How do I limit current? - Page 1
Forum section: electronics / projects

author: deadlock
author_rank: Contributor
timestamp: July 17, 2013, 09:02:59 pm
content: Hi,I have a project with a microprocessor with USB host interface. I want to connect a mobile phone to the USB host interface and therefor I need to provide the mobile phone with 5 volts so that it will start talking over the USB channel.I have done this with a TI REG710-5V chip (http://www.ti.com/product/reg710-5). Problem is that the mobile phone drives 30mA thru the chip and it will stop because of high temperature.I would like to limit the current in some way, but I need to keep the voltage at 5 volt because if it gets too low the mobile phone would not start communicating. Therefor a simple resistor would probably not do the job.Anyone have any ideas to help me? Maybe the REG710 chip is totally wrong?Jens

author: Bloch
author_rank: Unknown
timestamp: July 17, 2013, 09:27:23 pm
content: I think that the mobile phone wants more than 30 mABut you TI REG710-5V have a Iout (Max) 0.03So you will need at least 500 mA if USB 2.0 or 900 mA in USB 3.0

author: deadlock
author_rank: Contributor
timestamp: July 18, 2013, 03:32:44 pm
content: Well, I have runned the mobile phone from 2 small clock batteries in parallel. I dont intend to provide power to the mobile phone. I just want to signal to it that "Hey, the USB cable is connected, start communicating".To 30mA is perfectly enough. The mobile phone is happy and is communicating however the REG710-5V is getting hot and shuts down due to high temperature.I believe 100uA would be enough to get the phone communicating, but how do I limit the current to 100uA while maintainging the voltage?If I would put a resistor in between I would get a voltage drop which I dont wantKind regardsJens

author: Bloch
author_rank: Unknown
timestamp: July 18, 2013, 04:12:01 pm
content: Quote from: deadlock on July 18, 2013, 03:32:44 pmThe only way to limit the current is to reduce the voltage ! That is Ohm law.

author: deadlock
author_rank: Contributor
timestamp: July 18, 2013, 06:27:51 pm
content: Ok,Thanks for your help, sounds like it is not easy to solve this... Are there any other paths I could take?I want to give the phone 5 volt so that it starts communicating but I want a very low current so that it does not drain my battery.The phone will take everything I give it. if I give it 30mA it will charge slowly. If I give it 1.5A it will use it all for charging.In some way I need to give the phone a stable 5 volts, but only very limited ampere. Is this impossible?I tried now putting a 470R resistor on the V+ line on the USB cable. I meassured the voltage drop to 1.1 volt over the resistor so the phone is just getting 3.9 volt and not enough for it to start communicating. The amp dropped to roughly 0.01 though which is good.Could a current regulator diode help me?http://se.mouser.com/ProductDetail/Semitec/S-101T/?qs=wgO0AD0o1vtfHNu%2f2J%2fCdA%3d%3dWould it be a good idea to replace the REG710-5V with a REG710-5.5 volt and then I might get roughly 4.4 volt which will probably be enough? But the voltage drop on the resistor is dependent on the internal resistance of the mobile phone I guess. So if I change to another phone I might get very different results?Any more ideas?Thanks for allJens

author: Bloch
author_rank: Unknown
timestamp: July 18, 2013, 06:50:53 pm
content: Quote from: deadlock on July 18, 2013, 06:27:51 pmDid you read up on the USB standard.?http://en.wikipedia.org/wiki/USBThe USB 1.x and 2.0 specifications provide a 5 V supply on a single wire to power connected USB devices. The specification provides for no more than 5.25 V and no less than 4.75 V (5 V±5%) between the positive and negative bus power lines. For USB 3.0, the voltage supplied by low-powered hub ports is 4.45–5.25 V.A unit load is defined as 100 mA in USB 2.0, and 150 mA in USB 3.0. A device may draw a maximum of 5 unit loads (500 mA) from a port in USB 2.0; 6 (900 mA) in USB 3.0. There are two types of devices: low-power and high-power. A low-power device draws at most 1 unit load, with minimum operating voltage of 4.4 V in USB 2.0, and 4 V in USB 3.0. A high-power device draws, at most, the maximum number of unit loads the standard permits. Every device functions initially as low-power—but may request high-power, and gets it if it's available on the providing bus.Quote from: deadlock on July 18, 2013, 06:27:51 pmI cant help you before you learn to use OHM´s law

author: fcb
author_rank: Super Contributor
timestamp: July 18, 2013, 06:51:27 pm
content: I don't think a current limited solution will work.If it was iPhone, then just don't connect the D+/D- lines, it will charge at 50mA ish (certainly thats what I found).I would slowly cycle the power to the charging USB port, this will allow the phone to act normally. So if your phone normally charges at 900mA then a 1:30 duty cycle will give you an AVERAGE drain of 30mA. You probably want to experiment to see how long the phone takes to begin charging when you apply power.A 555 astable will just about do it with some medium size electrolytic's in the RC network - remember you don't car much about the frequency. A rough guess is 15 seconds on, 7.5 minutes off.Personally i'd do it with a PIC.

author: Gakex
author_rank: Contributor
timestamp: July 18, 2013, 07:21:32 pm
content: I'm kind of lost here, I always read or hear that the current don't mind if is more or enough in order to drive a system.So way limit the current?

author: Bloch
author_rank: Unknown
timestamp: July 18, 2013, 07:28:22 pm
content: Quote from: Gakex on July 18, 2013, 07:21:32 pmQuote from: deadlock on July 18, 2013, 03:32:44 pmI think he has them inseries. But they will fast run out off energy.

author: KedasProbe
author_rank: Frequent Contributor
timestamp: July 18, 2013, 08:57:55 pm
content: Isn't there a way to tell the device that the socket is already giving the max power (although it isn't)should make the device decide to drain less like current management in hubs.I don't know much about how this USB power management works but maybe you find a solution this way.

author: deadlock
author_rank: Contributor
timestamp: July 18, 2013, 09:18:38 pm
content: Bloch: thanks for your continious help. I have read the USB standard for USB 2.0.I know Ohms law, but to really understand it I guess will take me alot of time and practice in different scenarios. I guess in my case trying the resistor, I knew that I would get a voltage drop. I guess I should see the mobile phone as just another resistor adding it together with my 470R resistor. If the mobile phone has a resistance of 2280R I guess 20% of the voltage would be over my 470R resistor and the rest 80% would be over the mobile phone. Then I would also guess that the resistance of the mobile phone might be different depending on how much it is charged. Or am I thinking in the wrong direction here?Fcb: It is an Android phone. My aim is not to charge the phone. I just want to provide a steady 5 volt low current to it. But maybe this is not possible? Maybe it will be unstable if the phone is exhausting it?Bloch & Gakex: Yes I used them in series, they were almost depleted and together thy were very close to 5.0 volt. I used them for many days, the phone would not charge very much. there was even a message on the phone "Very low current, not charging" or something. But then again, that is what I want to accomplish. I dont want to charge the phone, I want to give it a very low current stable 5 volt reference so that it will start communicating. Unfortunately it needs the 5V to sense that there is something connected in the other end. Preferably I would cut the red wire of the USB cable, but that wont work.But maybe my setup with the TI REG710-5V is completely wrong for this?KedasProbe: I also thought about something like this, however I am not sure how to do it. I am using the FTDI VNC2 microprocessor for USB communication. I will dig into the documentation and see if there is something I can configure.Thank you all for your help and patience with me

author: TerminalJack505
author_rank: Super Contributor
timestamp: July 18, 2013, 09:30:40 pm
content: Does your phone support USB-OTG? Perhaps what you're trying to do is more suitable for it.In this case the phone will be the USB host and your device will be the slave.

author: deadlock
author_rank: Contributor
timestamp: July 18, 2013, 09:37:57 pm
content: It might support OTG mode. I have been thinking about trying that out.However I would prefer to use AOA if possible since support for OTG is still a bit low.Now I just read this on SparkFun about the IOIO:https://www.sparkfun.com/products/11343QuoteI think this is exactly what I want and they make it sound so easy, which I also thought it was in the beginning.Is it even at all theoretically possible to have a stable 5V 0.01A current if the mobile phone is trying to use 0.3 Amps ?Kind regardsJens

author: ovnr
author_rank: Frequent Contributor
timestamp: July 18, 2013, 10:02:22 pm
content: Quote from: deadlock on July 18, 2013, 09:37:57 pmNo. You cannot limit the current without affecting the voltage. However, if you do have your phone connected to a USB host controller, the controller can signal that the phone may not use more than a specified amount of current - but this will usually require a full-blown host controller. There is no analog solution that will allow you to retain communication with the phone while still limiting the current it is draining (to a value below 100 mA/500mA).If possible, OTG mode is indeed what you should be pursuing. If not, I don't see any easy way to accomplish what you are trying to do.

author: deadlock
author_rank: Contributor
timestamp: July 18, 2013, 10:07:41 pm
content: ovnr: Thank you very much for clearing this out for me.I will check if I can make the VNC2 instruct the mobile phone about the maximum current allowed.If that wont work I guess OTG is the only road aheadJens

author: experimentalist
author_rank: Newbie
timestamp: July 19, 2013, 12:27:10 am
content: Quote from: deadlock on July 18, 2013, 06:27:51 pmYou don't have to change the REG 5V to a 5.5V, just put another 470 resistor in parallel with the one you have now and you will get 4.45V. That might be enough, might not depending on if it is USB 2 or 3. If you want to current limit even if the load resistance changes you need to use a voltage regulator. Since the REG710-5V is a voltage regulator it should work fine.You just have to fine tune that resistance you have in series and don't be afraid to use more than one resistor combination in series or parallel to achieve exactly the voltage (and hence current) you are after.You could play with the series resistor you have and should get the results you're after. If 470 is giving you too low a voltage then you need to drop the resistance until you are in the voltage range you're after. This will increase the current, so you need to drop the resistance just enough and no more. You can do it by trial and error or calculate the resistor you need to get there.

author: deadlock
author_rank: Contributor
timestamp: July 19, 2013, 02:02:10 pm
content: experimentalist: Interesting, so you are saying that if I find the correct resistance I might both get enough voltage (not 5.0 but maybe 4.7 etc) and still have a very low current? I guess this would in that case be very specific to different mobile phones?Thanks for the ideas

author: davo
author_rank: Newbie
timestamp: July 20, 2013, 12:09:48 am
content: I'm not sure if it helps but have you thought about pulling the phone battery?If the battery is not there, it can't pull the current to charge it eh.ciaodave