On Nov 29, 9:42*am, MarkAren wrote:
Hi Joop,

Interesting circuit.

Where does the R1/D1/Q/.Q2 reference point connect to ?

I assume the parasitic diodes in Q1/Q2 form a current path ?

-Mark
....

The idea is that by using two MOSFETs in (anti)series like that, the
parasitic diodes can't conduct through from q1.drain to q2.drain.
However, they (the parasitic diodes) insure that q1.source=q2.source
is at most a diode drop higher than the lower of q1.drain and
q2.drain. Then if the photocoupler is "on", the gates are pulled up
toward close to the most positive voltage in the circuit, limited of
course by the zener diode to protect against gate-source breakdown.
In that state, both Q1 and Q2 are hard on. They are happy to conduct
as "on" FETs in either direction. If the photocoupler is "off," the
gate-source potential drops to essentially zero, and both FETs are
off, leaving only the back-to-back parasitic diodes.

Essentially the same circuit is used in certain optically coupled
relays, such as the Panasonic "PhotoMOS" relays, and in other places.
Although the "on" resistance is twice as high as a single MOSFET of
the same type, the circuit blocks both polarities when "off," and
conducts either direction when "on."

Cheers,
Tom