Jump to content

testing home-made generator


the guy
 Share

Recommended Posts

i am constructing a small home-made electrical generator as an experiment. it is very small and so will be low output.

 

what should i use to test it?

 

should i use a voltmeter or an ammeter? and where can i purchase one which will be sensitive enough?

Link to comment
Share on other sites

A voltmeter will show a voltage, whether or not it's connected to a circuit; and an ammeter will show a current only in a circuit. Most digital multimeters (they measure voltage, current, resistance, etc) should be sensitive enough for your use, and you can buy them in many stores. Can you give us an estimate of the voltage produced?

Link to comment
Share on other sites

i'm not sure about an estimate, perhaps you could help me with that?

 

basically the generator is a linear electric generator. the sliding magnet is small and cylindrical (1cm diameter, 1cm length). i don't know how many turns there will be in the coil but i aim for as many as possible.

 

the catch, however, is that the sliding magnet is not a permanent magnet itself. It is magnetized on demand by a close proximity external magnet (a super-strength neodymium magnet).

 

However i am no good at physics calculations, i was simply hoping to use a sensitive measuring device to see if anything was generated and how much.

 

but if you could help me with some predictions i would be very grateful.

Link to comment
Share on other sites

You can play with this, and leave H/m as is because this is the permeability of air . . .

 

You can calculate the inductive reactance by [math] X_L = 2 \pi f L [/math] using the L from the calculator. From this you can use [math] V_L = I_L X_L [/math] to calculate the voltage.

 

It's really easy stuff, don't tell yourself otherwise, it's basic multiplication . . . just plug in values for [math] I [/math] and [math] V [/math] that make you happy and take it from there!

Link to comment
Share on other sites

A couple of points I would mention.

 

 

Xittenn's formulae above are not applicable to the generation of electricity. They refer to the opposition to current flow presented by a pure inductor connected to an external source of alternating voltage. (or voltage across a pure inductor when alternating current flows through it).

 

 

The generation of voltage (e.m.f.) within the coil is dependant on the rate of change of magnetic flux and the number of turns the coil has. Note that this voltage is present (in fact at a maximum) when no current flows. So speed of movement , magnetic field strength, number of turns and length of the coil all come into it.

 

Now for some practical advice - Multimeters have a large range of scales. The safest way to ensure you don't damage the meter is to start with the largest scale and then work your way down through the ranges until you get a sensible reading. Just in case I forget, I always set my multimeter to the highest voltage scale when I have finished using it.

 

I think your idea of making it and seeing what comes out should be good fun - best of luck!

Edited by Joatmon
Link to comment
Share on other sites

Hence the link to the calculator . . . .

 

Calculate the inductance based on want . . . I couldn't find an emf calculator!

You don't need to consider the inductance of the coil. (or its reactance).

 

Consider a single loop with a magnet passing through it. An e.m.f. will be generated in the loop. The amount of e.m.f. generated will depend on the relative movement between magnet and coil. i.e. the rate of change of the flux linkage. Now consider a coil with many turns. e.m.f.'s will be generated in each loop and they will act in series with each other and thus add together.

 

http://hyperphysics....ric/farlaw.html

post-68560-0-82943600-1335016802_thumb.jpg

Edited by Joatmon
Link to comment
Share on other sites

i'm not sure about an estimate, perhaps you could help me with that?

 

basically the generator is a linear electric generator. the sliding magnet is small and cylindrical (1cm diameter, 1cm length). i don't know how many turns there will be in the coil but i aim for as many as possible.

 

the catch, however, is that the sliding magnet is not a permanent magnet itself. It is magnetized on demand by a close proximity external magnet (a super-strength neodymium magnet).

 

However i am no good at physics calculations, i was simply hoping to use a sensitive measuring device to see if anything was generated and how much.

 

but if you could help me with some predictions i would be very grateful.

 

 

You can play with this, and leave H/m as is because this is the permeability of air . . .

 

You can calculate the inductive reactance by [math] X_L = 2 \pi f L [/math] using the L from the calculator. From this you can use [math] V_L = I_L X_L [/math] to calculate the voltage.

 

It's really easy stuff, don't tell yourself otherwise, it's basic multiplication . . . just plug in values for [math] I [/math] and [math] V [/math] that make you happy and take it from there!

 

 

A couple of points I would mention.

 

 

Xittenn's formulae above are not applicable to the generation of electricity. They refer to the opposition to current flow presented by a pure inductor connected to an external source of alternating voltage. (or voltage across a pure inductor when alternating current flows through it).

 

 

The generation of voltage (e.m.f.) within the coil is dependant on the rate of change of magnetic flux and the number of turns the coil has. Note that this voltage is present (in fact at a maximum) when no current flows. So speed of movement , magnetic field strength, number of turns and length of the coil all come into it.

 

Now for some practical advice - Multimeters have a large range of scales. The safest way to ensure you don't damage the meter is to start with the largest scale and then work your way down through the ranges until you get a sensible reading. Just in case I forget, I always set my multimeter to the highest voltage scale when I have finished using it.

 

I think your idea of making it and seeing what comes out should be good fun - best of luck!

 

 

Last time I checked this is a Science Forum and I gave a practical scientific approach (hint, hint, wink, and stuff.) The approach given is simple and is used by many. There is a reason why there isn't an emf calculator, because it isn't a simple approach. If the OP wants to know how to use what I've posted to make his predictions, I am more than happy to help walk them through the process. If not that is quite alright as well.

Link to comment
Share on other sites

Last time I checked this is a Science Forum and I gave a practical scientific approach (hint, hint, wink, and stuff.) The approach given is simple and is used by many. There is a reason why there isn't an emf calculator, because it isn't a simple approach. If the OP wants to know how to use what I've posted to make his predictions, I am more than happy to help walk them through the process. If not that is quite alright as well.

 

I don't want to get into an argument but calculating the reactance of his coil and using current/voltage relationships for the coil will not help him estimate the expected output of his generator.

 

 

 

Link to comment
Share on other sites

I don't want to get into an argument but calculating the reactance of his coil and using current/voltage relationships for the coil will not help him estimate the expected output of his generator.

 

I did overlook the fact that my approach doesn't give enough information, this is still not a problem and really doesn't detract from my point (being there is nothing wrong with trying to do stuff with math.) I was too quick with my thinking and missed that we are left not knowing what either I or V will be and so we have no choice but to use the equation. Try correcting the problem next time or pointing it out, please stop shooting my replies down they are intended to open up the discussion!

 

Plug the current in until you get the equivalent B-field to your magnet, that was really tough . . . . . .

Link to comment
Share on other sites

Plug the current in until you get the equivalent B-field to your magnet, that was really tough . . . . . .

 

This does not help in any way with predicting the generator output. The B-field is a steady phenomenon and you suggest that driving current through the coil to produce the same B-field as the magnet will some how predict the generator output characteristics. You seem to suggest that the voltage connected to the coil and the current flowing through it will be the same as the voltage and current produced by the generator. This is (IMO) just not so. For instance the voltage coming out of the generator will depend on the speed of the magnet relative to the coil. How is this brought into your reasoning? Also when the coil is open circuit with no current flowing through it you will generate e.m.f. with no current. There are other problems that become apparent when you consider the effect of coil number of turns - I will explain these to you if you need.

Edited by Joatmon
Link to comment
Share on other sites

I hadn't seen this last post, it was added and changed or something . . . . .

 

The following is one example of the route that can be taken

 

step 1) define your the parameters of the solenoid

 

step 2) input the current until the B-field matches the magnet

 

step 3) input resultant inductance into the L parameter of [math] X_L = 2 \pi f L [/math] and set frequency of oscillation

 

step 4) solve for voltage [math] V_L = I_L X_L [/math]

 

step 5) adjust inputs or steps as necessary

 

Voltages and Currents are RMS!

 

** I don't suggest applying current to the solenoid, I suggest that passing a magnet through a solenoid creates a current.

Edited by Xittenn
Link to comment
Share on other sites

I tried to give you this as a personal message, but since I have been barred it has to be public.

 

I'll give you full marks for perseverance.

However, although I don't want to upset you, I will try to tell you what I see as the broken link in your analysis.

In step 1 you find the inductance of the solenoid, that's ok.

In step 2 you must be forcing dc current through the coil so it produces the steady magnetic field of the magnet. I don't see this as useful to your analysis.

In steps 3, 4 and 5 you do a good job of analysing how this solenoid will behave as (say) used as a load at the frequency you have set.

Now we come to the problem as I see it. The OP wants to know how much V he will get when he passes the magnet through the coil. It may be very little if the magnet moves slowly or be considerably greater if the magnet moves quickly. I don't see how you can give him this information from your analysis.

Furthermore, the coil will generate voltage, current flowing is dependent on any load connected to the solenoid. There isn't a constant V/I relationship independent of the load. As I have said - to me this is just a discussion between two people about a technical matter and I have never seen it as personal.

Jo.

Link to comment
Share on other sites

In step 2 you must be forcing dc current through the coil so it produces the steady magnetic field of the magnet. I don't see this as useful to your analysis.

 

No, I'm assuming that moving the magnet through a solenoid generates a current proportional to the magnets B-field and whose voltage is proportionate to the rate at which the magnet is oscillated through the coil in terms of frequency. I have no idea what you are suggesting by your comment. This is elementary E&M.

 

The OP is showing no interest, and no one else is interjecting comments. My logic is sound and I am moving on from yet another thread where I have presented an elementary process and where I have been refuted for doing so. Think about it Jo, as you are suggesting it there is no simple generalized solution, does this make any sense to you at all? It sounds like an excuse to not use the tools that you have been allotted in favour of doing things by hands on trial and error--a layman's excuse to avoid math and reason. This does not have to be addressed using calculus, the equations presented were designed to remove the calculus from the problem! My steps are my final answer . . . . .

Link to comment
Share on other sites

No, I'm assuming that moving the magnet through a solenoid generates a current proportional to the magnets B-field and whose voltage is proportionate to the rate at which the magnet is oscillated through the coil in terms of frequency. I have no idea what you are suggesting by your comment. This is elementary E&M.

 

The OP is showing no interest, and no one else is interjecting comments. My logic is sound and I am moving on from yet another thread where I have presented an elementary process and where I have been refuted for doing so. Think about it Jo, as you are suggesting it there is no simple generalized solution, does this make any sense to you at all? It sounds like an excuse to not use the tools that you have been allotted in favour of doing things by hands on trial and error--a layman's excuse to avoid math and reason. This does not have to be addressed using calculus, the equations presented were designed to remove the calculus from the problem! My steps are my final answer . . . . .

We shall just have to leave it at that then. My final say is that the maths is very simple and in accordance with the attachment. The formula does not need coil inductance or reactance - only the number of turns and rate of change of flux. I'm sad that this exchange got so acrimonious.

post-68560-0-64606500-1335209916_thumb.jpg

Edited by Joatmon
Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
 Share

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.