# What does the condition “f=100Hz” mean for the Gain Bandwidth Product?

In the LT6003 datasheet you can find the following information: How do I read the Gain Bandwidth Product? I mean there is a typical bandwidth of 2kHz under the condition of f=100Hz and I'm not sure what this condition could mean.

I also did a practical test with the LT6003:

• Simple buffer circuit
• Supply voltage of 10V
• Input signal: sine wave with 5V offset, 2Vpp

Interestingly the signal looks distorted for frequencies above of 150Hz.

## 2 Answers

It means that they test the gain at 100Hz, assume that it's slope is 20dB/decade at that point, and extrapolate the GBW product from there.

At 150Hz, the amplifier gain (and the loop gain in a buffer) is only a factor of 13 or so. That means that any inherent distortion in the amplifier is only divided by a factor of 13 -- and the whole notion of an op-amp is "overcome problems deriving from a crappy amplifier with tons of gain". "Tons of gain" varies by what you're trying to do, but I would say that $$\2^n\$$, where $$\n\$$ is the number of bits of precision you're interested in, is a good first approximation. So, if you're feeding a 16-bit ADC, you'd expect good performance with a loop gain better than 65000.

Conditions are the test conditions (which doesn't make sense in this case, because they tested it across many frequencies as shown in the Gain and Phase vs frequency plot on page 10 of the datasheet).

With a 2V input the max gain you could have without the signal railing out would be 5 (2V x 5 = 10V). I would also give the amp some head room on Vcc and run the rails higher at 10.2V because no amp is truly rail to rail.  • "which doesn't make sense in this case..." It makes sense if the graph is just a guideline, but the GBW product at 100Hz is what they intend to actually stand behind. – TimWescott Jan 24 at 19:44