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LM380N input protection

Ronald Jansen
Intellectual 520 points
Other Parts Discussed in Thread: LM380, LM384

Hello,

I am doing a design with an LM380N as a PC audio amplifier. The audio input connects to a jack plug input.

I would like to add some extra ESD protection. Usually I would clamp the input to the supply rails and/or use a zener, but I read in the absolute maximum ratings section that the maximum input voltage is +/-0.5V. This is really low.

How do you protect such an input without affecting the signal?

Thanks in advance for any help.

Ronald

  • 0
    TI__Guru**** 189399 points

    Hi, Ronald,

    This part is more than 20 years old, so we don't have a lot of data available on it any longer.

    With that said, I think that is supposed to say +0.5V above the power supply, and -0.5 below ground. That would make sense.

    -d2

  • 0 in reply to Don Dapkus
    TI__Intellectual 1240 points

    Hello Ronald-san,

    I'll interupt to your question about LM380N input absomax ratings. Yes, this device was released more than 20 or 30 years ago. If my memory is correct, the reason is as follow.

    When you look at the pg 2 on the datasheet, the input stage has PNP darligton configuration. The both inputs is taking vertical PNP type and the collector is connected GND which is substrate. If the input goes below GND or -0.5V, this vertical PNP's collector and base diode will turn ON and if large current flow, the device may get into latch up mode. So to suppress the latch up, the data sheet specified not exceed below -0.5V in negative side. For positive side, when you also looked at input resistor 150k ohms. This resistor is relatively high value. To meke this value, usualy it takes large area to achieve the resistor value. To minoimise the resistor area, it was taken pinched tresistor which has low break down voltage. usualy the break down voltage is similar to NPN Tr's base- emitter break down voltage which is around 6V to 7V. But signal synmetrical pointo of view, the datasheet specified +/-0.5V.  I'll suggest to put few k ohms resistor on signal input side to suppress input current in about less than 1 to 2mA.

    I hope this helps for your concern.

    Best regards

    T NAOKAWA

  • 0 in reply to Toyojiro Naokawa
    Intellectual 520 points

    Many thanks for your help.

    I did not notice that the inputs would clamp in a different way than I would expect. If the result of exceeding the ratings is that excessive current would flow, then it is relatively easy to prevent it.

    I will first clamp the input to the supply rails 0V, 12V. Then after the clamps I will add a 10K resistor as you suggested. This should limit the current to appr. 1.3mA.

    There is still one thing that concerns me. As this part is already very old, will it still be available for at least 5 years? I chose this part for a new design because of the low costs and good availability in low quantities. Also there are not many parts that can deliver more than 1W and do not require a heatsink. Most of the parts >1W are much more powerful and would require a heatsink.

  • 0 in reply to Ronald Jansen
    Intellectual 520 points

    Actually there is one thing I missed.


    The negative limit is rather normal. The positive limit would then probably be the breakdown of the 150K resistor. I assumed that limiting the current would do, but is this assumption correct or can the resistor also be destroyed when exceeding the voltage limit but with a low current?

    I would expect that exceeding the breakdown voltage would make it behave like a zener. In that case limiting the dissipation would be enough, but I am not sure if it works that way with a resistor.

  • 0 in reply to Ronald Jansen
    TI__Intellectual 1240 points

    Hello Ronald-san,

    Regarding positive side limit, Yes, your expectation is right. If the excess voltage is applied for ex, over 6V or 7V, the input resistor will start break down and current flow. this breakdown is zener break down but if you put resistor, even the input resistor had breakdown, it wouldn't flow excess current.  The current will be,  (Posistive voltage - Vz)/ resistor you put.

    Hope this helps for your understanding.

    Best regards

    T NAOKAWA

  • 0 in reply to Ronald Jansen
    TI__Intellectual 1240 points

    Hello Ronald-san,

    I have no idea about this device life. My understanding is that if customer in world wide base continue to use thisa device, then TI will continue the production.  This will be basic idea for continuing the products.

    Regarding alternative device, your desire will be 1W into 8 Ohms load and connection for the load is single ended. If your load connection is OK to take BTL connection, you will have 1W into 8 ohms load with 5V supply and without output cap. there are many devices I can recommend as follow. LM4951ASD, LM4990LD, LM4990MH,LM4990MM,LM4991LD, LM4991MA etc. If you still SE load connection, LM4950TS will be good though this device is stereo type, not mono.

    Best regards

    T NAOKAWA

     

  • 0 in reply to Toyojiro Naokawa
    Intellectual 520 points

    Hello,

    Actually I was looking for 3W, but I decided to settle for the 2.5W offered by the LM380. The application is an intercom system for noisy environments like underground parking lots. We want  to get some more volume than usual and a wider range than the 300-3000Hz usually specified for this. I was thinking about 150-200Hz. By my estimation it would require around 3W.

    Most parts I found will do around 1W, or 5W and more. There seems to be a gap in between. I did find other 3W parts, but not for an 8 ohm load.

    One of the things I like about the LM380N is the DIP14 package. It can dissipate heat from a relatively large package and also through 4 pins into the PCB.

    Best regards,

    Ronald

  • 0 in reply to Ronald Jansen
    TI__Intellectual 1240 points

    Hello Ronald-san,

    So you need 2.5W into 8 ohms load and prefered PDIP pkg.

    I checked TI products tree and found only 2 devices which are LM380N and LM384N. These devices has total 6 pins GND and 1 signal GND pin. Please check your operating temp range and device thermal resistance. If you want 2.5W, the supply voltage will be 16V. In this case max Pd will be arround 1.7W. On the other hand the required pkg therma resistance will be, (Tjmax-Ta)/Pd max, where Tjmax is 150 deg C, Ta is your max operating temp say 60 deg C. Then the required therma resistance will be about 53 deg C/W. So you may need to have some heat spread cupper area. Please refer to graph about power dissipation Vsambient temp.

    Best regards

    T NAOKAWA

     

  • 0 in reply to Toyojiro Naokawa
    Intellectual 520 points

    Thanks again for your input. I came to the same conclusion. As the power supply is 12V, I think there is no added value in using LM384. I did not yet think of the fact that it will not give me 2.5W at 16V. The latest test imply that the volume produced by the LM380N at 12V will be sufficient. So, it looks like I am lucky. Otherwise I would need a bridge configuration, which in turn will not allow me to use an 8 Ohm speaker without excessive dissipation.

    I am not too concerned about the dissipation of the current solution. The system will be used for speach. When someone is speaking 100% of the time still it will only have maximum dissipation  for 20% of the time, so we should be ok here. The PCB will give me about 4 square inches of cooling area. If all else fails a heat sink can be attached to the IC.