can please someone explain this circuit to me

[fala]

Hello, I have a question about a paragraph what I have read in datasheet of OP400. the complete paragraph has quoted here plus the corresponding figure, but for those who want the complete datasheet here is the link:
https://www.analog.com/UploadedFiles/Data_Sheets/3803980939304OP400_c.pdf

Quote OP400 datasheet from Analog devices:
DIFFERENTIAL OUTPUT INSTRUMENTATION
AMPLIFIER
The output voltage swing of a single-ended instrumentation
amplifier is limited by the supplies, normally at ±15 V, to a
maximum of 24 V p-p. The differential output instrumentation
amplifier of Figure 6 can provide an output voltage swing of
48 V p-p when operated with ±15 V supplies. The extended
output swing is due to the opposite polarity of the outputs. Both
outputs will swing 24 V p-p but with opposite polarity, for a
total output voltage swing of 48 V p-p. The reference input can
be used to set a common-mode output voltage over the range
±10 V. PSRR of the amplifier is less than 1 uV/V with CMRR
(G = 1000) better than 115 dB. Offset voltage drift is typically
0.4 uV/C over the military temperature range.

OP400 is a package of four OpAmps with just two supply voltage pins(±V) for all four OpAmps. I really don’t understand it, maybe because English is not my native language. How it is possible to have 48V pp voltage at output with just one ±15V supply. I am also don’t understand the circuit completely so I will be grateful if someone explain it to me so I understand how it works. I understand this much that first two amplifiers(u1, u2) just amplify differential signals because their non-invert pins are connected together by RG and for common mode they are just followers. But u1,u2,u3,u4 never cannot accept inputs beyond ±15 nor they can provide outputs beyond that, thank you in advance!

 

[shetoos_asu_65]

look i dont understand it but i will ask for it the faculity if u can wait 1 day it will b very good and i will answer if they understand it
sorry for not solving it now
bye

 

[fala]

of course I can wait, thank you for your kindness.

 

[elmolla]

Hi,
The OpAmps U1, U2 and U3 form an ordinary differential amplifier which is an amplifier with very high input impedance and very low output impedance; I think its familier, this part has output ±24 Vpp if the output is from U3 to ground.

U4 acts as an inverting amplifier with unity gain. ie. its output is shifted by 180° from the input or simply it inverts (multiplies by -1) the input. This way if the output of U3 is Vo, the output of U4 is -Vo relative to ground. Now, if we take the output between the terminals of U3 and U4 making U3 the positive terminal, the ouput will be Vo-(-Vo)=2Vo, this way the output will be in the range of ±48 Vpp.

In other words, the output is differential between two terminals, which are opposite in polarity, if one is 1V the other is -1V and hence the differential output voltage is 2 V. Hence if the maximum is 24Vpp for each OpAmp alone, the differential will have output between 48 Vpp

Added after 5 minutes:

Hi,
The OpAmps U1, U2 and U3 form an ordinary differential amplifier which is an amplifier with very high input impedance and very low output impedance; I think its familier, this part has output ±24 Vpp if the output is from U3 to ground.

U4 acts as an inverting amplifier with unity gain. ie. its output is shifted by 180° from the input or simply it inverts (multiplies by -1) the input. This way if the output of U3 is Vo, the output of U4 is -Vo relative to ground. Now, if we take the output between the terminals of U3 and U4 making U3 the positive terminal, the ouput will be Vo-(-Vo)=2Vo, this way the output will be in the range of ±48 Vpp.

In other words, the output is differential between two terminals, which are opposite in polarity, if one is 1V the other is -1V and hence the differential output voltage is 2 V. Hence if the maximum is 24Vpp for each OpAmp alone, the differential will have output between 48 Vpp.