DC supply without transformer

[electricsnake]

Dear All,

I found attached circuit that drop voltage V1 from 100 VAC to 5 V DC at the right C5.

I know that this circuit uses the Xc value of C1 to drop down the voltage and then fix with Z1 th output to 5 V at C5, however I still no well understand how it works.

In fact, is it C1 D1 a clamper?
Which is the function of R1?

In addition, I know that is C1 capacitance is too low, then we have risk that at the output of C5 the DC value is too low. But how to calculate which is the limit value of C1 that can ensure me 5 V?
In fact I need to make such C1 as small as possible due to space problem.

Many thanks in advance!

electricsnake

 

[duke37]

Any value of C1 should give you 5V, it will depend on load.

C1 has a fairly high impedance at your unknown mains frequency. Thus this controls current. When the current passes one way it takes current from D1 and in the other direction, it pushes it through D2.

This current is smoothed with capacitors and split between the output and the zener diode. Too much input current will fry the zener. Too little input current will not power the load sufficiently.

I think R1 is there to discharge C1 when the circuit is turned off.

The current with 1µf, 100V, 50Hz will be about 30mA

Remember that the circuit is live to the mains and so needs to be installed in a box to stop it and its load being touched.

 

[davenn]

electricsnake

what you are wanting to play with here is extremely dangerous!!

I hope your life insurance is paid up to date for when your family find you bar-b-cued

in the interests of safety, the thread is locked as we don't encourage this kind of activity

Dave