This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

  • TI Thinks Resolved

LMC6001: Why -IN is short-circuit with GND? How I broken it?

Part Number: LMC6001

I am designing a signal amplification circuit using LMC6001BIN.

the schematic is like this:

Before I test this current I test my LMC6001, It is ok. And I also check my circuit map and it is just like this schematic.When I power it on, I find the temperature is so high. And it is not work as what I want. So I test it and I find Pin2(-IN) is short-current with Pin4(GND). And I get signal voltage is about 8.5V .And I remove this chip and test it. And yes it is broken.

(Ip[current] on D1 is 0.7uA at most by the way.)

But I don't know why but I am sure I am wrong. So help me, Thanks? If the amplification ratio what I design is too high.

(I am a chinese so my english is not so good, I translate it to Chinese. Thanks)

我使用上面的电路去做一个光电二极管的放大电路。因为我的输入电流至多0.7uA,所以我设计这个电路端的电压等于Ip*R3*(1+R2/R1)

但是在我上电之后(尚没有给予光线照射),发现输出信号一直是8.5V左右,然后我察觉到芯片温度很高,所以我断电取下芯片测试发现LMC6001的pin2和pin4短路。

我个人觉得一定是我使用这个芯片没有注意哪些参数导致的。是不是我的放大倍数太大了?

而且我检查了电路确认电路和原理图是一致的。

请您给与帮助,谢谢。

  • Part Number: LMC6001

    I am design a signal amplifier circuit. But I don't know why my LMC6001 is broken. And here is the basic schamatic.

    I am afriad if I give a such large amplification.  Cause my signal Ip is really small, it is 0.6uA max. (Ip is the current of D1 and D1 is a photodiode)

     When I powen on my circuit just for a while. I find LMC6001 is really hot, and I don't get a right output. When I power off and take LMC6001 off and test it. I find Pin2(-IN) and Pin4(GND) is short-circuit.

    Please help me. Thanks.

    By the why, the chip is LMC6001BIN。

  • In reply to John Henry:

    John,

    First and most important is to faithfully follow safe ESD prevention procedures.
    Be sure that 12V supply doesn't spike above 15.5V during power up/down/running.
    Is this a new or old design? Measure ICC on a new device; is ICC high right from the start or does it increase later?
    0.7uA * 1Mohm * 21 = 14.7V , so yes the gain is too high. However that will not break the LMC6001.

    Regards,
    Ronald Michallick
    Linear Applications

    TI assumes no liability for applications assistance or customer product design. Customer is fully responsible for all design decisions and engineering with regard to its products, including decisions relating to application of TI products. By providing technical information, TI does not intend to offer or provide engineering services or advice concerning Customer's design. If Customer desires engineering services, the Customer should rely on its retained employees and consultants and/or procure engineering services from a licensed professional engineer (LPE).

     

  • In reply to Ron Michallick:

    Ron,

     Thanks very much. You are right about the very high gain.

     And this design  refer to the design in 《Photodioda amplifiers op amp solutions》,Jerald Graeme.

    And I use multisim to simulation this design, It is work well. 

    And I have modified my design. Change  R2  from 20K to 10K. And change R3 from 1M to 500k.(Actually, I have two different sense. another have a small If. So I design like this). It seem like ths didn't solve this problem.

    Yes, I am sure ESD is important. And I build this circuit by hand and I want to test it first and then design PCB file. Maybe I didn't check every net-connect on my circuit.

    I will let you now If I solve it.

    Thanks

  • In reply to Ron Michallick:

    Ron:

          I think I need tell U my solution.

          When I check my LMC6001, I find mark on this chip is NS not TI. So maybe this is an old product or fake one. Cause I haven't a new one  at hand. So I use LM358 to test this circuit. And It worked well. 

    And I have order some new LMC6001 and LMC6081. And I will test again.

          Have a good time. Thanks again.

  • In reply to John Henry:

    The NS symbol is still used on this device.

    Regards,
    Ronald Michallick
    Linear Applications

    TI assumes no liability for applications assistance or customer product design. Customer is fully responsible for all design decisions and engineering with regard to its products, including decisions relating to application of TI products. By providing technical information, TI does not intend to offer or provide engineering services or advice concerning Customer's design. If Customer desires engineering services, the Customer should rely on its retained employees and consultants and/or procure engineering services from a licensed professional engineer (LPE).

     

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.