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bad slew rate??

Mauricio Pagano
Prodigy 40 points
Other Parts Discussed in Thread: LM324, MSP430F1232

Hi,
to measure the settling time of a multiplexer (CD4051), opamp (LM324) and ADC, I assembled the following circuit. First select the input to ground (CH1) of the multiplexer and then select the entry with 50mV (CH0).
In the oscilloscope see the following signs:
ch2 -> pin 3 of CD4051 (multiplexer)
ch1 -> output of opamp That is connected to an MSP430F1232 channel ADC (10bit ADC,Rs:2Kohm, C:27pF).

The problem I have is that the signal have a great delay.!!!
The data sheet says that the LM324 opamp has a slew rate of 0.5V/uV much less what I'm seeing in the oscilloscope.

What hapened here?

thank,

regard,

(sorry for my the elemental english...)

Mauricio.

this pic is a zoom:

  • 0
    TI__Guru 63790 points

    Mauricio,

    Instead of probing ch2 -> pin 3 of CD4051 (multiplexer), use it to probe pin 3 of the LM324 (positive input of the op amp) or remove the R40 from your circuit (it is not needed) but it is most likely to cause a long delay while interacting with the common-mode input capacitance of LM324 and resulting in a long time-constant.

    Also, you seem to use +/-5V supply voltage while the PDS spec for slew rate of 0.5V/us is given only for +/-15V (see below)

     


     

  • 0 in reply to Marek Lis
    Genius 5430 points

    Mauricio;

    It looks to me like you are bandwidth limited, not slew rate limited. A much faster op amp is needed here-- or much less gain. You could get high gain bandwidth by using two op amps in cascade instead of one. Put each op amp in a gain of sq rt (total gain).

  • 0 in reply to Neil Albaugh
    Prodigy 40 points

    thanks for the responses,

    Marek I removed R40 and I did not get many changes.
    I think after doing some calculations, as commented Neil,  the problem is that this OpAmp have low bandwidth for my aplication.

    In PDS the unity gain bandwidth is 1.2Mhz, and with a gain at closed loop Gcl=50 then the BWcl=24Khz !

    The rise time (10%-20%) of the signal out (oscilloscope ch1) is aprox. 18uSeg, then as:

    rt=2.2*tau
    tau=8.18uS

    then fBW=1/(6.28*tau) = 20Khz  that freq.  is consistent with the calculated.

     

    best regard,

     

    Mauricio.