I was going through a circuit and got confused on a bridge part. I am not very good in circuit analysis. Can anyone guide me the working of attached circuit and what is the role of bridge in such kind of circuit. is this precision rectifier or schottky mixer or what. Thank you [![enter image description here][1]][1]
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\$\begingroup\$ Where did the circuit come from? \$\endgroup\$– Andy akaCommented Apr 30, 2019 at 8:52
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\$\begingroup\$ log amplifier circuit \$\endgroup\$– user777304Commented Apr 30, 2019 at 9:53
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\$\begingroup\$ Are you sure you copied the circuit correctly? It is possible to use a diode bridge in a log amplifier, but the way yours is wired, it simply functions as a linear amplifier. \$\endgroup\$– Dave TweedCommented Apr 30, 2019 at 18:15
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1 Answer
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The circuit as shown is not a log amplifier at all, but it does have different gain at different voltages, in a piecewise-linear fashion.
The 13k7 resistors establish a certain amount of current through the diode bridge — call this I0. The 989r resistor adds or subtracts current on one side of the bridge, and the same amount of current must flow on the other side (the side connected to the opamp input) — call this Iin.
As long as |Iin| is less than I0 the 989r and the 1k5 are effectively in parallel. In conjunction with the 1k on the right, these set a certain gain for the circuit.
However, as soon as |Iin| exceeds I0, note that two of the diodes in the bridge stop conducting altogether. This limits the input current to the opamp to I0, effectively eliminating the 989r from the input path. The 1k5 and the 1k now establish a new, lower gain for input voltages in this range.
Overall, the amplifier has an input-output transfer curve that is steep in the center and less steep at the ends. With the component values shown, the breakpoints occur at about ±0.84 V. I suppose you could consider this a crude approximation to a logarithmic curve.
answered Apr 30, 2019 at 18:54
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\$\begingroup\$ Can you suggest any topic through which i can understanding such circuit working. \$\endgroup\$ Commented May 1, 2019 at 12:09
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\$\begingroup\$ I believe we could call this transfer function a compressor since the gain is lower for higher input levels. \$\endgroup\$ Commented May 1, 2019 at 12:14
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\$\begingroup\$ How you calculate the ±0.84 V breakpoint \$\endgroup\$ Commented May 1, 2019 at 12:25
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\$\begingroup\$ Can you please explain how gain is lower for higher input levels. i don't know much circuit analysis \$\endgroup\$ Commented May 1, 2019 at 12:26
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\$\begingroup\$ please guide how you calculated the ±0.84 V breakpoints. \$\endgroup\$ Commented May 2, 2019 at 14:30