# Something simple...

Discussion in 'General Electronics Discussion' started by FrankyTee, Jul 30, 2011.

1. ### FrankyTee

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Jul 30, 2011
Newbie here...

I am not a electronician. I used to be a troubleshooting guy on CNC machines. So I am able to find the source of a problem and change the dammaged parts, but I dont have the theorytical knowledge to solve this little problem of mine. You might find this stupid or anything. I am trying to convert my 18V cordless weed eater to a 18V, with a cord weed eater.

The motor was originally powered by a 18v rechargeable battery with a capacity of 1700 mA. I have a transfo, 115V-28V with a capacity of 100va. The output is transit through a diode bridge and converts the power into dc. Since the output of transfo is 28V I need to lower it to 18V so I want to lower the voltage with the help of a potentiometer. The guy at the electronics shop sold me a 25k Ohm linear potentiometer and told me this should do the thing. But it is not working. The motor is not running at all. But if I bypass the pot, the motor starts to run. So I am sure the pot is the source problem. But I dont know if it is because it is defective or if the resistance is to high.

Can someone enlighten me? Any help whould be much appreciated.

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2. ### Laplace

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Apr 4, 2010
As a rough estimate let's say the resistor needs to drop 10 volts (28-18=10) at 2 amps. R=E/I=10/2=5 ohms. Your 25K resistance is a bit high. What I would try in place of the resistor is a string of power diodes in series, with each diode dropping 0.7 volt. You should not need more than 10/0.7=14 diodes, maybe a few less. Note that the power dissipation necessary to drop 10 volts at 2 amps is 20 watts. I don't suppose that 25K pot is rated for 20 watts either.

3. ### Resqueline

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Jul 31, 2009
How about ditching the full-wave bridge and using just a single diode to get half-wave rectifying and hence 14V DC average. No power being wasted as heat then.
A small capacitor can then be used to bring the voltage up to around 18V.
I'm not sure exactly how large it would need to be but a rough guesstimate is 470uF. You may want to try out different uF values to get the original (wanted) speed.
Some values to try is 220uF, 470uF, & 1000uF. You can connect them all in parallel if needed to get a higher capacitance.
Get ones with a large voltage rating, at least 50V, better 63V, or even 100V).
This better enables them to cope with a relatively large ripple current compared to their capacitance.
I also believe this half-wave scheme is simple & cheap and gives a much better speed regulation than using a resistor for example.

4. ### FrankyTee

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Jul 30, 2011
Thank you guys for your time.

One of my concern was the heat generated by the components as they will be enclosed in the original plastic battery case. I have also read somewhere that a pot was ineficient and resulted in a significant amount of power being wasted. And also I dont know the pot's rating on this mather. The transfo will also generate some heat so it might be a good idea to also try Resqueline solution.

I was wondering, will adding all these resistors, capacitors, diodes increase the current demand to the transfo? I am kind of limited by its capacity, and I have a smaller one that I also consider using.

Resqueline, I like the idea. Can you tell me the rating of the diode for this purpose? You also mentioned to connect the capacitors in parallel if I needed greater capacitance. What does the capacitance affect? Is it better to use 100V capacitors as opposed to 50V? Can you check the wiring schematics attached and tell me if it is how it is wired?

I have so many questions but I will start with this and do a little experimentations. I will probably go to the electronics store in montreal tomorow to purchase more st

Many thanks.

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5. ### Resqueline

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Jul 31, 2009
You can most likely use just one of the diodes in the bridge you already have there provided it has double the rating of the transformer. Connect to one of the AC's and +.
The capacitor fills in the voids between the half-periods. The degree of filling depends on the capacitance and so affects the averaged voltage. More cap. = higher V RMS.
See attached pic. The red area is supposed to equal the blue area. A steeper cap. discharge slope (the slanted line between 1st & 3rd half-period) will accomplish this.
Yes, 100V cap's will be "stronger" and will last longer than 50V cap's. Yes, your diagrams are correct. A 28V 100VA transformer is good for 3.57A of resistive load.
I notice you've drawn a center-tapped transformer however, is that what you have? If so you could use that to make 19V (with 7A capacity) straight w/o cap's.

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6. ### FrankyTee

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Jul 30, 2011
The transfo I have is really as drawn, with a center tap. But at the output I only measure 15V. How do I raise it to 18-19V?

7. ### Resqueline

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Jul 31, 2009
Oh, ok, that's also taken care of with a (bigger) capacitor (2200-10000uF, at least 25V). I see my statement above was confusing (I was thinking of the half-wave thing).
Being big it'll charge to the peak voltage of the rectified sine (being 1.414 times the measured RMS value) and pretty much keep there. You'll get 21V off-load.
You just connect the cap' & the motor positive wires to bridge +, and the negative wires to the center tap, leaving the bridge minus unconnected.
After you've measured the actual motor current draw (& voltage) under full load, we can decide if a smaller transformer (& capacitor) can do the job.

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8. ### Laplace

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Apr 4, 2010
Trying to visualize how the capacitor will interact with the back-EMF of the motor, I can't decide whether the capacitor will be a help or a hindrance. Rough approximations suggest that for a motor drawing several amperes of current, a capacitor would need to be sized in excess of several hundred thousand microfarads in order to have any noticeable effect. Alas, I do not have a test setup to measure what might be going on.

9. ### Resqueline

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Jul 31, 2009
The motor may have both an inductance and a back-EMF but its main power draw is surely resistive? I've never come across indications about any special considerations.
For most practical applications I'd consider a motor to be close enough to a variable resistance to apply the usual rules-of-thumb (until proven otherwise of course).

10. ### duke37

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Jan 9, 2011
The capacitor is charged every 1/120 sec. The rate of change of voltage on a capacitor is I/C so with 1A and 1mF, the voltage will drop about 8V, with 10mF it will drop less than a volt.
Things are quite complicated since the motor will only take current when the applied voltage is greater than the back EMF so it will run on pulses which will depend on the transformer and motor resistances. A capacitor will shorten the pulses supplied by the transformer and lengthen the pulses taken by the motor.

11. ### Resqueline

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Jul 31, 2009
Yes, that's all standard stuff, charging & discharging & timing & current pulses, but I don't see where it gets so complicated. Sure, the back-EMF changes with the mechanical load, causing a changing electrical load, so you have to make/accept some compromises, but this isn't exactly a rocket going into space.

12. ### poor mystic

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Apr 8, 2011
No-one has mentioned that the bigger the capacitor, the shorter the time windows during which it is being charged by the transformer will be, and the greater the peak current in the diodes.

13. ### Resqueline

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Jul 31, 2009
No, that's right, though it is inherent to the nature of the matter but is rarely a problem and rarely adressed in such rough & simple circuits as this. Most parts can handle it.
I felt it was not a natural issue of this thread and deemed it unneccessary to involve the o/p with that much detail, possibly confusing matters more than clearing them up.

14. ### FrankyTee

7
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Jul 30, 2011
Hello boys.

I have finally made some testing. I have purchased several capacitors last sunday. 220uF, 470uF, 680uF, 1500uF and 2200uF, all voltage ranging from 35V to 100V.

I have first tried this combinaison of caps, 2200uF connected with 1500uF in parallel (does this act as a 3700uF cap?), wired like Resqueline suggested, using one diode of the bridge, and using the center tap of transfo. Please review schematics as ref.

So If I understand the quote above, and correct me if I am wrong, if wired like this and if the motor is not connected, when I measure the voltage between + and - (see drawing) I should read 21V. If motor is connected, the voltage I will measure between + and - (measured RMS value) should be multiplied by a factor of 1.414 to obtain the peak voltage. This in mind, here are my results:

sec. of transfo => 16.2V
caps => 21.8V

Prim of transfo => 122V - 0.9A
sec. of transfo => 15.3V - 3.7A
caps =>13.3V - 2.9A

So if i understand you well, if I multiply 13.3V by 1.414 I get 18.8V peak. Is'nt that what I am looking for?acitor. The biggest I found was a 2200uF at 50V.

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15. ### Resqueline

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Jul 31, 2009
If you measured 13.3V dc at the capacitor under load then that is the voltage the motor gets. I had full-wave rectifying in mind for this configuration though, see pic'.
The peak voltage might've been 19V but the capacitor's unable to hold the charge. This is affected by 3 things: low capacitance, high current, half-wave rectification.

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16. ### FrankyTee

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Jul 30, 2011
Men I love this. It might not look like much to you guys but I am learnig a ****-load by doing this and thanks to you.

So I wired like you suggested Resqueline and wow, it made all the difference in the world.
I am not yet at 18V but it is getting close. The motor runs smoother, it feels like there is a lot less vibration. I get about 16.5V out of the capacitor at about 1.7A. It feels much better and more efficient.

So does this mean that getting cap with greater uF value, whould this get me closer to 18V and increase the efficiency of the control? This whould solve the issue of low capacitance right?

17. ### Resqueline

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Jul 31, 2009
With full-wave rectifying the time spans that the capacitor needs to fill in are much shorter so it becomes more effective (see previous graphs).
Yes, increasing the capacitance should increase the voltage under load (and may make it run even smoother).
Parallelling all the cap's you got there should get you 5070uF, a nice increase over 2200uF (or get another 2200uF).
I wouldn't use the term efficiency increase but the motor will run stronger, with less droop in rpm with load.
See attached graphs for an approximation of what happens as you vary capacitance and load current.
The full transformer voltage, half-wave rectifying, low-capacitance solution would have been far worse in this regard.
The diode drop solution would have retained a relatively "stiff" supply but would tax the transformer more and waste a lot of power.

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18. ### FrankyTee

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Jul 30, 2011
Hello again!

Hey I am sorry for the delay, I am on vacations and we are rather moving a lot. But I still manage to finnd the time to make some tests

I did like you suggested Resqueline, parallelling the caps with a tptal of 5370uF. Manage to get 16.8V. Then went out and buy two caps of 4700uf at 25V and connected it in parallel, could'nt get more then 16.8V.

Is there a point where increasing the uF of the caps does'nt change anything? Is it because the voltage rating of the caps is too low?

19. ### Laplace

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Apr 4, 2010
I would be interested in seeing test results for the voltage across the motor with no capacitors versus a full complement of capacitors (10,000uF). It seems clear how capacitors help with a purely resistive load but I still can't visualize how capacitors help with a motor.

20. ### poor mystic

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Apr 8, 2011
Work done by an electric motor is represented by a purely resistive element in the load seen by that motor's power supply.