Hi All,

I have a project of designing 220V, 5A AC to 230V, 3A DC converter. I thought of using Full wave bridge rectifier circuit to serve the problem.

The components which I selected are:
1:1 transformer - Isolation purpose
four 1N5406 diodes
Capacitor of 150uF, 400V (electrolytic) and load resistance 200 ohms.

I believe LC filter could provide better output instead of capacitor alone. But i am stuck with designing of Inductor value. Suggest the suitable Inductor value.

Also pls let me know whether this circuit configuration would actas the 220v AC to 230V DC converter.

Thanks in advance

 

Hi.Potentially a filter with an inductance is a superior filter. But not ofter used on account of expense. A lot depends on what the load actually is. You mention a 200ohm resistive load. If it is purely resistive why won't 220ac do. Or unfiltered dc straight from the rectifier.
If the supply needs to be filtered a 150uf may not do a lot. A rough rule of thumb is 1000uf per expected 1 amp load. Bear in mind that with no load (or a small load) the voltage from a supply like that will rise towards 1.4 x 220 = 308vDC which is very dangerous.
As for actually finding a choke rated at 3 amps (the core will saturate if it's not rated for the current you draw) it may be expensive and hard to find. My memory is a bit foggy now on this stuff, because of the use of switched mode stuff, but values between 10 and 20 henries come to mind. It will also be big and heavy. A choke input filter with a small permanent load will prevent the off load voltage rising, but 200v dc is also dangerous. Use ohms law to see what the volts drop across a choke will be due to its resistance and factor that into the design.

 

Instead of 1:1 isolation transformer you will want a step down transformer. The calculation goes like this:

220 VACrms * √2 = 311 VACpk
311 VACpk - 1.6V = 309.4 VDC

I think what you need is a 4:3 stepdown transformer

220 VACrms * .75 = 165 VACrms
165 VACrms * √2 = 233 VACpk
233 VACpk - 1.6 = 231.74 VDC

The 1.6V is the drop for the bridge rectifier.

 

Hi All,

I have a project of designing 220V, 5A AC to 230V, 3A DC converter. I thought of using Full wave bridge rectifier circuit to serve the problem.

The components which I selected are:
1:1 transformer - Isolation purpose
four 1N5406 diodes
Capacitor of 150uF, 400V (electrolytic) and load resistance 200 ohms.

I believe LC filter could provide better output instead of capacitor alone. But i am stuck with designing of Inductor value. Suggest the suitable Inductor value.

Also pls let me know whether this circuit configuration would actas the 220v AC to 230V DC converter.

Thanks in advance

The 600V 1N5406 is just barely the absolute bare minimum you should think about using.

I'd go for the 5408 rated 1000V - and still protect them from spikes with an MOV.

 

sorry about the load.. I am going to use this DC voltage to drive Power MOSFET which works as switch. The Vds handling of such MOSFET is 600V. So to provide such high voltage im developing this circuit.

 

Instead of 1:1 isolation transformer you will want a step down transformer. The calculation goes like this:

220 VACrms * √2 = 311 VACpk
311 VACpk - 1.6V = 309.4 VDC

I think what you need is a 4:3 stepdown transformer

220 VACrms * .75 = 165 VACrms
165 VACrms * √2 = 233 VACpk
233 VACpk - 1.6 = 231.74 VDC

The 1.6V is the drop for the bridge rectifier.

sorry i didnt understood 4:3 step down transformer calculations. can it be more explanative. also how to specify the number of turns

 

I did specify the turns ratio. It is 4:3, or 400:300, or whatever you like so that:

220 VACrms (on the primary) * .75 = 165 VACrms (on the secondary)
165 VACrms * √2 = 233 VACpk
233 VACpk - 1.6 = 231.74 VDC

and now you have your 230VDC plus a little bit. I don't understand, what could you not follow?

 

I did specify the turns ratio. It is 4:3, or 400:300, or whatever you like so that:

220 VACrms (on the primary) * .75 = 165 VACrms (on the secondary)
165 VACrms * √2 = 233 VACpk
233 VACpk - 1.6 = 231.74 VDC

and now you have your 230VDC plus a little bit. I don't understand, what could you not follow?

sry.. now I got the calculation.

 

sry.. now I got the calculation.

Np/Ns = Is/Ip, according to this formula, the secondary current is 6.66A, the diodes which I specified either 1N5406 or 5408 can handle only 3A. does diode need to be changed?

 

Hi All,

I have a project of designing 220V, 5A AC to 230V, 3A DC converter. I thought of using Full wave bridge rectifier circuit to serve the problem.

The components which I selected are:
1:1 transformer - Isolation purpose
four 1N5406 diodes
Capacitor of 150uF, 400V (electrolytic) and load resistance 200 ohms.

I believe LC filter could provide better output instead of capacitor alone. But i am stuck with designing of Inductor value. Suggest the suitable Inductor value.

Also pls let me know whether this circuit configuration would actas the 220v AC to 230V DC converter.

Thanks in advance

I am puzzled that I might be missing something here as I don't understand PapaBravo's advice..But here's what what I think.
Your are starting with a mains supply which is nominally 220v. For some reason you want a slightly higher DC voltage (230). In practice, given that the mains voltage may vary a bit I think you are right to say you need a 1:1 transformer - for isolation purposes, very sensible. As an isolation transformer may well have taps for different input mains voltages and the 210v tap (if present) might give a small step up. More ratio possibilities exist if you use the transformer in reverse. But it's unlikely that such precision is required unless you tell us why it's needed.
You mention a supply current of 5amps and a load current of 3 amps. In practice the rms current through the 1:1 transformer will be close toequal for the primary and secondary - 3 amps out requires 3 amps in.
Actually, if the load ie 200ohms (resistive) the current will be a tad over 1 amp.
PapaBravos calculation seems to be worked out on the peak voltage a capacitor input filter circuit will produce off load. In practice this will sag to nearer 220volts under load. The exact value will depend on the value of the capacitance, the actual load current and the resistance of the transformer windings.
The diodes you've chosen are, as has already been said, are too close to tolerance in this circuit. In practice you would be better off buying a bridge rectifier as a single unit. A quick look on eBay found a 10amp 1000v one for £1.68. Easier to mount if you are actually going to build this converter.
You want DC because you intend switching something (a resistive load of 200 ohms as far as we know ) with a FET. Such a load may not need smoothing and could be run from rectified DC direct. Thryristor control would also be a solution in this case.
PapaBravo or someone will correct me if I've misunstood you requirements or have just plain got it wrong.....

 

nice information

 

My original point was that getting to 230VDC from 310VAC(peak) is going to create a heat management problem if you draw any appreciable current. Eighty volts times 3 amperes is 240 Watt of heat you must get rid of. It is true that transformers may have taps on both sides to adjust the ratio of input to output. My idea is to arrange things so that the peak AC output is just a bit above the desired DC output. How much? That is a question for the TS/OP. But hey you folks can really do anything you want; you don't need my permission.

 

My original point was that getting to 230VDC from 310VAC(peak) is going to create a heat management problem if you draw any appreciable current. Eighty volts times 3 amperes is 240 Watt of heat you must get rid of. It is true that transformers may have taps on both sides to adjust the ratio of input to output. My idea is to arrange things so that the peak AC output is just a bit above the desired DC output. How much? That is a question for the TS/OP. But hey you folks can really do anything you want; you don't need my permission.

First off I would say again that if the load is resistive there is no call for smoothing (with the info we have at present ) so there is no requirement at all for heat management of the 80v drop. It doesn't exist. It's all a matter of analogue power supply design. The diodes barely conduct when the voltage rises to 310 dc and no current can be drawn without a ripple voltage appearing and the average voltage dropping.
Theoretically, if the diodes and transformer could supply huge short pulses and the capacitors were unrealistically big, the voltage may stay higher than 220 in the presence of a 1 amp load (let alone a 3 amp load).
But in the real world with typical value caps (1000uf for 1 amp load) there will be ripple and an average voltage of around 220 DC with ripple (power under the curve) to the load.

 

I am puzzled that I might be missing something here as I don't understand PapaBravo's advice..But here's what what I think.
Your are starting with a mains supply which is nominally 220v. For some reason you want a slightly higher DC voltage (230). In practice, given that the mains voltage may vary a bit .....

Somewhere I vaguely remember reading that the UK mains voltage specification was a bit of a fudge.

To fall into line with the rest of Europe; they didn't actually change the voltage they delivered - they 'modified' the + & - % limits to get away with it.