• State Resolved
  • Locked Locked
  • Replies 8 replies
  • Answers 1 answer
  • Subscribers 48 subscribers
  • Views 1531 views
  • Users 0 members are here
Support feedback
Options
Options
Similar topics

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.

LMH6624

Andres Torre
Prodigy 150 points
Other Parts Discussed in Thread: LMH6624, TPS60400, LM7705, LM6211, LM7321, LF347, OPA209

Hello,

I would like to know If I can use LMH6624 to convert a signal 0-10V to 0-5V. I would supply it with +12V-0V.

I have a sensor whose output is 0V to 10 V and I need to convert it to 0V to 5V because that is the input of the ADconverter.

I think I should make this circuit:



Uploaded with ImageShack.us

R1=1M and R2=2M

If someone knows a better circuit or OPamp to convert 0-10V in 0-5V with low noise and low consuption. please tell me.

than you very much

A. Torre

  • 0
    TI__Mastermind 22825 points

    Hi,

    The schematic you've proposed has several issues:

    1. The LMH6624 output won't be able to reach 0V. Even if LMH6624 were Rail-to-Rail output (which it is not), you'd still have issues because a RR output device still has ten's of mV of headroom to VEE (ground in this case) and won't hit 0V exactly.

    2. The resistors you've chosen are very large in Mohm range and this would have a detrimental effect on noise! There is also issues with Mohm resistor with a bipolar input device bias current and input impedance, etc.


    3. Your schematic actually gains up the input 10V to 20V (theoretically) instead of dividing it down to 5V

    You could use an inverting charge pump on your amplifier VEE pin to create a negative bias from a positive source (1.8-5.25V) so that your output can reach exactly 0V. Here are some possibilities:

    TPS60400, LM7705

    You would also replace the LMH6624 with a proper Rail to Rail output amplifier depending on your noise and speed needs. Here are some possibilities:

    LM6211, LF347, LM7321, etc.

    Hope this helped?

    Thanks,

    Hooman

  • 0 in reply to Hooman Hashemi
    Prodigy 150 points

    Hello,

    Thank you very much for the information, I have found it very useful.

    About the gain, I was a mistake writing the post, you are right about the resistors.

    This is my new circuit, I have followed your advices. My circuit would measure from 0-10V sensors and -5v-+5V sensors, depend on switch position. I need always 0-5V output for the ADconverter.

    What is your opinion about it? is it low consumption? could you help me to select the DC/DC +5v to 12V converter?



    Uploaded with ImageShack.us

    thank you very much

    A. Torre

  • 0 in reply to Andres Torre
    TI__Mastermind 22825 points

    Hi Andres,

    You don't need the +2.5V reference that you're generating with the lower OPA209! To accommodate -5V to +5V sensor inputs, SW1 left hand side should be tied to +5V supply directly, in order to shift the upper OPA209 output by 5V to 0V minimum. This also simplifies your design!

    Other than that, OPA209 seems to be a good choice (low 2.2nV/RtHz noise, low 150uV offset, RR output, input CM voltage range of 1.5V from either rail, low power). So, that should be fine. Make sure the 6.4V/us slew rate is not a limitation in your design. You may want to add supply decoupling caps close to the upper OPA209 supply pins.

    I did not thoroughly check your TPS60400 charge pump design (to create -5V from +5 source) but I assume you did that by consulting the datasheet. It seems to be done correctly.

    Thanks,

    Hooman

  • 0 in reply to Hooman Hashemi
    Prodigy 150 points

    Hi Hooman,

    thank you very much for your help,

    If I connect  +5V directly to switch left hand, output voltage increases 7.5V, so if input is -5V,  OPA209 output is 2.5V instead of 0V, and if input is +5V, the output is 7.5V instead of 5V. I think I should connect the switch left hand to 2.5V. You can see the simulation in the next picture:



    Could I use LM7705 to generate -0.23V and  use it to convert my signal sensor, instead of using TPS60400 to generate -5V? should be enough -0.23V OPA209 negative reference to do the conversion? or I need -5V to OPA209 negative reference.

    thank you very much for your help

    Andres

  • 0 in reply to Andres Torre
    Prodigy 150 points

    Hello,

    I have other question about getting 2.5 V reference, Which of following diagrams  is the best to get 2.5 reference voltage?

    you are helping me a lot.

    thank you very much.

    Cheers

    Andres

  • 0 in reply to Andres Torre
    TI__Mastermind 22825 points

    Hi Andres,

    I'm so sorry! You're correct in that you need the 2.5V reference to SW1 left hand side for -5V to +5V sensor input to be shifted to the 0-5V range! Sorry about my error.

    About using LM7705 to generate -0.23V instead of -5V negative bias: If you do that, your output swing range is Ok. However, since OPA209 input voltage range is 1.5V above VEE, you'd have issue with the input which swings all the way down to 0V. If you think the -0.23V negative bias is more beneficial for you (lower power, etc.), you would then need to replace the OPA209 with a device which its input voltage range includes the negative rail (or it is Rail to Rail input range). It'd be very difficult to hit all OPA209 specifications and to have input RR range! So, I recommend you stay with the negative bias scheme you have now!

    About generating the 2.5V reference: The zener diode scheme could possibly be noisier (needs decoupling cap to ground at + input of OPA209) and it could have a temperature gradient which you'd need to verify does not violate your system. The voltage divider scheme is a 2.5V reference which tracks the +5V supply by a 1/2 ratio. So, if you expect the 5V supply to vary appreciably, the voltage divider arrangement may provide some (not completely due to the 1/2 factor) temperature / variation immunity. Also, the voltage divider scheme is at a slightly higher noise gain (1.5 V/V) and it might be slightly more stable (although I don't expect that to be a factor).

    Thanks,
    Hooman

  • 0 in reply to Hooman Hashemi
    TI__Mastermind 22825 points

    Hi Andres,

    Please disregard my last post and use this one instead:

    ******************************************************

    Hi Andres,

    I'm so sorry! You're correct in that you need the 2.5V reference to SW1 left hand side for -5V to +5V sensor input to be shifted to the 0-5V range! Sorry about my error.

    About using LM7705 to generate -0.23V instead of -5V negative bias: If you do that, your output swing range is Ok. However, since OPA209 input voltage range is 1.5V above VEE, you'd have issue with the input which swings all the way down to 0V. If you think the -0.23V negative bias is more beneficial for you (lower power, etc.), you would then need to replace the OPA209 with a device which its input voltage range includes the negative rail (or it is Rail to Rail input range). It'd be very difficult to hit all OPA209 specifications and to have input RR range! So, I recommend you stay with the negative bias scheme you have now!

    About generating the 2.5V reference: The zener diode scheme could possibly be noisier (needs decoupling cap to ground at + input of OPA209) and it could have a temperature gradient which you'd need to verify does not violate your system. The voltage divider scheme is a 2.5V reference which tracks the +5V supply by a 1/2 ratio. So, if you expect the 5V supply to vary appreciably, the voltage divider arrangement may provide some (not completely due to the 1/2 factor) temperature / variation immunity. So, if your +5V supply is expected to vary, you'd get 1/2 that variation in your 2.5V reference and all of it at your ADC input. Also, the voltage divider scheme is at a slightly higher noise gain (1.5 V/V) and it might be slightly more stable (although I don't expect that to be a factor). The choice may come down to the relative accuracy of your +5V source vs. the relative accuracy of the 2.5V zener you choose. I'd pick the scheme with the lowest percentage variation.

    Thanks,
    Hooman

  • 0 in reply to Hooman Hashemi
    Prodigy 150 points

    Hi Hooman,

    Thank you very much for yours explanations, they are really useful for me.

    I think I will probably use the Zener scheme because I think it has the most stable output and it is more independent of +5V input (Because I am  not sure +5V are really stable) but I think this scheme is the most expensive because I will have to choose a  high accuracy 2.5 V zener.

    I was a mistake because I repeated the zener scheme two times in the previous post, the second circuit that I wanted to show to  you is the following one:



    thank you very much for your help, I am going to make all the circuits with the tree 2.5V referent circuits to compare them.

    I will show you the results.

    thank you again

    Cheers

    Andres