V and forward current is 10ma, what resistor do I need? I'm hoping this time I can get an answer and not a circle.
Click to expand...
Get another power supply.

images
images
 

KJ6EAD said:
Get another power supply.

So are you saying that its not possible to drive a 3.3 volt LED with a 3.3v power supply, does there always have to be a voltage drop in order for the calculation to work or and I'm not being rude so please do not take it that way do you simply not know?
Click to expand...
 

The calculation will work but the resistor value will be zero. The thing you probably aren't understanding is that an LED is not a static resistance. The forward voltage drop of the LED as specified in any data you have will vary even among LEDs from the same production lot. It will also vary with temperature and this is significant not just because of the ambient temperature but because when current starts flowing through the LED, it will self heat which reduces the effective resistance of the LED allowing greater current to flow, further increasing the heat, etc. in a self-perpetuating condition known as thermal runaway. This can quickly destroy the LED. You either need more voltage or a lower voltage LED or you need to accept the possibility of destroying the LED or having little or no control over it's current or brightness. You can throw a 10Ω resistor on it for a little insurance but then the LED may be too dim.
 
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