melting my bridge Recitifer

[deadcoffee]

Hi, i have a very simple project going on. On part takes 120v AC through a Transformer to 12.6vAC then a bridge Recitifer to get 12v DC - which then goes on to power the board. I need to take 12vdc from the recitfier and send it off to heat a thin nickle-crome filiment to red hot. (so i just have two wires going from the +/- ends of the rectifier to the this filiment

My problem is of course the filiment gets hot but then the rectifier smokes, putting a resistor in the loop gets the resistor hot or puts in too much resisitance so i don't get the Infrared red radiation i need.

if someone could help a newbie out and point me in the right direction, how to i isolate the filiment from the power supply, or how do i keep the current from feeding back.

Thanks in advance,

Ben

 

[Sebi]

Why need DC for a heating element?
If this is a resistor wire, need to know the ohm/meter value for current calculation.

 

[deadcoffee]

heating a filiment

Hey thanks for the quick reply, i have no good reason to heat the filiment from the DC power supply. I have the other electronics working fine and i'm just trying to figure out the best way to heat the filiment off the same power supply.

The filiment is a 1inch helically wound nickle-cromium wire which has about 0.558 ohms resistance. I could heat the wire using the 12.6vac from the transformer - my goal here is to get the filiment heated without melting anything or causing undesirable feedback.

 

[stevez]

Sounds like you have way too much load on the rectifier or you could use a heat sink (or both). Doing a little reading on ohms law might help here.

The filament is a resistor - apply a voltage and current flows - voltage divided by resistance gives current in amperes. 12 volts with a 1/2 ohm resistor will yeild a current of 24 amps but only if the supply can provide that - otherwise something has to give. Now, when the nichrome gets hot it's likely that the resistance will increase somewhat but to what extent I don't know.

Adding a resistor does increase the total resistance of the circuit and will reduce the current - how much depends on the resistor and the filament. Whether a filament or resistor - current flow thru a resistor yeilds heat. A small resistor will get hot and a larger one may only get warm - hot or warm being relative terms.

What you need to do - understand the current limits of your transformer and your rectifier. I'd put a fuse or circuit breaker in the secondary (12 volt side) so you don't ruin anything or start a fire. Next you need to understand more about the filament and how much power/current/voltage you need. Your transformer and rectifier capacity must equal or exceed the requirements of the load - the load is the filament and whatever else you have on the circuit.

 

[Klaus]

A neat way to infinitely regulate how hot your filament gets is to place a Variable transformer (variac?) ahead of your 120V AC transformer.
You thus adjust the *input* voltage to your transformer, which
carries low current, to get your variable output voltage for the heated wire.
You may or may not choose to leave the rectifier in place - I would leave it out unless there is a compelling reason to heat the wire with DC.

Variacs- if you don't own one already - are very useful for testing equipment. Keep in mind that they are not an isolated design so the output voltage has a direct connection to the input *mains potential!*
You may be able to get one second hand, they can be a bit pricey when bought new.
Klaus

 

[Platonas]

The current rating of your rectifier must be greater than the current supplied by the transformer. If for example the transformer is 120VA, then it can supply under normal conditions about 9A. So your bridge must be rated at least at 20A.
The above is just a rule. Your problem of overheating is due to the voltage drop in the diodes of the rectifier. You can choose a rectifier with lower forward volt drop for each diode. Otherwise, use a heatsink for the rectifier, a large one.
If we suppose a 10A load and a volt drop of .7V per diode then you have 10*4*0.7=28W of head on the bridge, so you have to cool it down with a headsink.
Regards
P.

 

[deadcoffee]

thanks so much for all the replies. here is where i stand. I have a big Tranformer converting 120v to 12vac and then into a bridge rectifier to convert to DC which is required by the rest of the board. The last step of finding a good way to generate IR radiation at 4.26 um wavelenghts.

The solution i have to to heat a nickle-crome filiment and filter it through a saphire lens which should work great. I'm still stumped on how to heat the filiment on using the original power source without draining the power from the rest of the circuit. I tried hooking the leads to the filiment to the 12vAC right from the transformer and noticed that the volts i'm getting from the rectifier would slowly drop as the filiment heated until i was at 0.

I've solved most of my overheating problems with a heat sink. My real challange is figuring out how to isolate the filiment from the rest of the circuit so i only draw the power i need.

Sorry if this is so newbie-like, i'm learning a lot but i'm a software engineer by day.

 

[Someone Electro]

whats the curent the trasformer is making?

why not use an IR LED to make the IR light?

To get tha curent mesure it whith an multimeter.Make shure if yours can mesure high curents if not you can fry it(mine has not problems since it can take up to 20A)

 

[Russlk]

The wavelength your are looking for is longer than available IR LEDs, so the hot wire is the way to go. Lets assume the wire is hot with 5 watts, then I = (P/R)^.5 = 3 amps. The voltage required is: I*R = 1.67 volts. You need to determine the actual power required, then make sure your components can supply the current and voltage (not too much voltage).