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Old 24th August 2008, 06:59 PM   (permalink)
Default which diodes to use

hello everyone this is my first post,
my problem:
it should be simple for you experts, i'm trying to run a 12volt dc fan
i'm using a small 100Ma 240v ac transformer converted to 24vac with 2 red and 1 white output wires. (should be 12v output)
when i construct a bridge rectifier with 4 diodes which type of diode should i use ????????????
ive succsefully done this before using old diodes i found laying around then when i bought 4x n4001 diodes , i couldnt get it to work, all my parts are from maplins i didnt realise there were so many diffrent diodes out there
any help would be appreciated and put me out my frustrating misery
thanks
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Old 24th August 2008, 07:10 PM   (permalink)
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Hi,

Your transformer sounds like it is center tapped (red=12v - black=0v - red12v) If you connect to a red and the black you will (from your description!) get 12vAC, If you use both reds and leave the black disconnected you'll get 24vAC.

If you have 2 1N4001 diodes handy then connect one to each red wire (Anode) then connect both diode's cathodes together and use this as your +ve then use the black center tap as your ground, This will give you a rough DC source but it's ample to run a fan on, This way you'll also get double the current from your transformer as you're using both 'windings of the transformer.

Paul.

When you used 1N4001 diodes and they didn't work then you must have connected something wrong, These are ideal diodes for what you need this time, I'm sure someone will correct me if I'm wrong but I would say these diodes are pretty much a standard when building a bridge rectifier with LESS than 1000mA load.
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Last edited by hackableFM; 24th August 2008 at 07:14 PM.
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Old 24th August 2008, 07:39 PM   (permalink)
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yes thats it as you described a centre tapped transformer
( red + white + red)
about the diodes thats exactly what i found out using a schematic from the web, so i soldered them in and used only the red wires not the white/black but it didnt work.

Last edited by fuseblower; 24th August 2008 at 07:45 PM.
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Old 24th August 2008, 07:43 PM   (permalink)
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oops, I misread your post, I said black instead of white...

Paul.
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Old 25th August 2008, 01:25 PM   (permalink)
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I've just been looking for the spec for a diode I plan to use in my next project, I accidentally stumbled accross the spec for a 1N4148 diode while I was looking, If you plan to draw a maximum of 100mA then this 1N4148 will suffice for your needs as it is rated at 150mA. I originally though it was rated at 100mA which is the reason I didn't mention it in the first place as I like to use components which are 'over rated' for the job.
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Old 25th August 2008, 01:49 PM   (permalink)
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being a total novice i didnt think you could use over rated components
such as having a transformer with output 100ma

for example if i had a bridge rectifier consuming 10ma and a fan consuming 100ma then being 110ma it wouldnt work or be sluggish

similar thinking is if my rectifier consumed 1.4v a diode ( 4x1.4 = 5.6) then that leaves me 6.4v to operate my fan so im wrong to assume this way
cheers paul
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Old 25th August 2008, 02:57 PM   (permalink)
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The rectifier doesn't consume any current, it drops a small voltage.

It doesn't matter if the power supply an output a higher current than the rating of the motor but the motor can't use more power than the power supply can safely provide or it might overheat.
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Old 25th August 2008, 05:24 PM   (permalink)
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Quote:
Originally Posted by fuseblower View Post
a 12volt dc fan
What is the rated current or power or CFM of the fan?
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Old 25th August 2008, 07:41 PM   (permalink)
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Quote:
Originally Posted by fuseblower View Post
being a total novice i didnt think you could use over rated components
such as having a transformer with output 100ma

for example if i had a bridge rectifier consuming 10ma and a fan consuming 100ma then being 110ma it wouldnt work or be sluggish

similar thinking is if my rectifier consumed 1.4v a diode ( 4x1.4 = 5.6) then that leaves me 6.4v to operate my fan so im wrong to assume this way
cheers paul
As Hero has said, the rectifier doesn't 'Use' any current, It does however 'loose' some of your volts..... If you feed the rectifier with exactly 12volts you loose approx 0.6volts through each diode in use at any given moment. During each half cycle you use 2 diodes hence 1.2volt loss on the diodes. So you will get 10.8volts from your rectifier with the 12volt ac input..... Now to try and explain the current, If you have a fan that requires 12volt @ 100mA and your transformer & rectifier are rated at exactle 100mA then all will work fine although your components will be working to their maximum ability. Now if you use a transformer and rectifier rated at 500mA or higher then everything will work the same BUT your fan will only pull 100mA through the rectifier (One Fifth) of the transformers ability, the other 400mA will be available to use if you connect more load but it will not be 'pushed' through the regulator. Try and imagine that you 'PULL' the power from the transformer that you need, the transformer doesn't FORCE the power through.

I hope this helps you understand. This doesn't apply all of the time but this is an instance where you can over rate components.

Paul
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Old 25th August 2008, 10:42 PM   (permalink)
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well iv'e just had a look at maplins website and it is rated at 200ma and 14.8cfm
basically i've made an incubator with a 60w bulb heat source and dimmer switch for controlling the temperature, and the fan is for spreading the heat around.
i would like to add or make a thermostat in the future for it.
ive found a thermostat at

http://www.maplin.co.uk/Search.aspx?...source=15&SD=Y

but it only works between 5 and 30 degress i need at least 100 degrees
so could i re-arrange or fit another type of thermistor or anything to meet my requirments??????
thanks for all your helps
cheers paul
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Old 26th August 2008, 02:14 AM   (permalink)
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Example for a mid-class center tapped transformer calculation rated: 12V/100mA

Omitting the center tap for usage

No load multiplication factor = 1.65 (no load output voltage 39.6VAC)

reduced by forward diode drop of 0.7V per diode active in a half cycle using a full wave rectifier (1.4V) = 38.2VDC. Peak DC voltage (no load) 38.2 * √2 = 54.01V

With the center tap used for full wave rectification the voltage values will be halved - not increasing the current output since 1/2 of the secondary winding is used per half cycle.

For a center tapped transformer use two-way rectification with the outer taps connected to the anodes of rectifier diodes, the cathodes being clamped together. Use the center tap for common ground.

Using 1N4148 is probably done by Chinese manufacturers. I'd recommend an 1N4002 for safety reasons.

Using overkill for components will prevent your personal overkill!

Boncuk
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Old 26th August 2008, 04:26 AM   (permalink)
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I like the practice of using things rated 50-100% more than necessary. Sometimes a component blowing up is a good way to find out when something is wrong. Using 1N4002 instead of 1N4007 in a bridge rectifier, for instance.
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Old 26th August 2008, 06:47 AM   (permalink)
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Quote:
Originally Posted by speakerguy79 View Post
I like the practice of using things rated 50-100% more than necessary. Sometimes a component blowing up is a good way to find out when something is wrong. Using 1N4002 instead of 1N4007 in a bridge rectifier, for instance.
... while an 1N4002 is rated for lower voltage. Who needs a rectifier diode rated 1,000V in an electronic circuit?
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Old 26th August 2008, 09:04 AM   (permalink)
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Quote:
Originally Posted by Boncuk View Post
... while an 1N4002 is rated for lower voltage. Who needs a rectifier diode rated 1,000V in an electronic circuit?
They look the same and cost the same in most component shops.

Which one will you get?
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Old 26th August 2008, 06:09 PM   (permalink)
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Quote:
Originally Posted by hackableFM View Post
As Hero has said, the rectifier doesn't 'Use' any current, It does however 'loose' some of your volts..... If you feed the rectifier with exactly 12volts you loose approx 0.6volts through each diode in use at any given moment.
That's true but don't be surprised if you loose 1.1V per diode at higher currents.

Quote:
During each half cycle you use 2 diodes hence 1.2volt loss on the diodes. So you will get 10.8volts from your rectifier with the 12volt ac input.....
AC is normally specified at RMS, 12V is 17V peak so you'll get 15.8V DC on the output.
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