Choosing Choke

[diaz028]

I would like to know which choke to use with a Dual Rectifier. I am building a clone, but I just need to know values for henry's and series resistance load.

thanks !

-D

 

[ray_moth]

A little more information would help. Will the filter be choke-input or capacitor input? What will be the output voltage and demand current on the filter? Will you be using a tube or SS rectifier? What value capacitors do you intend using?

 

[Geek]

Hi,

A rule of thumb... take operating voltage and divide that by operating mA and you have a choke value.

So if the circuit runs at 400V at 100mA, choke would be 4H (5H is close enough).

Cheers!

 

[diaz028]

A little more information would help. Will the filter be choke-input or capacitor input? What will be the output voltage and demand current on the filter? Will you be using a tube or SS rectifier? What value capacitors do you intend using?

Thanks, yes I will come up with values soon, I am still in the drawing process Didn't really think about all this yet, i'll draw it up and show here.

Hi,

A rule of thumb... take operating voltage and divide that by operating mA and you have a choke value.

So if the circuit runs at 400V at 100mA, choke would be 4H (5H is close enough).

Cheers!

Awesome, sounds good to me - i'll still post here to see what you think ! Thanks

 

[Tweeker]

Geek is giving a bear minimum there. This choke must be sized to maintain choke operation at idle and max load.

In choke input supplies the choke carries some ac current on top of the dc current.

 

[Geek]

Hi Tweeker,

Geek is giving a bear minimum there.

Yes, true enough. If you have a class AB amp, use the *minimum* expected current draw for calculations.

Cheers!

 

[cerrem]

The purpose of the choke, on a cap input filter, is to work with the following filter capacitor to form a LC filter....
DC current capabilty should be based on MAX DC current draw, to keep the choke from saturating....give your self about 20% to 30% headroom with the max DC current..
The LC filter is a DOUBLE POLE filter....this means it attenuates at -40db/ decade.....
Your main objective is to smooth the ripple...in most cases it will be twice the mainis frequency...so a 60 Hz mains will create a 120Hz ripple in the B+ that needs to be attenuated to an acceptable level..depending on application....
Example....
You have a 400V cap input supply...with a 5v rms 120Hz ripple on the first cap.... You next have a 5H choke and a 32uF cap.... 1/[6.28* (LC)^.5 ] ....You find that this LC has a corner frequency at 12.5Hz ..... So the filter has -40dB/ decade ...meaning that at 125Hz you will be attenuated by -40dB .... This is fairly close to saying 120Hz.... SO since -40dB is the same as having a 1/100 attenuation in voltage amplitude.... So 5V/100 = 50mV rms ripple after this filter stage.... Actually you would need to include the load current and place a Resistive element in the analysis since loading plays a role in filter performance...as well as damping, since you would like your filters properly damped from a step response in the system....
Depending on the type of load, the filter can behave differently..such as a Resistive load vs a Constant Current load and whether or not the load has a negative impedance...

Chris