Rectification circuit with AC filtering - understand this?

[ACharnley]

Hi,

Just trying to follow through the logic but hitting a barrier with C10. It's a rectification circuit with switched in impedance matching capacitors.

The skhottkys are rated at 40V, the capacitors are 35V, the C10 is oddly 16V - and has polarity.

K1-3 are relays. The TVS is 170V and must be the wrong value.

There's a device N$50 which I've not seen before.

The back-to-back capacitors switch in at different AC Hz and pass the AC while boosting the edges. It's a clever technique. If they don't get switched in at the correct frequency they become blocking. I assume therefore one of these is active at any point.

But C10, that's a curiosity. If the [DYN ~] left side is positive it would charge but the 16V rating seems to be limiting. If the left side becomes negative C10 is shorted removing the need for D7.

The TVS is too high but perhaps those filter caps are removing the circuit completely once a given frequency (which translates into a max voltage) is reached.

Can you do better?

Regards,

Andrew

 

[AnalogKid]

The circuit fragment in post #1 has no power source, no ground, no inputs, no outputs, no relay coils or their controls or circuits, and no clue what a N$50 is.

Your description gives zero information about what the circuit is, what its intended function is, and whether or not it works. "Impedance matching capacitors" - what impedance, what is a match, what isn't a match?

Can you do better?

ak

 

[ACharnley]

It works. Dynamo power converter. Gnd is at the bottom. DYN1&2 are inputs. AC variable frequency. Output is DC at the top.

It's not my circuit.

There is a shunt later on and I'll work out the max DC tomorrow. Explains the TVS.

 

[ACharnley]

N$50 I think is related to K1, it has a pin 1 next to it. I theorise if DYN2 goes positive K1 switches C10 out of circuit.

C10 must change the filter frequency of C14+C13 / C6+C7, though I'm unable to work out how.