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.

# TLV9004-Q1: Output resistance at DC conditions and how it should be accounted for in design

Part Number: TLV9004-Q1

Hi,

I am using the TLV9004-Q1 in a photodiode amplifier setup, and am not sure how to treat the output impedance under quasi-DC conditions. The datasheet states the amplifier has ~1200 Ohm output resistance at f = 100kHz and this drops to approximately 1100 ohm down to f = 1kHz. The plots do not show what happens at lower frequencies however. Furthermore, in the application section, a photodiode amplifier example is given using resistor values close to this 1200 ohm output impedance, but this is not taken to account in the design equations. I use this as an example to ask if the 1200Ohm output resistance is an important design condition to consider when using resistors in the 1e3 Ohm range with low frequencies, and what does the output impedance taper become at low frequencies close to DC? If it remains in this approximate value range, then it makes good sense to simply subtract this value from the resistor used for feedback.

Thank you

• Hey Austin,

I would leverage the 1 MHz, Single-Supply, Photodiode Amplifier Reference Design, the design doesn't cover the output impedance of the device in any of the calculations.
The output impedance mostly comes into consideration when trying to evaluate the circuit for stability, in this case the design covers a rate of closure evaluation for stability.

Furthermore, in the application section, a photodiode amplifier example is given using resistor values close to this 1200 ohm output impedance, but this is not taken to account in the design equations. I use this as an example to ask if the 1200Ohm output resistance is an important design condition to consider when using resistors in the 1e3 Ohm range with low frequencies, and what does the output impedance taper become at low frequencies close to DC? If it remains in this approximate value range, then it makes good sense to simply subtract this value from the resistor used for feedback.

I am not sure what is meant by this portion of the question...could you rephrase?
I don't see resistor values close to 1200 ohm output impedance:

All the best,
Carolina

• Hi Carolina,

My apologies, I meant to reference the current sense application, just in relation to that resistor. I am not targeting this design and only mentioned it as the design uses a 1.2K resistor and in the equations provided the 1.2K output impedance is not taken into account, so I was not sure if output impedance must be considered in low frequency applications when trying to be accurate. In the photodiode amplifier circuit I would agree the 1.2K is neglagible on 309k.

Thank you

• Hey Austin,

Understood, I don't believe the output impedance of the amplifier needs to be taken into consideration in the calculation for DC application. In the case of the current sense application seen in the datasheet, RG = 1.2k, is a coincidence that it matches the output impedance. There is a secondary reference design, Single-supply, low-side, unidirectional current-sensing circuit, that uses the same amplifier (dual channel) but has a different RG (in this case R2).

The design considerations in the circuit that is linked do not mention output impedance. However, as I mentioned earlier output impedance in consideration with the open loop gain and the load on the output of the op amp affects the phase margin (or the rate of closure) in consideration of stability.

We have a training on stability in TIPL: https://www.ti.com/video/series/ti-precision-labs-op-amps.html

We are also able to help with stability analysis through e2e if you share your schematic.

All the best,
Carolina

• Hi Austin,

I agree with Caro. The 1200R you mention is the open-loop output impedance. But the open-loop output impedance is not what you will see in a circuit with closed feedback loop. You will only see the closed-loop output impedance which is much smaller than the open-loop output impedance, at least at the lower frequencies. Divide the open-loop output impedance by the loop gain and you will roughly get the closed-loop output impedance. (Open-loop gain divided by closed-loop gain = loop gain.)

You can easily simulate the closed-loop output impedance by this test circuit, by the way:

austin_tlv9002.TSC

The "gain" dispayed on the y-axis means "Vout divided by IG1" which is the closed-loop output impedance. The unit is "Ohm", not "db" by the way. I better should have corrected it

You can see that at the lower frequencies the closed-loop output impedance is negligible and even plays no role at all at DC, as Caro already mentioned. But at frequencies higher than let's say 10kHz it becomes more and more relevant. This is no surprise for a 1MHz OPAmp. If you really want to handle 100kHz then you would want to choose a faster OPAmp, since a 1MHz OPAmp offers a gain reserve (loop gain) of only 20dB (10V/V) at 100kHz which may not be enough in a precision application.

Kai

• Thank you so much Carolina and Kai! This makes great sense now.