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TLV9061: whether can support to drive 50ohm load

Part Number: TLV9061


Tool/software:

Hi team,

My customers use TLV9061 as voltage follower. The simulation is as below:

But in the actual test, they found that when they added a 50ohm load, like R17, the output which is 5~6mV, can't follow the input which is below 100mV; when they keep the load open, the output is normal which is same as input voltage. So I am not sure whether the Zo influence on the output, which is just 100ohm. Can it support to drive 50ohm load?

  • Hi team,

    If TLV9061 can't support, could you please help provide some other candidate which can support 50ohm load?

  • Hale,

    The EC table spec only go down to 2k ohms, so 50 is a big reach. However the small signal keeps output current low. The +/-16.5mV signal is just 330uA so that should be OK. Are they happy with the simulation result?

  • But in the actual test, they found that when they added a 50ohm load, like R17, the output which is 5~6mV, can't follow the input which is below 100mV

    Hi Ron,

    The customers didn't get the expected output voltage which should be same as input voltage.

  • Hale,

    Great a real circuit exists. Show me the waveforms at IN+ and OUT pins.

  • Hi Ron,

    The input is 120khz ~<100mV PWM signal. And if with 2k load ,the waveform is as below: 

    the frequency is 120k.

    And if with 50ohm, it seems no output, just some noise< 10mV.

    If with open load, the waveform is as below:

    Do yo think that 2.2uF capacitor influence on the op-amp output signal?  The customers use it as block DC signal and the simulation has no issue. But the actual test has obvious issue. Could you help check and give your suggestion? How to change the circuit to realize the voltage follower function?

  • Hale,

    Once again I ask for the input signal at U3 IN+. I can't know which, if any, output is correct, if I don't know the input signal.

    The input is 120khz ~<100mV PWM signal.

    At the moment I don't care about the far left input signal. The e2e post is about driving the load. If OUT is same as IN+ then the op amp did its job.

  • I believe the problem you see is caused by the short-circuit current limiting the charging rate of 2.2uF capacitive load.  Since current through capacitor is defined by Ic = C*dV/dt, solving the equation for max slew rate: dV/dt = Ic/C = 50e-3/2.2e-6 = 0.023V/us where Ic is the short-circuit current and C your capacitive load.