Hi everyone, now I'm looking for the current transformer to provides the power in case of single-wire (Neutral missing)
so I unofficially call "CT power supply" circuit as attached files
Circuit spec:
- when Iprimary = 1A to 60A , Output must be 3.3Vdc (load: 5mA) for supply MCU

So I would like to know how to select CT burden for this circuit?

 

Hi,
As per your drawing the primary input ,not mentioned how much voltage.
While the secondary output must be 3.3vdc.

You are referring to a step down transformer and a full wave dc output.

I Do not see the neutral that is missing. Unlike you want dual 3.3v ac with a center
tap in the secondary winding. Then the neutral will appear and need to be grounded.

 

it would be best if you could show the whole schematic, instead of this snippet.

 

For 5ma minimum output at 1A primary current you would select a CT turns ratio of no more than 1/.005 = 200:1.
The transformer must not saturate with a 3.3V + 1.4V = 4.7V (say 5V) output.

At 60A the CT output current will be 60 * 5mA = 300mA, so you could use a 3.3V shunt regulator at the bridge output that can dissipate 3.3V * 300mA = 1W.

For the shunt regulator you could use a TL431 with a transistor booster to absorb the power, as shown below.
The V(batt) is the bridge output with a filter capacitor.
The PNP transistor needs to dissipate at least 1W, such as a TIP32.
(For the TL431, Vref = 2.5V)


That sound like what you are looking for?

 

Hi everyone, now I'm looking for the current transformer to provides the power in case of single-wire (Neutral missing)
so I unofficially call "CT power supply" circuit as attached files
Circuit spec:
- when Iprimary = 1A to 60A , Output must be 3.3Vdc (load: 5mA) for supply MCU

So I would like to know how to select CT burden for this circuit?

What is the turns ratio of your CT? Do you have a datasheet? You question suggests the output should be 3.3V, 5mA which would suggest a resistance of 660 Ohms?

 

sorry for my question that is not clearly
previous circuit, I bring it from other application note
so I try to write new circuit as attached files

 

sorry for my question that is not clearly
previous circuit, I bring it from other application note
so I try to write new circuit as attached files

You can use a 3.3V shunt regulator, as I showed in post #4, to replace both the 5V zener (which has to dissipate up to 1.5W) and the series regulator.

 

Hi,
As per your drawing the primary input ,not mentioned how much voltage.
While the secondary output must be 3.3vdc.

You are referring to a step down transformer and a full wave dc output.

I Do not see the neutral that is missing. Unlike you want dual 3.3v ac with a center
tap in the secondary winding. Then the neutral will appear and need to be grounded.

sorry for my question that is not clearly
previous circuit, I bring it from other application note
so I try to write new circuit as attached files

 

it would be best if you could show the whole schematic, instead of this snippet.

Thank you for you advice, I try to write new circuit as attached files

 

I think you question and post is clear.
Did you understand my reply?

 

For 5ma minimum output at 1A primary current you would select a CT turns ratio of no more than 1/.005 = 200:1.
The transformer must not saturate with a 3.3V + 1.4V = 4.7V (say 5V) output.

At 60A the CT output current will be 60 * 5mA = 300mA, so you could use a 3.3V shunt regulator at the bridge output that can dissipate 3.3V * 300mA = 1W.

For the shunt regulator you could use a TL431 with a transistor booster to absorb the power, as shown below.
The V(batt) is the bridge output with a filter capacitor.
The PNP transistor needs to dissipate at least 1W, such as a TIP32.
(For the TL431, Vref = 2.5V)
View attachment 143784

That sound like what you are looking for?

yes, this is what i am looking for. but how to select the CT burden(in VA/Ohm)?
thank you!

 

At least in europe there should never be a connection between L or N and PE at the load, so you would not get any current through the load and no power for your circuit.

 

I think you question and post is clear.
Did you understand my reply?

yes I uderstand your reply,it is very useful for me
Thank you!

 

At least in europe there should never be a connection between L or N and PE at the load, so you would not get any current through the load and no power for your circuit.

yes this is not normal connection, it is tampering of energy meter from customer side

 

The normal solution would be an RCD breaker, but I guess you have your reasons.
The burden resistor (which would be either across the CT or across the bridge output) doesn´t seem to be needed in this case, as the zener replaces it. Your circuit in post #8 should work, and I would remove R1 completely and connect the zener directly. Not that the zener will dissipate 1.5W at 60A primary current, so a transistor zener like in post #4 or a high power zener will be needed.