Best option to amplify current? ( From uA to mA) - Page 1
Forum section: electronics / beginners

author: RivaultUser
author_rank: Contributor
timestamp: January 08, 2013, 12:25:09 pm
content: Well, i've been trying to amplify uA current using darlington pairI need to get it to 3 - 5 V so it can make the relay worksThe input current will come from the magnet and coil that I've been working on ( Inductive electromagnet )The problem is, I'm limited with source & time. All I have is neodymium magnet & coil ( which I also need advices on it )I hope that I can create a current in uA range ( or is it only nA range???)And I need that current to make a 3V relay works,Any advice on the amplification?Here is my darlingtonhttp://imageshack.us/photo/my-images/547/dsc0272r.jpg/Do you think it will work?NOTE : don't pay attention to the previous comments, I made a mistake with my D pair before

author: firewalker
author_rank: Super Contributor
timestamp: January 08, 2013, 12:39:58 pm
content: Nope. Not even close.Alexander.

author: Kremmen
author_rank: Super Contributor
timestamp: January 08, 2013, 12:44:09 pm
content: You asked for it:http://www.ti.com/lit/an/slod006b/slod006b.pdfRead and understand - it is all you need for a long long time.

author: firewalker
author_rank: Super Contributor
timestamp: January 08, 2013, 12:49:09 pm
content: Quote from: Kremmen on January 08, 2013, 12:44:09 pmYep, excellent choice.Alexander.

author: IanB
author_rank: Super Contributor
timestamp: January 08, 2013, 02:17:03 pm
content: Why so complicated?If the need is to detect a current of a few uA and make it switch a relay then a simple transistor circuit is more than adequate. No need to go to op amps. If a single darlington transistor does not have enough gain then surely an extra gain stage would do it.To the OP: you have the right idea, but your circuit configuration is wrong. Search online for a few examples of using transistors to switch a relay with a small current. You should be able to discover how to make it work. The classic example is using a touch sensor with finger resistance as a switch. Using two or three transistors you can amplify the very small current conducted through the skin of your finger to switch on a light or a relay when you touch the switch pad.

author: Kremmen
author_rank: Super Contributor
timestamp: January 08, 2013, 02:24:15 pm
content: It could be argued that using an op amp is no more complicated than using a simple transistor in an essentially trivial case like this. Applying both is about equally complex or simple if yu are not really interested in the finer points of circuit theory.But putting that difference aside, i just answered the express question for links to op amp guidance. Mancini's text is second to none and in the intro there is also explanation of the simple transistor amplifier. That alone may do the trick + there is all of the rest as a bonus. Surely this is not the last circuit the OP will design...

author: Mechatrommer
author_rank: Super Contributor
timestamp: January 08, 2013, 06:03:52 pm
content: nice joke. the fun part is all consequtive transistor' bases are tied to the ground. i should have thought of this joke in the beginning.

author: IanB
author_rank: Super Contributor
timestamp: January 09, 2013, 03:32:38 am
content: Quote from: Kremmen on January 08, 2013, 02:24:15 pmYou did indeed, and I missed that question on first reading. Beg pardon.

author: RivaultUser
author_rank: Contributor
timestamp: January 09, 2013, 11:47:54 am
content: Quote from: Mechatrommer on January 08, 2013, 06:03:52 pmQuote from: IanB on January 09, 2013, 03:32:38 amQuote from: firewalker on January 08, 2013, 12:39:58 pmOh cmon mannn, this is not a joke.....I'm just a beginner that needs help !,I need to change the uA current to at least mA, and the method I know to amplify current is just to use Darlington Pair !!!Please, if you know how to do it please help me. I'm begging to you

author: baljemmett
author_rank: Unknown
timestamp: January 09, 2013, 12:25:38 pm
content: Quote from: RivaultUser on January 09, 2013, 11:47:54 amWell, use (or extend) a Darlington pair, then -- but pay attention to the details, as what you've drawn isn't going to work. You even quoted the post that points out why!

author: Mechatrommer
author_rank: Super Contributor
timestamp: January 09, 2013, 04:00:52 pm
content: Quote from: RivaultUser on January 09, 2013, 11:47:54 amsorry dude. 1st me myself is not very good at transistor design infact i still dont get it really dont get it, i still dont have rule of thumb for it, its been like 15years i'm tinkering with ee. recently i tried to make a logic circuit using bunch of it and my circuit went whacky i tossed the idea right away i just replaced them with relays which i've known for so long. 2nd i'm a teacher but the way you present your idea... i surrender, i surrender dude, i think you want to move too fast. do you have a "firm" idea of what V = IR is? what is ground what is potential to the ground? the mentioned obvious part is where your transistor's base tied to the ground, man i dont know transistor but at least i know what will happen when you connect the base to the ground. i suggest you start reading while tinkering at the same time, someone above has given you a very good reading material. go read it and have fun...

author: ftransform
author_rank: Frequent Contributor
timestamp: January 09, 2013, 04:19:36 pm
content: You can experiment using a high power amplifier too.

author: olsenn
author_rank: Frequent Contributor
timestamp: January 09, 2013, 05:11:13 pm
content: QuoteYour darlington pair should definately work; just use resistors to saturate the collector current where you want it to be. You can probably use a single stage common-emitter amplifier. If you're starting out with uA of current, do not bother placing a resistor at the base of the transistor (unless you're adding an input RC filter).It isn't required, but if I was designing this for TTL logic I would tie the output of the transistor(s) to an input terminal of an op-amp and tie the other input to a suitable voltage using a voltage divider. This will create a comparator. If V+ of the op-amp is 5V, and V- is 0V, it will keep the output at exactly one of these two values at all times. It also adds a buffer stage in case you want to add a filter or load something.

author: codeboy2k
author_rank: Super Contributor
timestamp: January 10, 2013, 10:50:46 pm
content: google a darlington pair and you'll see that the base(s) are not grounded, as you've drawn it.Also, you don't need so many stages. A single Darlington will work for 10's of uA.You can either use a packaged Darlington transistor, or build your own from parts you have lying about.Make sure the first transistor is higher gain than the second transistor in the pair. For example, a BC547B or C, which can have an Hfe above 400, 500 up to 800... and the second transistor should have higher current handling capability for driving the relay. If it's a small relay, than a 2N2222 is common, it can handle around 500ma. However, keep in mind that the actual choice of transistor is highly dependent on the the load relay you are using, how much current it draws, and what the voltage drops will be.As poster olsenn said, you will want to use resistors in the relay circuit, to ensure that the current is where you want it and the second transistor is saturated when it is on, so that you have very little power dissipation in the output transistor. When saturated, it is simply a switch and no power dissipation means no heat, and no need for a heat sink. This is important, because you have to make sure not to exceed the second transistor's power rating.Also, consider your voltage available, as well as the current. The Darlington pair needs above 1.4 volts to turn on (2 Vbe drops). If your source of current is also lower than 1.4 volts, then you should consider another approach, like a complementary pair (NPN-PNP) (needs 1 Vbe drop to turn on) or if your voltage signal is still lower than that, you will need a voltage gain stage first, using a common emitter amplifier.At this point, a single op-amp starts to look more promising. But you don't need an op-amp unless you also need the included voltage gain it provides.

author: RivaultUser
author_rank: Contributor
timestamp: January 12, 2013, 12:32:56 pm
content: Well, i've been trying to amplify uA current using darlington pairI need to get it to 3 - 5 V so it can make the relay worksThe input current will come from the magnet and coil that I've been working on ( Inductive electromagnet )The problem is, I'm limited with source & time. All I have is neodymium magnet & coil ( which I also need advices on it )I hope that I can create a current in uA range ( or is it only nA range???)And I need that current to make a 3V relay works,Any advice on the amplification?Here is my darlingtonhttp://imageshack.us/photo/my-images/547/dsc0272r.jpg/Do you think it will work?NOTE : don't pay attention to the previous comments, I made a mistake with my D pair before

author: codeboy2k
author_rank: Super Contributor
timestamp: January 12, 2013, 04:51:45 pm
content: without knowing anything about your magnet and coil, its hard to say what will work for you.have you measured the voltage and current from your coil and magnet pair? Chances are you are only getting a few uA and up to a few 10's or maybe 100's of mV.Note that your 5 transistors in series is not a magic current amplifier as you think it is. Each transistor adds 0.7 (700mv) input requirement. So your 5 transistors as you have drawn in a quasi-Darlington configuration (quintington??) the second transistor will not turn on until the first transistor sees (5 x 700mV) on it's base, or 3.5 Volts!Now, you say you need 3 to 5 volts to turn on the relay. ... that's easy, it's the final stage. But even a standard 2 transistor Darlington needs 2 x 0.700 volts to turn on, that's 1.4 volts on the base.If your coil output is less than 1.4 volts, it doesn't matter what the current is, it will never turn on a Darlington.If your coil output voltage is above 0.6 V and below 1.4 volts then you need a complimentary NPN-PNP pair, not a Darlington.see the first figure on the linked page :http://sound.westhost.com/articles/cmpd-vs-darl.htm#s1If your coil output voltage is below 0.6v, then you (a) need more turns on your coil (if you wrapped it) to produce a higher voltage, or (b) you need a voltage amplifier first, to bring the voltage up to some level that can turn on a transistor that controls a relay.

author: RivaultUser
author_rank: Contributor
timestamp: January 13, 2013, 09:13:19 am
content: Quote from: codeboy2k on January 12, 2013, 04:51:45 pmI think I have to prepare for the worst, can you give me more information on the voltage amplifier?

author: Mechatrommer
author_rank: Super Contributor
timestamp: January 13, 2013, 10:39:18 am
content: http://en.wikipedia.org/wiki/Current-to-voltage_converter

author: AlfBaz
author_rank: Super Contributor
timestamp: January 13, 2013, 11:32:16 am
content: Quote from: Mechatrommer on January 13, 2013, 10:39:18 amThat's weird Rin shouldn't be there should it? I always thought Rin was the source impedance

author: RivaultUser
author_rank: Contributor
timestamp: January 13, 2013, 12:23:24 pm
content: Quote from: Mechatrommer on January 13, 2013, 10:39:18 amAnd what is the type of the Op Amp that I will need???Op Amp is different with IC right?

author: Mechatrommer
author_rank: Super Contributor
timestamp: January 13, 2013, 06:11:33 pm
content: yup the Rin is the DUT circuit i guess as in...http://www.ecircuitcenter.com/Circuits/opitov/opitov.htmi havent done this only from reading, hope it helps.FWIW i found this as well while searching the matter. probably a good read to the OPhttp://www.ti.com/lit/an/sboa122/sboa122.pdf

author: Zero999
author_rank: Super Contributor
timestamp: January 17, 2013, 04:44:27 pm
content: Let's start from the beginning.What do you intend to happen when the magnet is placed near the coil? Do you want the relay to simply turn on or off? If so, it's simpler to use a reed switch to either directly drive the load or a relay switching the load.If you must use the coil. You need to be aware that the voltage/current will be proportional to the rate of change in magnetic field surrounding the coil and the polarity will depend on that of the magnet and whether it's being moved towards or away from the coil. This will mean that the relay will need to be turned on, say when a brief positive pulse is applied and off and a brief negative pulse is applied. A simple Schmitt trigger circuit will do this but if it's not sensitive enough, it will not detect the magnet when it's moved slowly but if it's too sensitive it will pick up noise.If the magnet is moved very slowly, the generated current/voltage may be too low to detect without picking up noise, in which case you need a way to detect a steady state field. The reed switch will do this as will a hall effect sensor. A coil could be used if an easily saturated magnetic core is placed inside the coil and the reduction in inductance is detected when the magnet is placed near the coil.

author: Mechatrommer
author_rank: Super Contributor
timestamp: January 17, 2013, 04:50:06 pm
content: ah good to hear from you Hero. you just gave him a food for thought