Elektor, Quiz problem - Page 1
Forum section: electronics / beginners

author: MikeK
author_rank: Super Contributor
timestamp: April 20, 2012, 11:12:38 pm
content: In the Electronics for Starters (4) column in the April 2012 issue there is a quiz. But I think there is a mistake for question 1. I can't get even close to the desired current of 350mA with any of the resistor values. Are the answers missing a decimal place?

author: IanB
author_rank: Super Contributor
timestamp: April 20, 2012, 11:20:04 pm
content: Can you post a summary of the question?

author: MikeK
author_rank: Super Contributor
timestamp: April 20, 2012, 11:34:44 pm
content: I apologize if this is a copyright violation...

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: April 20, 2012, 11:41:01 pm
content: Yes - they missed a DP - should be 0.47 ohms for 350mA

author: IanB
author_rank: Super Contributor
timestamp: April 20, 2012, 11:42:01 pm
content: Yes, the question has an error in it.If each LED drops 3.4 V that would make 10.2 V. If you then pass 350 mA through a 22 ohm resistor you would add on 7.7 V giving a total of 17.9 V. That is higher than the battery voltage.I guess the resistor choices should be 2.2 ohms, 4.7 ohms or 10.0 ohms. (Or not. Maybe Mike knows best...)

author: IanB
author_rank: Super Contributor
timestamp: April 20, 2012, 11:47:27 pm
content: Because 350 mA through 2.2 ohms at Rx would produce 0.77 V at the base of the BC547B, turning it on. As that transistor turns on it will pull down the base of the BD135, turning it off and reducing the current. The current through the LEDs will therefore be limited below 350 mA.With that wild stab in the dark out of the way, someone tell me the correct answer?

author: MikeK
author_rank: Super Contributor
timestamp: April 20, 2012, 11:52:42 pm
content: Since the problem stated that the base-emitter voltage would be 0.7V, I took this and divided by 350mA to get 2ohms. I just want to know the correct method for solving.My other method was to take the worst case battery voltage given, 11V, subtract the LED drops, and divide by the 350mA to get 2.2ohms.

author: olsenn
author_rank: Frequent Contributor
timestamp: April 21, 2012, 12:02:09 am
content: Here's a quiz for you: I'm thinking of a number between 1 and 3; what is it?

author: TerminalJack505
author_rank: Super Contributor
timestamp: April 21, 2012, 12:04:12 am
content: I think your first method for solving is correct. That's what I used, anyway.That's on of the classic current source circuits.The voltage drop across Rx is fixed at 700mV. So Rx = 700mV / 350mA = 2 ohms.

author: olsenn
author_rank: Frequent Contributor
timestamp: April 21, 2012, 12:05:36 am
content: QuoteThe correct answer is a pronoun, but I would have also accepted e.

author: MikeK
author_rank: Super Contributor
timestamp: April 21, 2012, 02:01:20 am
content: I wonder how Elektor is going to handle it, then. Because that question is part of a quiz contest. Everybody loses!

author: amspire
author_rank: Super Contributor
timestamp: April 21, 2012, 02:37:12 am
content: Quote from: MikeK on April 21, 2012, 02:01:20 amI guess if you wanted a 2.2 ohm resistor but you had to choose between a 22 ohm, a 47 ohm and 100 ohm resistor, then 22 ohms is your best choice.Or even better, select all the options and put them in parallel to give you 13 ohms.