limit current of USB power to 2.5A - Page 1
Forum section: electronics / beginners

author: purezerg
author_rank: Contributor
timestamp: August 08, 2019, 03:26:26 pm
content: Hi, ok let me explain.I have a camera that uses firewire 12v to power. it states that it uses 10w. but the problem is it is 12v.I am thinking of using USB power to boost from 5v to 12v. I have tried those USB boost modules. it seems fine.the question is how to build a circuit to limit it's current draw to 2.5-2.8aI have tried hooking it up to a 5a 5v power supply. it seems for the first few mins it draws as high as 3.2aI am hoping to power the laptop via the USB powered port.the problem is the port shuts down at 3.0A

author: james_s
author_rank: Super Contributor
timestamp: August 08, 2019, 03:36:23 pm
content: If you limit the current you are supplying then you won't have 5V anymore, you don't get something for nothing. You have to get the load to draw less current.

author: ejeffrey
author_rank: Super Contributor
timestamp: August 08, 2019, 04:03:43 pm
content: There are basically 3 possibilities:* your input voltage is < 5V. This is actually pretty common, at the very least you get IR drop across the cable wires. So you may have only 4.75 V at the input under full load which will increase the current draw requirements. Measure the voltage at the input to the boost converter and if it is too low try a shorter or heavier gauge USB cable.* your device power draw is > 10 watt. You can easily measure this with an external 12 V supply.* Your converter is too inefficient. 10 watt is 2 amp at 5V. If your converter is 90% efficient that would be ~2.22 amp input draw. If 75% it would be more like 2.66A. If the input voltage is 4.7 and the converter efficiency is 70%, then you would need to supply 3.1 A. You should be able to make or buy a boost converter with >85% efficiency, but you could easily be doing a lot worse than that.Figure out which of the above is your problem or problems and go from there. If the problem is your converter efficiency that is easy to fix. If it is the output voltage of the port or the peak draw of your load, you are probably screwed.

author: purezerg
author_rank: Contributor
timestamp: August 12, 2019, 06:14:30 am
content: 1) i have tried with a 12v 3a power supply. it can draw as much as 17w if there are batteries that are not fully charged inside the battery.2) without any batteries inside. it will draw 9w3) with full charged batteries, it will also draw 9w.4) when the camera is active, it will draw 12-14w5) i do know that 12v @ 0.5A will also work because most macbook pro firewire provides only 0.5a. the powerspike is handled by the batteries.6) the droop measures as low as 4.91v at the end of 4.5m cable @ 2.5a7) i dont have a 12v dummy load to measure the efficiency of the boost module. but i do have 2 modules. the good stuff and the cheap stuff. both exceeds 3.0Ai have a LMxxx something module that bucks. i tried a 12v3A to 5v. the good module draws as high as 3.2-3.3a. the cheap module draws 3.6a.for those wondering. the camera is a phaseone medium format camera.I am wondering if I do a 1.8ohms 8x 2w parallel circuit. would it work?

author: Siwastaja
author_rank: Super Contributor
timestamp: August 12, 2019, 07:01:45 am
content: Well, if your camera takes up to 17W, even if you had 100% efficient converter, that would be 17W/5V = 3.4A at 5V.Even with your 14W number, that would be 14/5 = 2.8A at 5V.No converter is 100% efficient, so it's more than that in reality.Your camera needs whatever power it needs; if you won't supply it, it likely stops working. You can't supply your required 14W or 17W from 5V 2.5A, because 5V2.5A is only 12.5W.If you limit the input current 5V 2.5A by whatever means, the camera stops working because it can't get the power it needs to operate. If you add resistors, you are wasting even more power from your already negative power budget!Your options are modifying the camera to be more energy-efficient (quite a big project!), or to accept the fact you need a higher power source - if it has to be 5V, it has to be way more than 2.5A. Assuming your 17W number is correct, and assuming 85% converter efficiency, you need 4A at 5V.Your camera just is too power-hungry to be powered from most USB chargers. You need to come up with something else.

author: magic
author_rank: Super Contributor
timestamp: August 12, 2019, 07:35:44 am
content: eBay has DC-DC converters with current limit. Set one up for 12V/1A, done.

author: Siwastaja
author_rank: Super Contributor
timestamp: August 12, 2019, 07:42:32 am
content: Quote from: magic on August 12, 2019, 07:35:44 amBut then the camera requiring 17W (1.4A at 12V) will just drop the supply voltage and die.Current limit won't magically generate power.

author: magic
author_rank: Super Contributor
timestamp: August 12, 2019, 08:22:45 am
content: OP says its internal battery can handle transient peaks if the PSU sags.Of course OP may be wrong, but no harm in trying. That's his only hope anywayAnd by the way, if you use eBay junk switchers to power a camera which costs 10000x more, consider adding a beefy 13V transil to the output. Just sayingI am speaking from experience. Some of those modules can sometimes overshoot on power on. Who knows, perhaps on sudden recovery from current limiting too. Generally, the quality of all those eBay PCBs isn't stellar and neither is their QA.

author: purezerg
author_rank: Contributor
timestamp: August 12, 2019, 12:47:27 pm
content: thx guys. yeah I understand your worries.apparently during the image capture the sag is taken care of by the the battery. the 12v1A is just a means to slow charge the camera when it's not capturing the image.I am not expecting magical extra power but rather just hoping that i could find a way to sustain 12w to the camera and cut off at 12v1a or thereabouts.currently I am overdoing things by supplying a 12v3a into the firewire port while connecting via a type-C(laptop) to the (type-B)camera.the camera being firewire has the advantage of handling a wide voltage of 7-24v. hence the inaccurate 12v rail is adapted according by the camera internally.

author: Zero999
author_rank: Super Contributor
timestamp: August 12, 2019, 07:21:34 pm
content: If the camera will work from 7V to 24V, connect it to a bench power supply and vary the voltage from 7V to 24V.Does the current change?What voltage results in the least power draw, i.e. P = V×I?Design the boost converter to give that output voltage.Ultimately you can't get power from no where, so if your camera requires more, than the USB port can supply, then you're SoL. If it doesn't need to be operated continuously, then you could charge a battery and power the camera from that, but the average power will still be under what the USB port can provide.

author: Kasper
author_rank: Frequent Contributor
timestamp: August 13, 2019, 12:01:25 am
content: Quote from: purezerg on August 12, 2019, 06:14:30 amI struggled a bit reading your posts. If you describe where the current measurements are happening, it might be a bit clearer. Keeping in mind current in and out of switching regulators is not the same.1, 2 and 3) Looks like the battery charger was drawing 8W. Could draw more or less depending how full the battery is.4) camera active: 14Wtotal power 8W + 14W = 22WTo limit the current draw, you could try to disable the charger when camera active.You could also control a switching regulator between your power source and the load. Measure the current draw and use that to limit the output of the regulator. When load draws more current than your limit, the voltage to load should decrease. Sounds like load is fine with that, if external power is too weak it takes power from battery instead. To test that, power it with variable voltage supply, adust the voltage and see what happens.

author: Kasper
author_rank: Frequent Contributor
timestamp: August 13, 2019, 12:05:12 am
content: Quote from: Kasper on August 13, 2019, 12:01:25 amIf you try setting external supply voltage to less than battery voltage, measure the current to make sure your battery isn't sending current to your external supply.

author: purezerg
author_rank: Contributor
timestamp: August 13, 2019, 07:01:24 am
content: "total power 8W + 14W = 22W"no no no, that's not what i mean.idle = ~3wcharging battery = ~5wcamera in operation = ~6w-10wI unfortunately dont have a bench power supply. I'm not as hardcore as you guys.based on what I am gathering from you guys, it's a mission impossible? so far most of the comments are towards giving the camera as much power as it depends.the camera takes power from the battery when capturing the image. but when idle, it uses the extra power to charge the batteries.the whole pet project is the give it more power via the firewire(12v) rails, to charge it when it's idle.

author: purezerg
author_rank: Contributor
timestamp: August 13, 2019, 07:05:35 am
content: Quote from: Kasper on August 13, 2019, 12:05:12 ami doubt that will happen.** extra bit of info. the battery is 2 cell @8.4v full charge.the camera has got 2 of these batteries. each battery is basically a 2x NCR18650B.USB BC1.2 is only able to charge 1 of the battery. due to the fact that the USB is able to charge 1 of the battery, i believe there is a gate there to prevent backflow.firewire 12v is able to charge both batteries.firewire spec doesnt have a specific voltage range. i connected the firewire to a macpro and measured the voltage rails and used a clamp to measure the amps. the max was 1ai BELIEVE the camera can handle a wide range of voltage. but wondering further from 11-13v is anyone's guess.

author: Kasper
author_rank: Frequent Contributor
timestamp: August 13, 2019, 03:22:34 pm
content: Could use a switch to disconnect external power while camera is active