電流問題

whisky1314

我岩岩諗住試下個電流
分別做左3個唔同既情況

1)9V電,750R電阻,1粒食人魚LED　　　　　　得出結果:6.22(於DCA:20m制)
2)9V電,750R電阻,2粒食人魚LED(並聯)　　 　得出結果:6.27(於DCA:20m制)
2)9V電,750R電阻,2粒食人魚LED(串聯)         　得出結果:2.92(於DCA:20m制)

我諗落有D奇怪
於DCA:20m制時度到6.22,咁樣應該係話電流係6.22mA(其實呢電流我係唔多識睇,如果有錯請指正)

於CASE1度,得出電流係6.22
於CASE2度,得出電流係6.27(但係依家呢個CASE比CASE1多左粒LED,仲要係並聯,咁電流唔係應該係CASE1的兩倍嗎?)
於CASE2度,點解2粒LED既電流仲細過1粒LED

raymondchan338

The reason is that this circuit is an active circuit. The rule for passive circuit did not apply. You may consider the voltage across the LED is constant when the voltage reached a certain voltage. In this situation, either for 1 LED or 2 LED, the voltage across the LED is the same and as a result the total current is more or less the same.

lamtszhin0528

你試下Step2個陣....每粒LED用獨立750R 量一次  睇睇有無分別

whisky1314

回復 2# raymondchan338


    下..
因為並聯既關係
LED既P.D.一樣,呢點我係明既

但係總電流一樣
如果咁講既話

用一粒LED又係用咁多電?
用十粒LED又係用咁多電?
咁樣應該係唔岩嫁阿

應該用得愈多就愈耗電

whisky1314

回復 3# lamtszhin0528


    STEP2?姐係CASE2?

姐係分兩次度一粒LED既電流?

raymondchan338

回復 5# whisky1314

If you need to double the current when connecting 2 LEDs, you should connect a 750R resistor in series to each LED and connect to the 9V battery (as suggested by upcupcupc in 4#).

lamtszhin0528

xxxxx [--750R--LED---]
VCC---                     ----A Meter----Gnd  
        [--750R--LED---]


咁

Leopard

睇圖樓主case2好似係[750--(D1//D2)]?
想要current變成2倍的wrong concept不是電阻與二極管的問題
(750+R1//R1)怎=[(750+R1)/2] ?

whisky1314

回復 4# upcupcupc


   
如果咁講既咁
控制電流既只係電阻
而LED唔會控制到電流?

我係咁諗既
係CASE2
一粒LED要6.2mA
咁兩粒應該係要12.4mA