Hi,
I have two questions about the graph(Figure-24) in the LMV342 data sheet.
Could you please see the following question?
Could you please teach me?
Thanks in advance,
Toshiya
This thread has been locked.
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.
Hi,
I have two questions about the graph(Figure-24) in the LMV342 data sheet.
Could you please see the following question?
Could you please teach me?
Thanks in advance,
Toshiya
Hello Hiromatsu-san,
We do not have any information about the exact circuit used to develop Figure 24 in the LMV342 data-sheet. Commonly, the inverting configuration is used for these measurements. Practical values for RF and RG are 100kΩ and 1kΩ, respectively, for a closed-loop gain of -1000V/V (60dB). The minus sign indicates the output signal inversion.
The main points regarding Figure 24 is how the closed-loop gain rolls off after 2kHz, and how the capacitive load at the output affects the phase margin. The plot depicts how the phase margin deceases as the capacitance at the output is increased. If the phase margin decreases to zero and there is still closed-loop gain remaining the amplifier becomes an oscillator. This is because the feedback signal is in phase with the input signal.
Regards, Thomas
PA - Linear Products Applications Engineering
(on behalf of SLL)
Hello Thomas-san,
Thanks for your prompt response and explanation.
It is helpful for me.
Well, about the value of RF and RG.
If gain is -1000V/V, I think RF=1000kohm and RF=1kohm or RF=100kohm and RG=100ohm.
Which is correct?
Could you please teach me?
Regards,
Toshiya
Hello Toshiya,
You are correct my values result in a gain of -100V/V, not -1000V/V. I slipped a decimal place - sorry about that. You likely won't see much difference in the AC response if you decided to use either 1MΩ/1kΩ, or 100kΩ/100Ω for the RF/RG ratio. The feedback resistance RF interacts with the operational-amplifier input capacitance and that does add phase shift to the feedback loop. Using a larger value RF introduces the phase shift at a lower frequency and can degrade the phase margin which is displayed in Figure 24. Therefore, you may want to use the 100kΩ/100Ω values. But again I don't think you'll see a big difference in the responses.
Regards, Thomas
Hello Thomas-san,
Thanks for your prompt response.
I understand your explanation.
Regards,
Toshiya