Hello,
I have a question about parallel connections. In the picture, I have two LEDs. But the red LED has a forward voltage of 2V, and the blue one has a forward voltage of 3V. If I connect both in parallel to a 12V battery, what voltage will I measure at both LEDs? I read online that the voltage is divided equally. Does that apply? And does that also apply to resistors?
Thank you!
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You probably got that wrong. The voltage is NOT divided into parallel circuits (but the current intensity when this is limited). The voltage is divided into series circuits because the voltage is the energy per charge and it has only one way!
However, a simple series connection is not sufficient, because the LEDs together use only 5 V.
Even if you bring down the voltage, you need a protective resistance that is standard for LEDs anyway and usually limited to about 20 mA! At 20 mA and 12 V a 350 Ω resistance would be required!
Thanks for the detailed answer. Yeah, I’m just standing on the hose. In the series connection, I imagine that each component (LED, resistance.) Energy required. And if each LED has to “adjust” one after the other, each one gets a share, but if I measure between the minus of the battery and the end of the LEDs in the series connection, we get the voltage more and more the more I go in the row, because every energy takes off. In parallel, I don’t understand that yet. That I can divide up, but why do you not measure 2V and 3V before and after the LEDs as in a series connection (once assumed the battery would have 5V)?
Sry, but with your sketch you might be 10, because you already have signs in the 5th/6. Class has AND there are own symnoles for LEDs.
In a parallel circuit, the partial voltages are the same. The voltage is measured as energy difference PRE and HINTER to the component (actually the potential difference). However, PRE-&HINTER is the same in a parallel connection with all components! The voltage is determined by the voltage source! The voltage on the components is ‘required’ (and is not ‘supplied’!) because they do not run properly at too little voltage and break quickly at too much voltage.
May I add a little bit to the answer? If the LEDs both have a flow of 20 mA, from which we start, the tactics with the series connection is incredibly much better!
On the left we have a resistance in front of the red LED, 10 volts drop at 0.02 ampere. these are 0.2 watt or 200 milliwatts of power loss!
In addition, we have a resistance in front of the blue LED at which 9 volts drop, which would then be 0.18 watt or 180 milliwatts, thus making 380 milliwatts, which are heated in a senseless manner.
Let’s compare this to the right circuit! only 7 volts drop at a resistor. 0.02 Ampre are 0.14 Watt, i.e. 140 milliwatts. these are 240 milliwaatts about a quarter watt less than on the left.
Doesn’t sound much, but if we operate the circuit on the battery, the reduced current extends the life of the battery to the double!
The circuit doesn’t work!
You put the cathodes of the LED on the plus pole of the battery, so the diode paths are connected in the blocking direction. It’s dark…
that will not work, both LEDs will burn
I’m just talking about principle.
you have on both LED 12 V, both burn through
Thanks for the info.
With almost empty 1.5 V round cells, I could also operate red and green LEDs. However, the rated voltage is somewhat higher, while the operating voltage of old batches can also be lower. With very fresh battys, I had bad luck and the same LEDs have been burned! Standard are therefore protection/prevents!
Something surprises with your 1.2 V: Theoretically, optical LEDs should be designed more than 1.6 V, because this also corresponds to the photon energy of red light! For 1.2 V, if IR LEDs are rather designed.
But as I said…for me, red/green LEDs glow from 1.2 V! They are only designed for more! But only with resistance.
In my Comments was about whether LEDs must in principle be operated with pre-resistance, even if the voltage is tuned to its desired value, in my case 1.2 V. That was new to me, and I wanted to get some more information that I got.
the basic problem (the actual question) were not 1.2V but 12 V
which results in a difference by factor 10
Thank you for your detailed comment. I had operated (in the context of student experiments) LEDs on laboratory power supplies or NiCd Accus and had no problems. Maybe I just had luck?
The problem with LEDs is that, with an extremely low voltage change, they already undergo a high current change. These have no linear resistance characteristic. The values also vary due to production. If an LED is very close to the Tolleranz at the limit value, even at exactly 1.2V a higher If can flow, as eig permissible. In the case of LEDs, the current is regulated by an additional protective resistor, in order to avoid precisely this case. It can therefore happen that an LED is overloaded even at 1.2V and breaks down significantly faster. Too high power at LEDs is safe death. Just under the admissible If it lasts forever. If you look at the STrom voltage characteristics of LEDs, you quickly realize that a Δ0,1V in the nominal range already has a great influence on the flowing current, in some cases almost double.
Could be anything… you can be lucky…
So, I have 1.2 V LEDs that I easily operate on 1.2V Accus.
Only if the current is limited to 20 mA. Normally you take a little more AND a protective resistance!
I cannot operate a 1.2 V LED on 1.2 V?
Also at 4 V both run through because you only use LEDs with protective resistance! You can conduct a lot more current at the operating voltage than you can use yourself! Without resistance, they usually burn immediately, or very easily if you don’t just take an almost empty battery!
And at 4 V operating voltage?
Between plus and minus, you always have 12 volts here, as you draw this. that’s just like on the mehfachplug strip. you also have 230 volts on each socket.
In your sketch you would drop over both LED 12V
You would measure 12V on both LEDs. And the LEDs would be scraped in a fraction of a second because they burn through.
If you turn on it so directly, you will measure exactly 12V and the LEDs are over.