Need help finding the current in a basic circuit - Page 1
Forum section: electronics / beginners

author: Evy
author_rank: Contributor
timestamp: August 15, 2017, 11:32:38 am
content: Here is the circuit I need help finding a current in. I am seraching for the current through RL.RL= 1kVCC= 12VVEE= -12VI2= 10mAR1+R2>> RL(they are much bigger than RL)There is almost no current through R1and R2The voltage across RLis 12V, right? Then current must be 12V/1k= 12mA. And then there is of course the current source that is drawing 10mA current from the load.Is the total current through the load then 2mA?

author: ptricks
author_rank: Frequent Contributor
timestamp: August 15, 2017, 11:43:04 am
content: The current through the load doesn't change .RL is basically a load connected between VCC and ground, everything else in the schematic doesn't matter.

author: ironmonkey
author_rank: Contributor
timestamp: August 15, 2017, 11:44:28 am
content: Short answer: The current source is not drawing current from the load. Try to figure out this firstEnviado desde mi MotoG3 mediante Tapatalk

author: Evy
author_rank: Contributor
timestamp: August 15, 2017, 12:39:34 pm
content: Is it drawing it from the VCCthen? If that is the case then the current is 12mA, right?

author: Benta
author_rank: Super Contributor
timestamp: August 15, 2017, 02:02:49 pm
content: Yes.

author: Vtile
author_rank: Super Contributor
timestamp: August 15, 2017, 03:18:30 pm
content: ** In these kind of situations it is typically beneficial to redraw the circuit. Ie. for me it is many times hard to understand intuitively these electronic pull-up/pull-down style of drawings, not to mention trying to remember what the Vee,Vdd etc. did mean. Common rule through electrical drawings is that higher voltage potential is top of the drawing, but there is exceptions. The ideal sources can sink or source infinitely of current or voltage to satisfy the given voltage or current, in real life such things hardly exists and there is always some parallel or series components in them.Edit1.Ugh* Goofed up the image..Removed.** Edit2

author: Ratch
author_rank: Regular Contributor
timestamp: August 15, 2017, 07:05:01 pm
content: Evy,Obviously, Vcc is supplying 22 ma. 12 ma will be present in RL, and 10 ma in the current source.Ratch

author: Zero999
author_rank: Super Contributor
timestamp: August 15, 2017, 09:11:09 pm
content: Is this part of a larger schematic? If so, please post the rest of it. Otherwise, it doesn't make much sense. What's the purpose of I2?Is it a voltage regulator circuit, with VCCbeing the output voltage?

author: Benta
author_rank: Super Contributor
timestamp: August 15, 2017, 09:48:50 pm
content: QuoteI don't think so. It seems Evy is attending an electronics course and wants us to do his/her homework.That's my perception.

author: Zero999
author_rank: Super Contributor
timestamp: August 16, 2017, 11:09:41 am
content: Quote from: Benta on August 15, 2017, 09:48:50 pmHere most schools/colleges don't resume until September and presume that's the case for most of Europe, so there's still plenty of time to get it done.There's also a difference between providing help and support and doing all of the work.

author: Evy
author_rank: Contributor
timestamp: August 16, 2017, 02:08:56 pm
content: Quote from: Hero999 on August 15, 2017, 09:11:09 pmThis is part of a bigger schematic. This is what you get when you analyze the two scenarios of which the one is to assume that the transistor in the second stage is a short circuit fromthis thread. Didn't want to make this thread too long so I stripped the circuit down a bit and removed the unessential parts. Its not the same problem but similar and has other values. I had to calculate where the output clipped in an output stage. In the key of the book the author wrote that the current had to be infinite (when transistor is a short circuit) which I didn't understand and hence, created this topic. I can post the whole problem if someone is interested.I2 is the biasing source for the transistor in the second transistor stage.

author: Evy
author_rank: Contributor
timestamp: August 16, 2017, 02:21:20 pm
content: Quote from: Benta on August 15, 2017, 09:48:50 pmI am sorry that I, somehow, led you to believe that I wanted you to do all the work for me, which is not the case. I just want helpful tips and your valuable insight on how I could solve my problems when I get stuck.Also, these problems are not part of my homework. But even if these problems were part of my homework/ school assignment I would be allowed to post these in this forum as long as I provide enough information on how I tried to solve them.NEWBIES PLEASE READ BEFORE POSTINGQuote from: EEVblog

author: Zero999
author_rank: Super Contributor
timestamp: August 16, 2017, 02:22:23 pm
content: Quote from: Evy on August 16, 2017, 02:08:56 pmActually it would be much easier to follow and you'd get more relevant replies, if you kept it to one thread per topic/circuit, rather than splitting up into many. Creating too many threads, means people lose track of what's been said elsewhere and it stops one from seeing the bigger picture. This seems to be a common error by new people asking for help.Do all the circuits you've asked for help with recently, refer to the same book? Ideally it should all be in one big thread.

author: Evy
author_rank: Contributor
timestamp: August 16, 2017, 02:28:03 pm
content: Quote from: Hero999 on August 16, 2017, 02:22:23 pmYes. But this one is a different problem (but similar) and I don't need help with the same thing as the other thread. It would be hard for other people with same problem to get any valuable information in the future when they find the problem on this forum, if I had everything in one giant thread. Other, similar, forums usually have 1thread / problem.

author: Zero999
author_rank: Super Contributor
timestamp: August 16, 2017, 03:46:58 pm
content: Quote from: Evy on August 16, 2017, 02:28:03 pmI'm not aware of any such rule. All the questions you have are linked, therefore they should be kept together. The function of one part of that circuit depends on another part. It's important to understand how it all fits together.