I want to power an 3.2V LED (5730 SMD LED) as draws 150mA. I'm going to use one 18650 battery without built in protection circuit.
Can i just place an 2.5V Zener diode between battery and the LED?
Zener diode will "shut down" with voltage decreases under 2.5v?
I'm going to use an buck/boost converter.
Draw the schematic on a sheet of blank printer paper, take a photo and post it. (include the converter in the drawing)
From your description I would say no.
Why do you want to do this ?
I can do that later.
But i meant 2v led in my first post.
I do not want led to discharge my 18650. Therefore in my head i think an 2.5v zener diode cu5s the power when the voltage to,led goes under 2.5v. ?
Couldn't answer that without a schematic showing how it's connected.
As you can see, that's not how zener diodes are usually used
zener diode
The zener voltage is usually tapped off the cathode. You're talking about putting it in series. (so there would be
no zener voltage and I believe it would still conduct but maybe not the way you think)
Putting a zenr in series should drop the zener voltage across the zener resulting in a lower voltage at the anode. Is that what you want ?
Do not do that! You need a resistor in the series circuit to
limit the current that will flow.
Do not do that! You need a resistor in the series circuit to
limit the current that will flow.
This is why I keep asking for a schematic.
Zener diode will "shut down" with voltage decreases under 2.5v?
I don't think so.
If you don't, then you should NOT power the led
without monitoring the battery voltage frequently until you learn the average battery life until it reaches the minimum operating voltage.
I do not care about the money. The charge controller mentioned above costs more than the battery it self. It,s not an gold coin, it's a... battery.
But what happend if i tok one 18650 battery without built in sircuit protection, solder one wire to + and -, and solder and 2V red led with 20mA in other end. Before the led i also solder in an let's say 2.8V zener diode. (And off course an resistor for limit the current.
Wil the diode stop the power if the battery goes under 2.8V?
Wil the diode stop the power if the battery goes under 2.8V?
I don't think so. It will drop 2.8V across the diode so the led will see 4V (18650 minimum charged voltage) -2.8V =1.2V.
If connected to DC in forward bias, it will have an approx. voltage drop of 0.6v, where resulting voltage will be approx. (supply voltage - 0.6).
If connected to DC in reverse bias, it will not conduct until breakdown voltage is reached, where resulting voltage will be approx. (Supply Voltage - Vz), where Vz is the diode’s breakdown (avalanche) voltage.
In your case "Supply Voltage=18650 voltage (4.0 to 4.3V) (3.7V is NOT the operating voltage, it is the nominal voltage, or basically , the minimum voltage BEFORE charging)
In your example, Vz=2.8V
Thus, (Supply Voltage - Vz) = 4V-2.8V=1.2V.
I want to power an 3.2V LED (5730 SMD LED) as draws 150mA.
Now, it the current limiting resistor is chosen for R = (Supply Voltage-V)f/ILed(max) =0.8V/ 0.150A =5.3 ohms (round up to 5.6 or round down to 5.1)
Thus, (Supply Voltage - Vz) = 4V-2.8V=1.2V .
Since 1.2V < Vf , the led will NOT conduct or turn on.
Okay, but can someone show me a schematic that protects my 18650 battery againt over discharge with use of an zener diode in the simpliest way?
Who said there is ?
I'd like to see it.
Where did you get this idea ?
Is that smoke I can see??
Man, you have good vision
Just remember it all runs on smoke. Let the smoke out and it stops.
I'm here to learn, and asks questions. I'm an beginner.
And i do not care about letting out smoke. It's like that i learn. But i do not have time to sit in my workshop all day, and therefore i asks questions, read the answers, and try things out when i finally got to my workshop.
I know this is not rocket sience for you guys. But for me it kinda is. For some people i can see this is an very hard thing to understand.
It's just an handfull people in here that actually explain things in the way i understand it. larryd is one of them.
You guys now what you talk about. No doubt about that. But to explain the knowledge in a way that a beginner understands it it's not everybody strongest side.
Of course, it dosent help that my technical english is bad either.
I'm understand the basic of electronic (generally), but sometimes i'm got stucked. Hard for me, walk in the park for people in here.
And yes, i know i can buy ready made modules that protects my battery and all that. But i also want to learn how thing works. Isn't that what Arduino is all about?
jremington:
Battery protection is not an Arduino problem. It is a common sense problem.
I'm posting in "Generall electronic" It is not allowed to asks questions about that in here?
It is a common sense problem.
Common sense for you, not for me.
If you are serious about learning, then why didn't you read Reply #9 ?
What does this mean ?
"If connected to DC in reverse bias, it will not conduct until breakdown voltage is reached, where resulting voltage will be approx. (Supply Voltage - Vz), where Vz is the diode's breakdown (avalanche) voltage."
raschemmel:
If you are serious about learning, then why didn't you read Reply #9 ?What does this mean ?
"If connected to DC in reverse bias, it will not conduct until breakdown voltage is reached, where resulting voltage will be approx. (Supply Voltage - Vz), where Vz is the diode's breakdown (avalanche) voltage."
I replyed in post #10 when saying "Okay". That means that i have read it.
I also aksed if someone could give me an simple example circuit when a Zener diode being used. After that i just get replies about smoke, how bad vison i have, question that ask where i got the idea from etc.
The answer to your question is in that paragraph.
There are only two possibilities.
No. I can't for the above reasons.
If it's reverse biased, it drops the battery voltage
by the zener voltage. If the remaining voltage is
1.2V and the led is a 3.2V led, will it turn on ?
You may not have to worry about your battery because the led isn't going to turn on. So tell me
how a zener diode protects a led that can't turn on because the available voltagevis 2V less than the
minimum led voltage ?
I don't have the answer. I told you all I can about that. You're on your own now. I got nothing.