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Tool/software:
Hi team,
I am using TLV9041 in my design (see the attached schematic).
TLV9041 is used as a non-inverting amplifier. The input to the Op amp is a 0-3.3V 1KHz 5% duty signal directly from an MCU GPIO pin.
What I don't understand is why the current consumption is a lot higher than what is mentioned on the datasheet.
Please see the attached images where I capture the current data.
In my experiment, I left the output open, i.e. V_amp is not connected to anything.
We can see that the average current is nearly 2mA.
Please help me understand what could be the issue.
If you need further information, please let me know.
Thanks.
This circuit tries to force the output to a voltage higher than the supply. This saturates the output stage, which indeed increases power usage.
What is the purpose of this circuit? You say it is an amplifier, but with input voltages of either 0 V or 3.3 V, the output will also be 0 V or 3.3 V. There might be better circuits or devices, depending on what you're trying to do.
Hi Nhan,
Interesting, 2mA is far too high! You mention using a 3.3V PWM signal feeding into the OPA, but this would not be good to use as this will cause the OPA to saturate the output due to the high gain configuration.
GV/V= Rf/R1 +1 = 31V/V. OPA has a 3.3V rail and cannot output VOH of 31V/V * 3.3V
What was your input when running this Iq test? I suspect this difference in expected Iq from measured Iq is due to the OPA being non-linear. You can even see this in the Iq graph when the Iq drops to normal conditions when the output is presumably in linear operation.
What is the goal of this circuit? Are we trying to buffer a PWM signal?
Thanks,
Jacob
Thank you for your responses, Clemens and Jacob!
Thanks to you now I know about op-amp's saturation causing high supply current.
About my application:
The input signal in my application is actually in the 0-200mV range (it will be ambient light dependent). I need to amplify it into the 0-3.3V range.
The reason I choose to set the gain so high (gain = 31 in the schematic) is because if the max output voltage is < 3V it won't be very useful for the subsequent processing circuit, which is a missing pulse detector. With a very high gain, although the op-amp will be saturated but the output voltage's amplitude will be what I want. Below is captured waveforms:
My application will have low power consumption requirement. Therefore I think I should try to avoid the op-amp saturation as much as possible.
What should I do to avoid saturation, if I still want to have a very high gain?
Thanks.
If you are comparing the peaks against a fixed (or somewhat dynamic) reference voltage, then you can use a comparator like the TLV7031 instead.
Thanks, Clemens. My input will be pulsing (see the cyan waveform in previous figure) so I can't use a comparator here. I also don't get what you mean by "dynamic reference voltage".
If the reference voltage would not have to change quickly, you would be able to use a DAC.
What are you doing with the V_amp signal? How do you differentiate between good and bad peaks?
Nhan,
How low power does this have to be?
I tested your circuit and confirmed in lab that this behavior exists.
Notably, this Iq change is quite a bit lower when the OPA is configured in less closed loop gain.
The best way to solve this problem involves avoiding this saturation condition. The OPA is becoming unhappy as the input pins are being pulled apart due to the output clipping. Perhaps a unique external diode clamping structure on the input stage of the amp could prevent the inputs from pulling significantly apart from one-another.
Can you change to a different OPA? I tried OPA310IDCKR and the Iq in saturation is about 370uA and Iq normal is about 170uA with the feedback elements. I suspect this lower saturation Iq is in part due to OPA310's unique ESD diode structure. Would this Iq be tolerable in your system?
Please let me know what you think.
Thanks,
Jacob