# Failed to build Instrumentation Amplifier

I'm trying to build an instrumentation amplifier, following the schematic attached below. I got two LM358P chips, signal generated by a computer sound card and an oscilloscope. The schematic and overall assembly attached below.

I got two types of resistors here. The blue ones are 1KOhm, the orange are 3.3KOhm. So that Rgain and R2 are 1kOhm, R1 and R3 are 3.3kOhm. I was trying to get the connections clear on the photo, and here's the best I could get. According to formulas on Internet, I should get amplification of around 10 times. Or at least any. But what I see on the scope is the same signal with values below zero being cut (ch1 is output, ch2 is input).

What's even more confusing, if I don't use the second stage of the amplifier and monitor the signal before R2 - it goes even more weird.

I assume I'm taking measurements wrong. Or my assembly is wrong. Could someone explain what is going on and why this whole thing does not work?

• How is the circuit powered? May 23 at 10:59
• 5v dc from a lab power supply May 23 at 11:00
• If the signals go negative, and your op-amps are powered with only positive, then the negative half cycle will be cut off. May 23 at 11:16
• That makes sense... But can it affect amplification? (absence of it? ) May 23 at 11:22
• Yes because all the voltage below the CM input range is not seen. May 23 at 12:00

There are a few issues you have to solve regarding the first circuit. For the second, I will only comment that it seems that you have shorted the opamp output to ground (via the oscilloscope ground clip), but it may be parallax error in trying to understand the circuit from the picture.

1. To see positive and negative voltages at the outputs (including the internal nodes), you need a symmetric power supply

2. If I applied correctly the information on your text to the schematic, there is gain in both stages, so:

$${A_v} = \left (1 + {2 R_1 \over R_\mathrm{gain}} \right ) {R_3 \over R_2} = \left (1 + {2 \times 3.3 \over 1} \right ) {3.3 \over 1} = 25.08$$

1. From the datasheet, the opamp inputs should be within this range:

Which means 0 V to 3 V with your current power supply. You have not shown the input in the schematic, but if there is no DC offset, the negative part of the input is lost (but also some of the positive part - see below).

1. Another problem is that the outputs (including the two opamps in the first stage) should remain within:

Again, this must also be valid for the three opamp outputs. This means that the lower opamp at the input is basically dead (its output maximum value would barely reach the valid range).

If the differential input is 50 mV (peak) the peak-to-peak output voltage will be aprox. 2.5 V. This doesn't leave much room unless you can apply a DC voltage to the input and output signals, which could be something like this:

It seems it would be much easier to use a symmetric power supply.

• Yes, this setup works fine. I still have lot of concepts to understand, and this gives me a point to start off. Thank you! May 24 at 10:08