feedbackresistorssample and hold
What role does the 30k resistor play in the LF398?
How would that resistor eliminate any offsets introduced by the op-amps in a circuit like this one:
On this explanation, I don't get this line:
These diodes then require the 30k resistor to avoid overloading the
output amplifier.
A peak detector is something of a sample and hold that samples all the time, and holds the peak:
Follow the input with this, and connect the output to your ADC. Make C significantly larger than the capacitance used by your ADC's sample and hold, or follow it with a buffer, so the voltage across C doesn't sag as you read it.
Trigger an ADC reading just after the event happens, and when it's done, close the switch to reset the peak, or make the switch a resistor forming an RC time constant significantly longer than the time it takes to measure the peak but significantly shorter than the interval between peaks to accept a little error but avoid the need to reset the peak detector.
I've seen some comments here stating that the pulldown is needed to keep the transistor turned off, instead of floating, or for noise reasons. The base pulldown does in fact help keep the transistor off, but no one has answered why. This is why I chose to answer, rather than extend the comments. User alexan_e is thinking correctly about it, so I'll elaborate here.
There is a Miller capacitance between the collector and the base of all BJTs. Most designers know all too well about the MOSFET's Miller Capacitance and forget that the BJT has some as well. The BJTs Miller capacitance provides a leakage path from collector to base, injecting charge carriers into the base region, which can be amplified by the BJT's Hfe (gain). This allows noise current to flow from collector to emitter. The inclusion of the base pulldown will provide a path to ground to discharge the Miller capacitance and keep the BJT hard off and noise free.
Best Answer
When the sample switch is open, there is no feedback around the first buffer, which means that any voltage difference between its inputs will cause its output to saturate. The diodes prevent this from happening, which would slow it down. They keep the output voltage of that buffer within ± one diode drop of the signal input.
However, when the sample switch is closed, you want feedback around both buffers in order to cancel out internal offsets caused by the switch and/or the second buffer. In this mode, the feedback insures that both inputs of the the first buffer are equal, and the outputs of the two buffers are (nearly) equal. The diodes do not conduct in this mode.
The resistor is required in order to isolate the outputs of the two buffers from each other when the switch is open. Each buffer has a low-impedance output, and if they were connected directly together (even through the diodes), large amounts of current would flow. The resistor limits this current to a value that does not affect the accuracy of the second buffer's output.