Hi,

I found this circuit on one of op amp's application notes and I'm not sure why there's a diode bridge at the output of an op amp.
I initially thought the diode bridge circuit acts as a clipper and will limit Vout to about +/-5V. But when I ran simulation Vout actually rose above +/-5V (although distorted).
Can anyone explain why it's there and how it exactly works?

Thank you!

 

Can't see any obvious reason for the bridge, unless it's for gating the opamp output (by reversing the two supply rails to the bridge) to a following stage? Unlikely, though.

 

I think it may be producing a small dead band around zero volts output. Possibly so a control system did not hunt arround the zero output position. If you gave more information on the topic of the application note it might help.

Les.

 

It generates a clipped output at about 1 diode drop below the bridge supply voltages if you add a load to the output (try 50kΩ)

 

Hi,

I found this circuit on one of op amp's application notes

What did said application note say was the purpose of said circuit?

It's very unusual for an app note to just have a schematic without at least a meaningful name associated with it.

 

You might not be seeing the forest because of the trees. if the 5V supply was elsewhere and at 0V, it could protect the downstream device.

 

What did said application note say was the purpose of said circuit?

It's very unusual for an app note to just have a schematic without at least a meaningful name associated with it.

It didn't really have much information. It was just one of many circuit diagrams in their note

 

It generates a clipped output at about 1 diode drop below the bridge supply voltages if you add a load to the output (try 50kΩ)

Thanks crutschow! Yes, adding a load creates a clipped output