# Understanding an oscilloscope schematic

Discussion in 'General Electronics Discussion' started by darksize, Oct 26, 2013.

1. ### darksize

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Oct 26, 2013
Hi all,

I'm looking at this schematic: http://www.rlocman.ru/i/Image/2011/01/06/XPROTOLAB_sch.jpg

channel 1 and 2 signals are taken as input by amplifiers u3a and u3b, the output of these non-inverting amplifiers are followed each one by a network formed by two resistor and one capacitor (for channel 1the network formed by R1,R4 and C2).

Can someone explain me the function of these network at the output of the opamps?

Thank you very much in advance!

2. ### Harald KappModeratorModerator

11,440
2,627
Nov 17, 2011
R1 R4 make a voltage divider. Since R4 is connected to +2V (not, as you'd expect to GND), this circuit also level-shifts the bidirectional signal (+/-) from the opamp U3 to a unidirectional signal. E.g. *-2V become 0...+2V. This is required for teh A/D coverter of the ATMEGA which probably doesn't operate on negative input voltages.
The capacitor removes high frequency signals for antialiasing.

3. ### KrisBlueNZSadly passed away in 2015

8,393
1,271
Nov 28, 2011
R1 and R4 are there to add a 1V offset to the output of the op-amp so the signal going into the MCU is centred around 1V, the half-scale voltage for the ADCs.

The op-amps are DC-coupled buffers with total gains of 0.2. When the input is floating or grounded, their outputs will be at 0V. The ADC in the MCU is operating with a VREF of 2.0V from U3C so its input range is 0V~+2V. The two 10k resistors shift the no-signal voltage at the ADC inputs from 0V to +1V. They also attenuate the signal by a factor of 2, making the total gain from input to ADC equal to 0.1. With the ADC input at 2V p-p the voltage range at the input connectors is ±10V.

The capacitors provide a simple single-pole low-pass filter. The -3dB frequency is calculated as f = 1 / (2 pi R C). For R = 5000 ohms (a voltage divider using two 10k resistors has a Thevenin equivalent resistance of 5k) and C = 100 pF, this is:

f = 1 / (2 pi 5000 100e-12)
= 1 / 3.1416e-6
= 318,310 Hz.

This will be a simple anti-aliasing filter to reject high-frequency components in the input signal that would cause aliasing due to the ADC sampling rate.

4. ### darksize

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Oct 26, 2013
thank you very much guys!!!!