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TLC7524,GAIN in Unipolar Operation

David Garcia1
Prodigy 50 points
Other Parts Discussed in Thread: TLC7524

I'm designing this schematic about TLC7524, Unipolar Operation 

Could you clarify this information about GAIN adjustment: datasheet pg7  "RA and RB used only if gain adjustment is required" but not specify anything

  • Input  250=> D7...D0 [1111 1010]
  • Vout 10V
  • Vref=  -1.5V
  • R19=  732Ohm??
  • R21 = 5K Ohm

Is Correct this assumption?

Vout= -Vref * R21/R19 * (D7/2+...+ D0/256) = 1.5 * 5000/732 * 250/256 = 10V

 


Thanks a million

  • 0
    TI__Genius 10615 points

    Hi David,

    The gain adjustment is usually used in order to improve the gain error of the DAC. RA allows for negative trim (attenuation), while RB allows for positive trim (gain). The size of RA and RB, depends on the input impedance of the REF pin and the RFB pin. The gain of MDAC relies on the input of the REF pin to be matched to the output of the RFB pin. Therefore by introducing a series resistance at these pins, you can modify the gain of the MDAC. The input impedance of the REF pin will vary from 5 to 20 kΩ, so a 2 kΩ will be able to make small adjustments which is all that is needed for gain error adjustments. You can assume that the gain of the DAC will be equal to REF_input_impedance / RFB_input impedance. Since these are very well matched it is usually very close to 1. If we include the RA and RB terms, it will look something like this: DAC_gain = (REF_input_impedance + RA) / (RFB_input impedance + RB)

    For the positive gain that you are trying to achieve, it would be necessary to replace RB with a potentiometer so you can manually adjust the gain. RA is not necessary in this case because you are only trying to increase the gain. Since the output of the RFB pin can be anywhere from 5 to 20 kΩ, I would suggest assuming 20 kΩ since it would be the worst case scenario. You are trying to achieve a gain of roughly 8 times, so using a 160 kΩ potentiometer for RB should give you enough gain.

    David Garcia1 said:

    Is Correct this assumption?

    Vout= -Vref * R21/R19 * (D7/2+...+ D0/256) = 1.5 * 5000/732 * 250/256 = 10V

    You were very close, but the R21 and R19 relation is not quite that simple, hopefully my explanation above was clear enough. If not, please do let me know and I will try to explain it a little differently.

  • 0 in reply to Eugenio Mejia
    Prodigy 50 points

    Thanks so much Eugenio your explanation is  FABULOUS

    I noticed a minor typing error "DAC_gain = (REF_input_impedance + RA) / (RFB_input impedance + RB)"  instead of :

    DAC_gain = (RFB_input impedance + RB) /  (REF_input_impedance + RA)

    to clarify:

    => DAC_GAIN  = ( RFB(5K to 20K) + RB(Potenciometer 160K) ) / ( REF(5K to 20K) + RA (0Ohm)

    Gain = 8 When RFB and REF (5K) =>  RB = 35K ; RA= 0 

    Gain = 8 When RFB and REF (20K) =>  RB = 140K ; RA= 0

    Best Regards,

    David Garcia

  • 0 in reply to David Garcia1
    TI__Genius 10615 points

    Hi David,

    David Garcia1 said:

    I noticed a minor typing error "DAC_gain = (REF_input_impedance + RA) / (RFB_input impedance + RB)"  instead of :

    DAC_gain = (RFB_input impedance + RB) /  (REF_input_impedance + RA)

    Thank you very much for the correction. You are absolutely right, the correct equation is DAC_gain = (RFB_input impedance + RB) /  (REF_input_impedance + RA)

    David Garcia1 said:

    => DAC_GAIN  = ( RFB(5K to 20K) + RB(Potenciometer 160K) ) / ( REF(5K to 20K) + RA (0Ohm)

    Gain = 8 When RFB and REF (5K) =>  RB = 35K ; RA= 0 

    Gain = 8 When RFB and REF (20K) =>  RB = 140K ; RA= 0

    This is correct.