Connect with us

LED Charging Indicator

Discussion in 'LEDs and Optoelectronics' started by jonn, Dec 12, 2012.

  1. jonn

    jonn

    7
    0
    Dec 11, 2012
    Hello I am new to this forum,I know very little about electronic circuits but the very basics.
    I apologize for joining and immediately asking for assistance.I have searched online but due
    to my limited knowledge I have come up empty handed.

    I am designing a dash panel for a snowmobile and it includes a power port(think cigarette lighter socket)This port will be used as an outlet aswell as a charging port for the snowmobiles battery.The port will be wired directly to the 12v battery's positive and negative terminals.What I want to do is have an LED light up when the charger is plugged in and recharging the battery but not when something being powered by the battery is plugged in.

    Is there a simple circuit that can achieve this?

    John.
     
  2. GreenGiant

    GreenGiant

    830
    6
    Feb 9, 2012
    You may be able to do that with a few diodes, being that current will be flowing into the unit when charging and out of it when draining.

    You should be able to use 2 diodes (one before and one after the LED) directionalized for current to flow when charging and not while draining, this would be a circuit in parallel with the actual power connection so you dont have to rely on the diodes and LED to pass the current needed for charging the battery, or using the device.
     
  3. Harald Kapp

    Harald Kapp Moderator Moderator

    9,293
    1,884
    Nov 17, 2011
    The diodes are a viable solution if you can live with the voltage drop.
    If you can`t, you can use a small shunt resistor (e.g. 0.1 Ohm). Sense the voltage across the resistor with a sense amplifier (OpAmp) and use it to control an indicator LED.

    What is your charge current?
    If you design the resistor such that R=Vled/Icharge you can connect the LED directly parallel to the resistor. But again you have a voltage drop here (Vled). You may also experience a problem if the load current is much higher than the charge current, because in that case the reverse voltage on the LED may be too high. In that case the diode solution or a sense amplifier will be better suited.

    Harald
     
  4. jonn

    jonn

    7
    0
    Dec 11, 2012
    Hello and thanks for the replies.
    GreenGiant I think I have an idea of what you are saying but might be way off.The image attached is what I got from your advice.I don't know if this is what you meant but it did not work.

    Herald your suggestions have exceeded my very basic knowledge.I know nothing about opamps or shunt resistors or how that circuit would be wired.
    The charge current is either two or four amps,if only one of the two can be used I would go with two amp.I don't think there would ever be a load greater than two amps.

    John
     

    Attached Files:

  5. Harald Kapp

    Harald Kapp Moderator Moderator

    9,293
    1,884
    Nov 17, 2011
    I don't think your circuit reflects GGs idea. But I have to admit that I don't really understand how he has the diodes connectd to detect the direction of current flow without puitting the diodes in series with the current flow.
    Your diagram, however, will have the LED constantly off because you connected the diodes anode to the batterie's cathode and vice versa. But even if you turn around alll diodes, the LED will be constantly on. There is no way of detecting the direction of current flow.

    At 4 A the power lost in a series diode or is at least 0.6V*4A=2,4W. That requires some cooling. A currentg sense resistor is the better solution. Look at this circuit. This circuit is meant to measure the current Irs. Connect the batttery to the right, the port to the left. Then if Irs>0 you are charging and Vo >0. If Irs<0 then you are discharging and Vo<0. Add an LED with series resistor from the outputto GND. Increase R2's value until you get a reliable indication of current flow.

    Harald
     
  6. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    8,393
    1,266
    Nov 28, 2011
    Allegro make some neat little ICs that can detect the direction and magnitude of current flowing through them, with almost zero shunt resistance. The current actually flows THROUGH the IC.

    http://www.allegromicro.com/en/Prod...ifty-Amp-Integrated-Conductor-Sensor-ICs.aspx

    The ACS714 looks pretty good. It measures up to 50A in either direction. If you need more than that, they also make a range with 200A maximum.

    http://www.allegromicro.com/en/Prod...p-Integrated-Conductor-Sensor-ICs/ACS714.aspx

    It's packaged in a 8-pin SOIC so you will probably want to use an SMT-to-THT adapter board to support it physically and enable you to connect to it. It needs a +5V supply.

    Its output is nominally VCC/2 at 0A current flow. You will need to use a comparator with a presettable thresold to detect when the device's output goes above some nominal threshold. It may be difficult to adjust it to detect a low charge current, for example, less than 5A, because the device's output will be quite close to VCC/2 for a low current.

    These parts are available from Digikey, and probably Mouser, for about USD 4 in one-off quantities.
     
  7. GreenGiant

    GreenGiant

    830
    6
    Feb 9, 2012
    Yea that was a brain fart moment for me, it was early in the morning and I was thinking about it quickly...

    Glad you guys cleared it up, my mistake haha
     
  8. jonn

    jonn

    7
    0
    Dec 11, 2012
    Thanks for the ideas.
    The circuit Harald is suggesting is something I can do.I just have no idea what the values of the resistors and shunt should be or what op amp to use.I am assuming RS is a shunt.Maybe somebody could give me an idea of what they should be or help me to learn how to calculate them.

    John
     
  9. Harald Kapp

    Harald Kapp Moderator Moderator

    9,293
    1,884
    Nov 17, 2011
    Look at the MAX4372 datasheet. That is the IC proposed on the website I originally linked. Starting on page 6 there is a detailed circuit description with equations how to chose the component values.
     
  10. jonn

    jonn

    7
    0
    Dec 11, 2012
    Am I understanding correctly that with the MAX4372 all I would need is the correct sense resistor,the led with resistor and a 0.1uf capacitor.

    Sorry if I'm way off here.Unless it's something I can actually touch or take apart I tend to have a very hard time understanding without being shown or having it explained in detail.

    John
     
  11. Harald Kapp

    Harald Kapp Moderator Moderator

    9,293
    1,884
    Nov 17, 2011
    Look at table 1 in the MAX4372 datasheet. For a full scale output voltage of 2V at a load current of 1A a sense resistor of 100mOhm and a gain of 20 is given. A gain of 20 means you use the MAX4372T.
    Since your min. charge current is 2A, you can reduce the sense resistor to 50mOhm.
    At 4A this simply means your output goes to 4V instead of 2V.

    Connect an LED with series resistor to the output of the MAX. Assuming you have a red LED (Vled ~ 1.6V) and your LED operates between 10mA and 20mA current, the series resistor is R= (2V-1.6V)/10mA= 40Ohm. Use 47 Ohm as a standard value.
    If the output goes to 4V, the same circuit will produce a current of I=(4V-1.6V)/47Ohm= 51mA. That should be acceptable for the LED.
    If you are worried about the difference in LED current, you can use a simple LED driver as shown in figure F on this site. Replace the finger ( :D) by a series resistor from the output of the MAX into the base of the Transistor. 4.7kOhm is a good value.

    Now the only thing you have to observe is the polarity such that the circuit lights only when charge current is flowing, not while discharging.
     
  12. jonn

    jonn

    7
    0
    Dec 11, 2012
    Thanks alot Harald I appreciate your help.
    The LED I'm using is 2.1 volts(Digi-Key #492-1540-ND)I'm not sure it would work with the 2 volt output.
    Would using MAX4372F with a 50mOhm give me 10 volts at 4 amps and 5 volts at 2 amps.Also will the sense resistor need a heat sink.
     
  13. Harald Kapp

    Harald Kapp Moderator Moderator

    9,293
    1,884
    Nov 17, 2011
    maybe yes, maybe no. It will possibly light only very dimly.

    The MAX4372 will give 5V at 2A for a 50mOhm sense resistor and go to 10V for 4A.
    I suggest you use the driver circuit I linked. At 5V...10V otput of the max, use a 10kOhm series resistor into the base. The series resistor for the LED needs to be adapted to the supply voltage of your circuit. At (probably) 12V, the resistor drops ~10V (12Vsupply-2Vdiode), thus R= 10V/Iled, where Iled<30mA. Set Iled=20mA and you get R=10V/20mA=0.5kOhm Use e.g. 470Ohm from the E-standard series.

    The sense resistor will dissipate P=I^2*R=(4A)^2*50mOhm= 16*0.05W= 0.8W. Use a 1W resistor and you will probanly not need a heatsink.
    Observe, however, that the power may rise if your discharge current is higher. Then you'd need a heatsink or switch to the MAX4372H which has twice the gain of the MAX4372F. Then reduce Rsense to 25mOhm, thus halving the power loss.
     
  14. jonn

    jonn

    7
    0
    Dec 11, 2012
    I have assembled current sense and LED driver circuits .The problem I'm having is there is constant power from the out on the microscopic MAX4372.Does anybody have any ideas why this is happening?I did not use an actual sense resistor,I used a thin film resistor,is a sense resistor different from normal resistors?
     
  15. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    8,393
    1,266
    Nov 28, 2011
    A thin film resistor will be fine.

    Can you post the exact schematic of your test circuit?

    The MAX4372 has an input offset voltage of up to 1.3 mV in either direction which will translate into an output offset of up to 130 mV for the MAX4372H.

    You haven't given much information on your requirements or component choices.

    Which MAX4372 are you using?
    What is the output offset voltage you are seeing?
    What resistance is your shunt resistor?
    What is the maximum current you intend to measure?
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-