Is Digi-Key the Best Source for Parts Information and Ordering?

[superkaho]

When the rotor is turned, it is common to measure the voltage by multimeter.

However, there is a problem that I cannot fully understand.
When I directly measure the voltage by connecting the two terminals of generator, it is found that the measurement is not same as the voltage across a load(say 50ohm) in series with the generator, why?? Which reading is more reliable?

 

[vk6kro]

Do you mean like this:

[PLAIN]http://dl.dropbox.com/u/4222062/generator%202.PNG

The generator has internal resistance due to the resistance of the windings and the resistance of the brush contacts.

When current flows through this internal resistance, voltage drop occurs and this reduces the output voltage.

So, you would measure a lower voltage across the resistance than if you removed the resistance.

Both measurements are OK. The open circuit voltage is higher but fairly useless as there is no power being supplied by the generator.

 

[superkaho]

Do you mean like this:

Both measurements are OK. The open circuit voltage is higher but fairly useless as there is no power being supplied by the generator.

Yes, you are right. The cirucuit diagram is very clear, thanks.

But I still have some further questions...
Why the open ciruit voltage is useless when there is no power supplied by generator?
Moreover, if I want to measure the power output of the generator.
Is it correct to measure both the voltage and current across the resistor?
And then P = VI ??
How should I connect the circuit when measuring the current?

 

[vk6kro]

Yes, you are right. The cirucuit diagram is very clear, thanks.

But I still have some further questions...
Why the open ciruit voltage is useless when there is no power supplied by generator?
Moreover, if I want to measure the power output of the generator.
Is it correct to measure both the voltage and current across the resistor?
And then P = VI ??
How should I connect the circuit when measuring the current?

The open circuit voltage is useless if you are trying to get power out of the generator, because without current there is no power.

You can measure the power without measuring the current.
The current is voltage divided by resistance.
If you know the voltage and resistance, Power = Voltage times (Voltage divided by resistance)
or power = (voltage squared ) / resistance

For example... if your generator produced 5 volts across 50 ohms,
the power would be
5 times 5 divided by 50
or 0.5 watts

 

[superkaho]

Many thanks, I get it!~

 

[superkaho]

The open circuit voltage is useless if you are trying to get power out of the generator, because without current there is no power.

You can measure the power without measuring the current.
The current is voltage divided by resistance.
If you know the voltage and resistance, Power = Voltage times (Voltage divided by resistance)
or power = (voltage squared ) / resistance

Suppose there are eight mini turbines each genearate voltage with 1.5V, if I want to measure the total power produced by them.

Should I connect them in parallel or in series circuit with the 50ohm load?
Which method is more accurate?

If they are in parallel, the voltage across the load would be around 1.5V instead of 1.5V*8. Should I also measure the current in that case?

If they are in series, the voltage across the load would be around 1.5V*8, can I calculate the power directly by the "power = (voltage squared ) / resistance"?

i.e. Total Power = (1.5*8)^2/50 = 2.88V??
Would it not reasonable to have 64 multiple of power generated by 8 turbines?

 

[vk6kro]

Suppose there are eight mini turbines each genearate voltage with 1.5V, if I want to measure the total power produced by them.

Should I connect them in parallel or in series circuit with the 50ohm load?
Which method is more accurate?

If they are in parallel, the voltage across the load would be around 1.5V instead of 1.5V*8. Should I also measure the current in that case?

If they are in series, the voltage across the load would be around 1.5V*8, can I calculate the power directly by the "power = (voltage squared ) / resistance"?

i.e. Total Power = (1.5*8)^2/50 = 2.88V??
Would it not reasonable to have 64 multiple of power generated by 8 turbines?

If the internal resistance is small compared with 50 ohms and you only use a 50 ohm load, then the power in the resistor would be greater if you use the generators in series.

You can deduce the current if you know the voltage and resistance. Resistors do change resistance if they get hot, but not by much usually unless the resistor is too hot to touch.

If you do want to measure the current, I have modified the diagram in the earlier post. You could put a current meter in the position marked "A". As always, you have to cut the wire and insert the current meter between the two open wire ends.

The most power a generator can produce is when the load resistance equals the internal resistance. So, you may find that 50 ohms does not dissipate the maximum power your generator can provide.

Would it not reasonable to have 64 multiple of power generated by 8 turbines?
Don't forget that each generator would be producing more current if the generators are in series.

Individually, with a 50 ohm load, they produce 1.5 volts / 50 ohms or 0.03 amp.

In series, they produce 12 volts / 50 ohms or 0.24 amps. So, each generator has to produce this extra current and this may become important when you allow for the internal resistance.

You have to be careful that the current is not enough to damage the generator.

 

[superkaho]

If the internal resistance is small compared with 50 ohms and you only use a 50 ohm load, then the power in the resistor would be greater if you use the generators in series.

For those 8 mini-turbines, only 2 or 4 of them are operated at the same time. Would the other turbines become motors and dissipate power from the 50 ohms resistor?

It is known that the internal resistance of each generator is around 14ohms.
How should I connect them such that the maximum power is obtained from the reisistor??

If I connect them in parallel, should I add a diode along with each generator??

 

[vk6kro]

You probably have enough information to work these problems out for yourself.

The internal resistance of a motor/generator appears in series with the voltage it produces as a generator.

So, if you short-circuited the generator (ie put a very low resistance load across it) while it is producing 1.5 volts, it will produce a current of about (1.5 volts / 14 ohms) or 107 mA.

If you connected a 50 ohms resistor across it, the current will be:
1.5 volts / (14 ohms + 50 ohms) or 23 mA.
The power in the 50 ohm resistor would then be:
I^2 * R = 0.023 amps * 0.023 amps * 50 ohms ... or 26 mW.

If you put them in series the voltages add, but so do the internal resistances.

Putting a diode in series with a generator has the disadvantage that there will be a voltage drop of 0.6 volts across the diode, even if there is only a small current flowing. If you only have 1.5 volts, this is a huge loss of output voltage.

I will leave the other parts of your question for you to work out as they depend on the exact behaviour of the motors.

 

[superkaho]

Putting a diode in series with a generator has the disadvantage that there will be a voltage drop of 0.6 volts across the diode, even if there is only a small current flowing. If you only have 1.5 volts, this is a huge loss of output voltage.

Thanks for your reminder~~

Suppose each generator can prodcue the emf with 2V.
Which circuit design below would be better for maximizing the power output of 50 ohms??

Each two generators are placed in four different directions. (i.e. East, South, West, North)
Only one side of generators are operated at a time, the rest will consume power if all generators are connected in series...
Therefore, apart from putting a diode, is there any better solution to avoid current passing through other motors(not operating generators)??

Circuit 1

Circuit 2​

 

[vk6kro]

The second circuit would produce 30 mA into the 50 ohm load (1.5 V+ 1.5 V- 0.6 V(diode)) / (50 + 28 ohms) = 30.7 mA, so it is a little better than before.

Could you have a wind vane that pointed into the wind, with the generators mounted on it, so that all the generators work at the same time?

 

[superkaho]

The second circuit would produce 30 mA into the 50 ohm load (1.5 V+ 1.5 V- 0.6 V(diode)) / (50 + 28 ohms) = 30.7 mA, so it is a little better than before.

Could you have a wind vane that pointed into the wind, with the generators mounted on it, so that all the generators work at the same time?

Since the wind is artificially produced by eight fans outside a square, each side have two fans. (East, South, West, North). Each time only one side of the two fans are switched on. So in order to capture the maximum amount of wind energy, I can only place them face to each fan with the least distance allowed. However, as they are facing with 4 directions, it is difficult to let them work at the same time...any suggestion for the circuit design?

By the way, in order to measure the internal resistance of motor/generator, is it accurate to measure it by just connecting the two terminals with multimeter?

 

[vk6kro]

The brush contact resistance will vary when the motor is turning, but just measuring the resistance with a multimeter is good enough for this purpose.

You can get some improvement by using Schottky diodes. In the single generator parallel case, the current in the load went from 14 mA to 22 mA just by changing to Schottky diodes (from Silicon diodes). I used 1N5817 diodes in a simulation.

With two generators in series and one diode, the current in the load goes from 28 mA to 35 mA by using a Schottky diode.

I tried using saturated Silicon transistors, but I couldn't get as low a voltage drop as using Schottky diodes. This is where you connect a small resistor between collector and base and then connect the positive voltage to the collector and take the output from the emitter.

 

[superkaho]

You can get some improvement by using Schottky diodes. In the single generator parallel case, the current in the load went from 14 mA to 22 mA just by changing to Schottky diodes (from Silicon diodes). I used 1N5817 diodes in a simulation.

With two generators in series and one diode, the current in the load goes from 28 mA to 35 mA by using a Schottky diode.

Thanks for your information, you are really great!
Since I asked some salemans yesterday, but they just led me somewhere with plenty of electronic elements...such as resistors, diodes, transistor, IC etc...
but what I learned the characteristics of diode is only about its unidirectional property and the cut-off voltage.

How should I ask a salesman to buy a diode with the lowest voltage drop??
what is the voltage drop of 1N5817? is it lower than 0.6V??
I found another diode called 1N60P, is the voltage drop of this diode is 0.24V??
Below is the specification of it:
http://www.weclonline.com/downloads/pdf/33-27-0060.pdf"

 

[vk6kro]

The 1N60 will not handle enough current. See the forward continuous current rating of 50 mA. This is too risky. These are Germanium diodes and they do have a good low voltage drop.

Here is a graph showing the voltage drops across a Schottky diode (top graph) and a silicon diode (bottom graph):
[PLAIN]http://dl.dropbox.com/u/4222062/Schottky%20and%20silicon%201N5817%201N4148.PNG

You can see that at 40 mA current, the Silicon diode has 3 times as much voltage drop as the Schottky.

Buying them may be difficult. This may not always work, but watch the counter for a while and see who the other sales clerks go to when they have a difficult question. Then wait for THAT guy.
Failing that, get a list of their diodes and look them up on Google.

 

[superkaho]

oh...this curve of the diode is really cool!
but is there any other diode that can be interchanged??
because I just asked the store by phone and they said they do not have 1N5817.
Instead, they only have 1N5819...Could you suggest some other diodes that can serve the same purpose?

 

[vk6kro]

I have the data for the 1N5819. It is almost identical to the 1N5817 except that it has higher reverse voltage ratings...40 volts for the 1N5819, vs 20 volts for the 1N5817. So, it is a better diode.

At a current of 40 mA. it should drop 253 mV, which is exactly the same as the 1N5817.

This is only a simulation, so individual diodes may vary, but it looks promising.

 

[Fish4Fun]

A great source for parts information as well as ordering is:

www.digikey.com

They have thousands of diodes in stock, as well as any other part you might need. There are several other suppliers as well, but digi-key, IMHO, has the best site for researching parts. They are are rarely the "cheapest", but they are typically a "one-stop-shop".

Good Luck!

Fish

 

[Fish4Fun]

A great source for parts information as well as ordering is:

www.digikey.com

They have thousands of diodes in stock, as well as any other part you might need. There are several other suppliers as well, but digi-key, IMHO, has the best site for researching parts. They are are rarely the "cheapest", but they are typically a "one-stop-shop".

Good Luck!

Fish