Driving LED

[FreshmanNewbie]

I have an RJ45 connector connected to a PHY output.

This PHY output actually drives an LED which is internal to the RJ45 connector.

But that LED is not visible to the end user due to mechanical constraints.

I want the same GPIO output from the PHY to drive another LED that is discrete (not present in the RJ45 connector).

Can I connect an LED on the same line (in parallel) which also goes to the LED that's present in the RJ45 connector?

If not, can someone help with an alternate solution?

 

[D.A.(Tony)Stewart]

No it is not OK to assume it will work until you figure out the V,I R requirements. The LED with the lowest Vf may illuminate, but what is driving it?

 

[KlausST]

Hi,

you´re not an electronics newbie anymore.

You already know that electronics is all about V and I.

Now we have three devices involved:
* PHY output: What is it´s output specification?
* LED in connector: What is it´s specification?
* your "external" LED: What is it´s specification?

Every one who wants to design this circuit needs these values.

* If you want to be sure, there is no way around.
* but if you want trial and error: then you don´t need us.

Thus my recommendation:
* collect the informations
* do your own calculations
* in case you´re in doubt: Show your calculations and all the specifications.

Klaus

 

[FreshmanNewbie]

Hi,

you´re not an electronics newbie anymore.

You already know that electronics is all about V and I.

Now we have three devices involved:
* PHY output: What is it´s output specification?
* LED in connector: What is it´s specification?
* your "external" LED: What is it´s specification?

Every one who wants to design this circuit needs these values.

* If you want to be sure, there is no way around.
* but if you want trial and error: then you don´t need us.

Thus my recommendation:
* collect the informations
* do your own calculations
* in case you´re in doubt: Show your calculations and all the specifications.

Klaus

The PHY output spec is :

Voh = 3.3V-0.2V = 3.1V
(Ioh) = -12mA

LED Series resistor is 2.2k.

Hence, the LED current = 3.1V / 2.2k = 1.4mA.

And the LED block spec is

Max Vf of LED = 2.8V & Iled = 20mA.

But since the I(oh) = -12mA @ 3.1V.
But the max Vf of the LED is 2.8V. Since, 2.8V < 3.1V, there will be more current (higher than 12mA) from the PHY out pin which will reduce the voltage from 3.1V to 2.8V. Hence, the LED in the LED Block anode will glow with lesser brightness since the required ILED will not be exactly 20mA, but lesser than 20mA. Am I right?

Hence, I think the PHY output should be OK for the individual LED as well as the LED anode block, right?

 

[KlausST]

Hi,

I see you are not aware about the real situation.

Do yourself a favour and draw a realistic schematic of your circuit. All your circuit. Hand drawn is O.K..

Klaus

 

[danadakk]

Quick answer, trial, put a 1K R in series with LED and try it.

I concur its best to know what the PHY driving capability is and all the specs of it and LEDs
involved, but just give it a try. If you use a high efficiency type LED,. only takes low current
to get good brightness,.

LED color matters :

Regards, Dana.

 

[D.A.(Tony)Stewart]

The PHY output spec is :

Voh = 3.3V-0.2V = 3.1V
(Ioh) = -12mA

LED Series resistor is 2.2k.

Hence, the LED current = 3.1V / 2.2k = 1.4mA.

And the LED block spec is

Max Vf of LED = 2.8V & Iled = 20mA.

But since the I(oh) = -12mA @ 3.1V.
But the max Vf of the LED is 2.8V. Since, 2.8V < 3.1V, there will be more current (higher than 12mA) from the PHY out pin which will reduce the voltage from 3.1V to 2.8V. Hence, the LED in the LED Block anode will glow with lesser brightness since the required ILED will not be exactly 20mA, but lesser than 20mA. Am I right?

Hence, I think the PHY output should be OK for the individual LED as well as the LED anode block, right?

Your analysis is wrong.

You indicated the source is 3.3 has resistance that drops -0.2V/-12mA = 16.7 ohms .
The LED needs 2.8V to draw 20mA but if 2,2k is added in series between 3.3V source and 2.8V LED then the current me be 3.3-2.8= 0.5V divided by 2.2k = 0.23 mA but then the LED Vf will drop some 10% .

Due to low voltage source, in order to draw 20 mA you need 3.1V-2.8V = 0.3V/20 mA = 15 ohms, but with the large tolerances on both the source and Vf, this will not be accurate.

I suggest you choose 150 Ohms and be happy with 2 mA.