Calculating a needed resistance (18.2v to 15v [AC])

[hofmann]

I need to convert 18.2v AC into 15v AC using a resistor in series.

Is there an online calculator or an App (great business idea )  I'm terrible at maths...

Thank you!

[danadak]

If you do not care about regulation then just a series dropping R works as long
as the load current does not change.

E = I x R, R = Edrop / Iload, E the drop you are trying to achieve. All values RMS or Peak
or Average, but all the same category.


Regards, Dana.

[hofmann]

If you do not care about regulation then just a series dropping R works as long
as the load current does not change.

E = I x R, R = Edrop / Iload, E the drop you are trying to achieve. All values RMS or Peak
or Average, but all the same category.


Regards, Dana.

Thanks Dana, but this is confusing to a maths illiterate dummy like me 

Can you show me the exact calculations? 18.2v to 15v =how many ohms?

I found this http://www.gtsparkplugs.com/Ohms_Law_Calc.html but don't know how to use it.

Thank you

[Cliff Matthews]

4 to 6 pairs of parallel back-to-back diodes in series might be an option. But as the diode junctions heat-up, the Vf drop (per diode pair) will go down.. so you'll want to know the anticipated current to avoid too much variance.

[danadak]

R = Edrop / Iload

You know the drop you require, you have a load that daws a known current........

E is in Volts, I in Amps, R in Ohms.


Regards, Dana.

[danadak]

Use a 3.3V zener, then load current changes will not affect Vdrop unless
I load goes to 0 or becomes so high it burns zener out.

Pick a zener with a dissipation rating of Vz x Iload. There will be thermal
design considerations for this as well. Same as R, Presistor = Vdrop x Iload.


Regards, Dana.

[hofmann]

Use a 3.3V zener, then load current changes will not affect Vdrop unless
I load goes to 0 or becomes so high it burns zener out.

Pick a zener with a dissipation rating of Vz x Iload. There will be thermal
design considerations for this as well. Same as R, Presistor = Vdrop x Iload.


Regards, Dana.

Thanks Dana, so use a 3.3v Zener? By the way the device draws 1 amp. What does this mean? dissipation rating of Vz x Iload

Thanks again

[digsys]

errr because it is AC, you will need 2 Zeners in parrallel, opposing polarities. 3.3V x 1A = 3.3W
That's a fair bit of heat, even shared, so 5W may be a little light. If you can get 10W Zeners, or ones in a TO220 tab and heatsink them. all the better.

[Cliff Matthews]

The zener will conduct as a normal diode in the opposite direction, so they should be rated at 2.7v and series connected.

[Cliff Matthews]

8 x 4amp recifier diodes (like MUR420's) should be cheaper than the zener solution, though.. (about $3 USD)

[danadak]

Thanks digsys and Cliff for pointing out the obvious, eg. AC,
I temporarily lost my marbles on this one.


Regards, Dana.

[hofmann]

Thanks cliff and digsys 

I still want AC out the other side. Will these diodes create DC?

8 x 4amp recifier diodes (like MUR420's)? This is the answer? like this? http://au.mouser.com/ProductDetail/Vishay-Semiconductors/MUR420-E3-54/?qs=cfR2GKjf2RmZMQwrlOf4Ew%3D%3D
How do I connect these 8 diodes?

[Cliff Matthews]

No they will pass current on both half's of the cycle, so it will still be AC, albeit with some zero crossing parts missing as the diodes begin to conduct on each half of the waveform (could be filtered a bit with a 50-cent cap if it's for an audio circuit).

[IanB]

Thanks Dana, but this is confusing to a maths illiterate dummy like me 

Can you show me the exact calculations? 18.2v to 15v =how many ohms?

I've never understood how people can be so proud of being incapable, and not be embarrassed about it 

Elementary arithmetic is no harder than speaking or writing English. You apparently can speak English, so you can do the calculations.

You should curl up and die of shame creating a post like this.

[hofmann]

Thanks Cliff

How do I connect these 8 diodes?

This is the circuit;





Using 240vAC (Australia) I have installed this http://80.93.56.75/pdf/0/5/5/2/1/05521864.pdf but instead of the 15vAC its spitting out 18.2vAC.

Ian. You should go eat shit. Troll. This section is for beginners.

The device needs 15v

Thanks Cliff 

[Brumby]

By the way the device draws 1 amp.

This was the critical piece of information missing from your original question.  With it, this question can be easily answered.  Even so, there is one further important detail that must be clarified - that this 1 amp will be constant.

Ohm's Law will give you the answer.  The formula is  V = IR  Volts = Current x Resistance  **

So, since the Voltage will be the voltage across the Resistor, V = 18.2 - 15 which is 3.2V
The formula now looks like this: 3.2 = 1 x R
R is obviously 3.2 ohms

If you want to get the generic formula for working out the resistance, then you just apply a little algebra.  Take V = IR and divide both sides by 'I'.  this gives you  V/I = R.  Plug in the numbers above and R = 3.2/1 ... same answer!

Now we know the resistance, we have to make sure we get one that can withstand the power it will need to handle.  This is done using another formula - which is just as simple as the one above....
P = VI  where P is the power (in Watts) with V and I being the same as we have just used.
Substituting for V and I (plugging in the actual numbers) we get:
Power = 3.2 x 1  which is 3.2 Watts

What does this mean? dissipation rating of Vz x Iload

We just worked it out.  Dissipation just means 'the amount of power that has to be got rid of'.

So, you will need a resistor with a value of 3.2 ohms and a power rating of at least 3.2 watts.  The resistance value needs to be 3.2 ohms (or close enough) for the voltage drop to be correct, but its power rating can be more.  Choose a resistor with a lower power rating and it is going to heat up far too much and it will fail.



** You can blame the French for 'I' representing Current - but don't get too worked up about it ... between Current and Capacitance, one of them had to lose out.
Also, just to add confusion, 'E' or 'U' can sometimes be used for Voltage.

[IanB]

Is no one going to point out that the transformer is expected to produce 18 V and that no circuit modifications are needed?

[Cliff Matthews]

You will not need To do anything at 18volts AC. All regulators in your circuit can handle at least 30v DC.

BTW, don't start off on the wrong foot here. All forums have sardonic humour. and.. Ian is not a troll, he is one of the most respected and super helpful members on this forum. Period. 

[hofmann]

You will not need To do anything at 18volts AC. All regulators in your circuit can handle at least 30v DC.

BTW, don't start off on the wrong foot here. All forums have sardonic humour. and.. Ian is not a troll, he is one of the most respected and super helpful members on this forum. Period.

Thanks Again Cliff!

Ok so my circuit is good without dropping the voltage to 15vAC before my PSU MOD. 

Cliff, having been a member of a few electronics DIY forums there are always trolls. They are a pain in the ass. You may believe this about Ian but unfortunately he has proved your view of him wrong.
He is being unhelpful and disrespectful. Period. Regardless of his desired humor.
Sorry to disagree with you here Cliff. If Ian wants to continue to be "one of the most respected and super helpful members on this forum" he should show some respect to beginners also. 

Again Cliff. Thank you sir  I appreciate your help.

[Brumby]

Is no one going to point out that the transformer is expected to produce 18 V and that no circuit modifications are needed?

I answered the original question.  The application circuit was not published until I was about to post.  In reviewing the subsequent posts, I decided to still post my response - and then come back with a further comment on the application .... but Cliff beat me to it:

You will not need To do anything at 18volts AC. All regulators in your circuit can handle at least 30v DC.

I would also support this statement as well ...

BTW, don't start off on the wrong foot here. All forums have sardonic humour. and.. Ian is not a troll, he is one of the most respected and super helpful members on this forum. Period. 

Even if it came off as abrupt, you must not let your fear of maths get the better of you - you are going to NEED some mathematical competency if you want to go further in electronics.

[hofmann]

Thanks all

[Brumby]

You may believe this about Ian but unfortunately he has proved your view of him wrong.
He is being unhelpful and disrespectful. Period.

If you are basing your opinion on one post of his, then you are not being particularly fair.

We have had more than one instance of people coming here and posting stupid questions and 'playing' with people who have spent a lot of time trying to help them.  These are the trolls we have to deal with - and sometimes the differentiation between them and real beginners will take a few posts....

[hofmann]

Thanks Brumby. 

I can only form an opinion based in interaction. I don't have a crystal ball or the time to brows through all of Ians posts.

As far as 'playing stupid' I think we are going into paranoid territory here. 

I am a genuine beginner electronics DIYer and I thank you all for your help.

Enjoy our day/night 

[Brumby]

Don't think I was accusing you of 'playing stupid'.  Sorry if I gave you that impression. That was in reference to people who deliberately put on an act to just cause trouble.

True beginners have the problem of sounding the same at the outset - but it might take a few posts for the difference to become clear.

[hofmann]