Doubts about voltage-dropping resistors and LEDs
An USB programmer for PIC, dsPIC, AVR, serial memories
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Sirs, explain to me, stupid, how can LEDs (D1-D2) work in this programmer circuit with such a high resistance of voltage-dropping resistors R8-R9? After all, with a normal typical current of 20 mA LEDs, a voltage of as much as 44 volts should drop on each voltage-dropping resistor with a resistance of 2.2 kOhm, according to Ohm's law! (V=IR=0.02A2200ohm=44v) Where does such a high voltage come from in this circuit? In my opinion, the outputs of RA1-RA2 cannot have a voltage of more than 5 volts, which means that, for example, for a white or blue LED (on which the voltage usually drops 3.2 volts at a current of 20 mA), 1.8 volts must be dropped on the voltage-dropping resistor, and if Mr. Ohm was right, then this resistor musts have a resistance of 90 ohms (R=1.8v/0.02A=90ohm). Yes, I understand that in Mr. Ohm's time there were no electronic programmers yet, so Ohm's law may not apply to their circuits. However, putting aside the jokes, I would like to know where the error in my reasoning lies?
Sorry for my bad English - it's not me, but the electronic translator to blame :)
What tells you that the current is 20mA?
I=(V-Vled)/R=(5-1.8)/2.2k=1.45mA
The passport data of the LED, as well as those measured by me in a practical scheme: The current is 20mA with a voltage drop of 3.2 volts.
Any LED is a bipole with a certain IV curve, like this:
As you can see it keeps a significant forward voltage even at 1.5mA.
Grazie mille per la spiegazione così visiva. I dati che hai fornito corrispondono molto bene ai risultati che ho ricevuto: ho misurato specificamente la corrente attraverso alcuni LED che ho a varie tensioni di alimentazione - e si è scoperto che molti LED iniziano a brillare già a una corrente di 0,8 mA! Ora tutto è diventato chiaro per me. Grazie ancora per aver risposto alla mia domanda.