I'm looking at a pretty simple overvoltage/transient protection circuit designed to protect users but I don't understand what the transformer is doing.

The circuit is designed to protect the output from seeing mains or transients generated by mains (2kV+) as the output is technically user accessible.

What I don't understand what the transformer does given its a 1:1 ratio. Surely the mains/transients will get to the output before the fuse blows?


simulate this circuit – Schematic created using CircuitLab

Extra: From my own reading online, it looks like it provides galvanic isolation, but I still don't understand the concept of this keeps a user safe.


Mains transformers protect in two ways:

  • By providing isolation
  • By limiting energy with a saturable core

Transformers usually have a core made of a material that has a saturation point (like ferrite), this limits the magnetic energy that can be transfered from one coil to another. This is nice because if there is a short on the secondary, it won't get the full power of AC mains. The transformer should have a power rating, like 24VA, this lets you know how much power can bet transferred before saturation (or heating).

enter image description here
Source: https://www.quora.com/What-is-the-core-saturation-of-a-current-transformer

The second thing a transformer does is provide isolation, which means no DC current through the core. This is nice for saftey reasons because only AC currents can get through the core, spurious or unwanted DC signals are blocked. It also means that the secondary side can be grounded to the chassis, in the event of a fault the current flows into the chassis and out to ground instead of creating a potential.

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    \$\begingroup\$ Sort of, but I don't agree with "this limits the magnetic energy that can be transfered from one coil to another". The core does not see how much current is being transfered, only how much it's excited due to voltage-time area on the primary due to Vt=NAB. In this particular case, said limitation in voltage-time area works to you advantage along with leakage inductance, resistance and your hopefully saturated core will shunt a lot of current in the primary. \$\endgroup\$ – winny Mar 15 '19 at 15:42

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