Looking for a 300W .7 ohm resistor?

[things]

We are doing some zinc plating, but the power supply supplies more current than we need, We are trying to "split" the current, but are going to need something like a 300W resistor. Anyone know who stocks these, do they even exist??? I realise there will be alot of waste heat, but i dont think there is any better cheap way to do it.

 

[Pommie]

To dissipate 300W in a 0.7Ω resistor requires 15V across it. What voltage is your supply and how are you intending to split it.

Mike.

 

[Diver300]

https://uk.farnell.com/652611

There is a range of those, so you could parallel a 1 hm: with a 2.2 hm: (which would only need to be 100 W)hm: to get 0.7 hm:

 

[stevez]

While wire isn't free you might consider that copper wire does have resistance - other less suitable conductors have greater resistance. Heaters are more or less power resistors.

A shunt is added to an ammeter to extend the range and can often be made from a coil of wire - so long as you stick with direct current. In your case you might have some long extension cords around that you might use. What you need to do is select a wire that provides the resistance without getting so hot it causes a problem or melts. I did a quick check and you'd need 277 ft of no 14 wire to get 0.7 ohms. Using both conductors (series, not parallel) of some extension cords might get you what you need.

You might take a look at McMaster Carr - they handle a lot of stuff. I purchased some heating elements at very low cost.

As Diver300 already suggested - multiple resistors in parallel might be a reasonable approach.

Auto headlamps in parallel are another possibility.

 

[things]

The power supply is rated at 12V @ 25A. We need to dissipate about 18A of that. If i were to get a big resistor, would it require any sort of special cooling?

 

[stevez]

The cooling needs of the "big resistor" are likely to have been established by the manufacturer. Some resistors have enough surface area so that natural convection (just letting the air circulate around it) is sufficient. Sometimes forced air cooling is required. It all depends on how hot it can get - again, mfr would indicate that.

Quite often the ratings for such items is described - and possibly under ideal conditions. As your situations depart from the ideal you might need to de-rate the capacity of the resistor. Example - the ambient temperature in your situation might exceed the mfr's specs - you'd need to compensate for that.

 

[JimB]

The power supply is rated at 12V @ 25A. We need to dissipate about 18A of that. If i were to get a big resistor, would it require any sort of special cooling?

Hold on, I think that I detect a misunderstanding here.

Just because a PSU is rated at 25A, does not mean that it will always supply 25A to the load. The load will take what it wants, you dont have to "dissipate" 18A (I assume to make the total load 25A).

What current does your plating bath take when connected to the PSU?
7amps?
If 7A is what you need, job done, you dont need to waste the other 18amps to warm up the room.

However, if your plating bath takes more than 7A, you need a resistor in series with the supply, not a resistor in parallel to "split the current".

JimB

 

[Pommie]

Hold on, I think that I detect a misunderstanding here.

Just because a PSU is rated at 25A, does not mean that it will always supply 25A to the load. The load will take what it wants, you dont have to "dissipate" 18A (I assume to make the total load 25A).

What current does your plating bath take when connected to the PSU?
7amps?
If 7A is what you need, job done, you dont need to waste the other 18amps to warm up the room.

However, if your plating bath takes more than 7A, you need a resistor in series with the supply, not a resistor in parallel to "split the current".

JimB

The reason I asked the earlier question is because I thought plating supplies were fairly low voltage. A resistor that would drop 15V at the stated wattage seemed odd. In reality, they probably only need to drop a couple of volts at 7A and a 50W resistor will easily handle it.

I thought they wanted to split it between two baths.

Mike.

 

[JimB]

The reason I asked the earlier question is because I thought plating supplies were fairly low voltage. A resistor that would drop 15V at the stated wattage seemed odd. In reality, they probably only need to drop a couple of volts at 7A and a 50W resistor will easily handle it.

I thought they wanted to split it between two baths.

Mike.

Another fine example of a badly stated problem, ambiguous and open to several different interpretations.

JimB

 

[chemelec]

The power supply is rated at 12V @ 25A. We need to dissipate about 18A of that. If i were to get a big resistor, would it require any sort of special cooling?

A power resistor like that really isn't practical for Electro-plating.
The Current you require for electro-plating needs to be adjustable as the current required varies somewhat, depending on the Area of the object being electro-plated and also the PH of your plating solution.

If your power supply is a Transformer type of power supply, Put a Variac on the input of it to give you really good control of the output.

If not, Put a "PWM" circuit on its output to control the Duty cycle to control the current.

 

[Areal Person]

TEST: I'm trying to post but it did not work

 

[Pommie]

TEST: I'm trying to post but it did not work

Yes it did.

Mike.

 

[RODALCO]

A variac would be the way to go at the primary side of the supply.

It is a lot more efficient and full voltage control is available.

If a variac is not available or too expensive, do primary resistance control, use a lampbank to control the transformator.

A lampbank can be made up of 5 x 100 watt lamps which can be parrallelled up by means of switches to give a range of 100 - 500 Watts in series with the primary winding of the supply TX.

Regards, Raymond

 

[things]

jimB, the bath draws more than we want it to, thats why we need to waste some current. Don't get me wrong, i know about this stuff, its just that i'm having a hard time finding something that will dissipate 18A without vaporising, and itsnt going to take up the whole room

 

[Boncuk]

How about some experimenting?

A 0.75 square mm cable can carry a load of 1.500W max (already getting warm) take off the insulation of the cable (sorry, I don't know the English term for it, but it consists of many wires). Split the bunch to half and wind them on a beer bottle the way they won't short. Then connect the electrolytic bath and measure the current flow. Adapt it by jumping from one winding to the next.

If the current is OK, don't change anything on the windings. Fix them with good glue (epoxy) on the bottle and hide it in some place. Applying PWM for electrolytic processes is not recommendable. The plating won't be nice and shiny but rough and ugly. Try it with copper on iron. It really doesn't look too good applying to much current like it is the case using PWM. It's the peak current which counts for plating.

Your voltage seems a little high to me. Applying chrome the nominal voltage is 1.5V. You don't want to boil the material in the bath. I use 2.5V max for double sided PCBs and 1A current flow per 10X10cm of surface.

Regards

Hans

P.S. If you want to do plating extraordinarily well apply reverse pulses from time to time. Reverse pulsing prevents an extra deposit at drills and holes.

 

[pfofit]

We are doing some zinc plating, but the power supply supplies more current than we need,

Well, the "power supply" supplies what the load wants.

The power supply is rated at 12V @ 25A. We need to dissipate about 18A of that

As has been alluded to, just 'cause a supply is rated at 25 A, does not mean it will supply that unless the request is there.
My cars speedometer says it can do 280KMs/hour, but..... it ain't been nowwhere near that cause my foot and brain controls it.

jimB, the bath draws more than we want it to, thats why we need to waste some current.

Instead of wasting current, why don't you regulate it?

So, facts please:

1. How much current is being drawn ?

2. What do you want to limit it to?

3. What is the PS output voltage doing when the load is connected?

cheers

 

[dknguyen]

The power supply is rated at 12V @ 25A. We need to dissipate about 18A of that. If i were to get a big resistor, would it require any sort of special cooling?

THe power supply can supply up to 25A but it won't do that unless the load resistance is low enough. Since you are using a voltage source (and not a current source) you don't "dissipate 18A of that". You "dissipate X volts of that". Adding a resistor in series with the load makes the resistor take up some of the voltage being output from the power supply which reduces the voltage on the load which reduces the current.

So you would need to know the resistance of your bath and if it varies (since it is a bath with possibly changing conduction levels and volumes) then a simple resistor won't do. You would need to adjust the voltage based on the current the power supply is outputting (turning it into a current source).

Since your description is incorrect, it is difficult for people to recommend stuff to you since they don't know if you just understand your problem but just worded it wrong (ie. you experimentally determined the bath was drawing too much current) or if you calculated it wrongly (in which case everything is outright wrong and you need to revisit the whole thing again).

BTW, a 0.7R resistor will only dissipate a maximum of 205W with 12V across it (P = V^2/R). So I think you misunderstand it outright. THe power supply will try and keep the VOLTAGE constant and "adjust" the current to do so (within it's limits). It won't try and keep the current constant (unless it is a very special power supply and you would know if it was). The 25A rating is a MAXIMUM rating. And the 12V rating is what the power supply tries to keep constant.