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I have come across this circuit on a mock paper, which is asking me to find the overall voltage gain of the circuit, it is a bit of a scary looking circuit and i'm not quite sure where to start, any help would be massively appreciated!

enter image description here

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    \$\begingroup\$ Cropping the image would probably be a good start. \$\endgroup\$ – uint128_t Mar 23 '17 at 0:09
  • \$\begingroup\$ You don't need to make the image bigger, you need to make it smaller and eliminate the white space \$\endgroup\$ – Voltage Spike Mar 23 '17 at 17:43
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Don't be scared about this rather simple circuit. C9+R3+R12 and C5+R13+C8 is just a high pass filters. R1+R12 and R7+R8 form voltage dividers to offset X3 (non-inverting amplifier) and X1 (inverting amplifier) respectively. C3 and C7 give a little bit of integrating abilities to them. R3+R12 also forms voltage divider that will affect voltage gain. Q1 almost do not make any voltage gain because it just an source follower. So, overall gain will include voltage divider (R3+R12), gain of non-inverting amplifier (X3) and gain of inverting amplifier (X1) if signal lays in pass zone of all filters:

G = [R12/(R3+R12)] x [1+R5/R4] x [-R9/R6] ≈ -4604

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  • \$\begingroup\$ Thank you for your reply, would that answer be in db? Also I am trying to work out the voltage at the inputs and outputs of each of the op amps, so far I have got 5v/R1+R12 X R12, which gives me 2v and the same for X1, so that is 2v at + input on both op amps, so i'm guessing that means the - inputs are also 2v? I then rearranged the formula on x3 to get 2V/RF X RT, which is 2v/10k X 2210K = 442V. Am I on the right track with this or have I got muddled up somewhere? EDIT; I've since realized it's not in db, and to use 20log to convert it to db! \$\endgroup\$ – Jdoe Mar 29 '17 at 15:24

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