How to power up a hair trimmer: NiCd vs. NiMH battery

Hi guys!

Thank you all for the much needed help.
I did the measurements, and it seems that the mystery is about to be solved.
I will try to do some experiments and will present you with a full report tomorrow.
I only hope that my experiments will not kill the motor.
Wish me luck.

Hi,

Well good luck, and i cant wait to hear what you find out. This has been a little bit of a mystery so far

Hi, guys!

I followed your advice, MrAl, and that proved to be a real good one.
So, I believe, the mystery is resolved, though the problem still remains to be solved.

Here it is:

- the NiMH batteries that I used,
**broken link removed**

- the trimmer, it's a Philips Norelco brand, I added a second switch (black one), so that it could be powered from an external source
**broken link removed**

- trimmer, blades detached,
**broken link removed**

- trimmer inside, I had to use a saw to get there,
**broken link removed**

- the original NiCd battery,
**broken link removed**

- the same battery form another side,
**broken link removed**

- this is how I connected the NiMH for the test,
**broken link removed**

- and finally, these are the data.
**broken link removed**

As you can see from the data, the load resistance is VERY low, and it decreases even more when the motor gets stuck (or is stopped manually). The total resistance for NiMH setting is on average 0.2 Ω higher than that for the NiCd. This is most likely because of the connections and maybe the switch. So, creakndale in post № 4 was absolutely right. I just could not imagine that such a small resistance can make such a huge difference. Well, shame on me. As you can see, when the current is relatively low, this additional resistance does not affect much the power, delivered to the motor. The situation changes dramatically when the blades get stuck. Now NiCd has the power to move them, but NiMH (in this particular setting) has not. Because of 0.2 Ω additional resistance !!!!!!!!! If I connect the NiMH directly where NiCd connects, it performs just as well. So, it is not the battery to blame, but the way it is connected.

Now about the problem. I still want to make this trimmer to be able to work from both internal source and an external one. Because no internal source can last long enough, eventually, it would be very convenient in such a case to be able to switch to another source, lest to go for 10 hours half-trimmed . Replacing the battery inside every time it discharges is a pain in the leg.

External source means connector and a cable. This would amount to about 0.3 Ω additional resistance. The only way I see to increase current is to increase voltage. But with batteries this can only be made stepwise, that is to use two batteries in series.

I tried that, but so far did not succeed. I tried to use a voltage divider circuit using an additional wire resistor. I tried to make it 0.4 Ω, but eventually it turned out to be 0.8 Ω and it did not work. It is hard to adjust such a low resistance to obtain necessary current and voltage on the load. Besides, at the current of 2 A the wire resistor burns with a very intensive smoke (it is surprising that the motor does not). So, this way is not going to work.

Thus, now my question is: how can I construct a circuit (preferably a simple one), which is powered by two NiMH in series, and would produce 2.5 A and around 1.2 V on the motor, without setting fire to my apartment ?
Would a voltage regulator work ? Something like LM338K or such?
I will appreciate any ideas.
And thank you all for the help!

Of course your Mickey Mouse alligator clips have resistance!

Also, your power calculations are completely wrong.
If the Ni-MH battery has a current of 2.0A and its voltage is 1.2V then its power is 2 x 1.2= 2.4W, not 1.6W.

audioguru said:

Of course your Mickey Mouse alligator clips have resistance!

Also, your power calculations are completely wrong.
If the Ni-MH battery has a current of 2.0A and its voltage is 1.2V then its power is 2 x 1.2= 2.4W, not 1.6W.

Click to expand...

Hi, audioguru!

I believe most of this 0.2 Ω resistance comes from the connector and maybe some from the switch. I tried it without alligator clips, the wires were soldered together - performance was just as bad.

I calculated the power on the "motor". So if the total resistance is 0.6 Ω, the "motor" resistance is 0.4 Ω, and the "connections and wires" resistance is 0.2 Ω, then the voltage on the motor would be 1.2 x 0.4 / 0.6 = 0.8 V. Of course these values are approximate.

Hi GreenBear,

Ahh, the problem is getting more apparent now...
That little tiny connector is made for currents of maybe 100ma or less. It's made for charging the cell, not for powering the shaver fully. Ditch that thing and you'll be on your way.

The next experiment should be to solder in your battery holder wires directly to the shaver, not to the connector. That should make a big difference. You can make a couple small holes in the case to run the wires out neatly.

If you dont like wires hanging out with a battery holder too, you might consider installing a heavy duty connector...one that is made for 5 amp or greater current levels. The connector body does not have to fit inside the shaver itself, it can be epoxied to the outside of the case with the wires poking in through the small holes.

Hi, MrAl!

Thanks for your help. I tried to attach the NiMH directly to where the NiCd did, and it worked fine, just as well as the NiCd does. So it is obviously this little resistance that is lost on connections and wiring that causes the problem. I would never think of that if not you prompted me to find it out directly.

As to the 5 amps connector, I thought these things were meant for high voltage. Would it have a sufficiently low resistance? Say, if the voltage is as high as 200 V and the current is 5 A, then the load is 20 Ohms. In such a case a 0.2 Ohms connector resistance would not make much difference, as it may cause only a voltage drop as big as 1%, not 30% as in my case.
It is hard to measure such a low resistance reliably, yet it may prove critical.

I am starting to think that maybe the best way for me is to use a power supply, and then I can use it together with a cable and connector. Say, if supply can provide 1.6 V at 2.5 A, then even with a voltage drop on connections and wiring the motor would still get enough power. But most likely I will have to build such a supply. I have a spare transformer, I believe it will do. I think I just need some minor parts like a rectifier, capacitors, resistors and a voltage regulator.

Do you know a good place to ask circuit design questions?

Hi again,

You're welcome. A connector such as a molex with pins like those used for the computer hard drive power connector should work very nicely. Until you get one you can run a permanent set of wires i guess, say 16 gauge or better. I used 16 gauge for running my 'cordless' power drill and it gives me a little (needed) voltage drop. You might need heavier though.

Here's an idea:
If you build a variable voltage power supply that can handle at least 3 amps, what you could do is measure the voltage at the shaver while you adjust the power supply voltage. When you get the right voltage, you're good to go. Probably get away with 16 gauge wire this way too.

We design circuits like this all the time here. There are three terminal regulators that can handle 3 amps so your task is a simple one, unless you want to go with a more efficient switching power supply in which case you can probably use an ordinary 9v, 1 amp DC wall wart.

Im sure there are people here that could point you to a power supply and connector too, but if no one replies i'll post something myself for you.

QUESTION:
What kind of DC wall warts do you now have or can get? Or alternately do you have any power transformers around?

Hi, MrAl!

Thanks for the help, again.

Having considered my options, I believe that probably building a new power supply would be the most cost-effective.
Besides, I always wanted to build one .

I have a transformer like this:

**broken link removed**

And I am thinking about assembling something like this:

**broken link removed**

I do not understand a few things, though:
How to calculate the max. current that such a circuit can provide?
How to calculate what current the transformer can handle without damage?
Will 17 V input be enough for the LM338 voltage regulator?
And, how to make this circuit protected form short-circuit accidents?

Can you please provide me with an advise or direct me to a source where I can find this information?

For an output of only 1.6V then the 16VDC from the rectified and filtered transformer is much too high since the heat (voltage across the regulator times its current) will fry the regulator. The regulator will be much cooler if the transformer voltage is only 5.0VAC.

The current from the transformer must be high enough to provide 1.6V plus 2V at the peak voltage of the transformer (7.1V) before the full wave bridge 2.0V voltage drop. So if you want 3A at 1.6V then the 5V transformer must have a power rating of 3A x 7.1V= 21.3W which is a 5V transformer with a current rating of 21.3W/5V= 4.2A.

If the output of the regulator is shorted then it will limit the current to 7A if it has enough heatsinking. If the regulator gets too hot then it will shut-down until it cools then it will beging working again.
It is not recommended to let a regulator heat very much, shut-down and cool then heat very much again over and over because the thermal cycling shock will kill it.

GreenBear said:

Hi, MrAl!

Thanks for the help, again.

Having considered my options, I believe that probably building a new power supply would be the most cost-effective.
Besides, I always wanted to build one .

I have a transformer like this:

**broken link removed**

And I am thinking about assembling something like this:

**broken link removed**

I do not understand a few things, though:
How to calculate the max. current that such a circuit can provide?
How to calculate what current the transformer can handle without damage?
Will 17 V input be enough for the LM338 voltage regulator?
And, how to make this circuit protected form short-circuit accidents?

Can you please provide me with an advise or direct me to a source where I can find this information?

Click to expand...

Hi again,

Oh yes, that circuit looks good except for the input voltage as audioguru pointed out, if it is that high you'll heat up the LM338 too much with a low output like 2v or so.
There are other ways around this if you dont have any other transformer, such as using a whole bunch of series 3 amp rated diodes, but that's up to you. The diodes drop the voltage so the LM doesnt have to handle all the heat.

You should really use 3 amp diodes for the rectifier too, and make sure your transformer can handle the output current up to say 2.6 amps.

Hi guys!

Thanks for help!

I thought about what you said, and, sad as it is, it appears the transformer that I have probably will not do for such a power supply. It appears it can barely handle 15 W (I was unable to get the exact rating, but this figure seems reliable). I saw many power supply circuits where exactly this transformer was used, but all those were for low current, like 500 mA.

I will try to see if I can get a better transformer, like the one audioguru has suggested.

In the meanwhile, yet another option is to use two AA NiMH batteries in series (since I usually trim with vigor, a cable connection is preferable to an option where a battery is hanging on the trimmer).

But I am at a loss here.

What circuit can I use to convert 2.4 V input into 1.6 V output at 2.5 A ?

The resistor voltage divider will not do - it is hard to adjust such a low resistance, misadjustment may kill the motor, and way too much heat.

I thought about two diodes in series, like this:

**broken link removed**

But, while the output is more reliable in terms of voltage and current, still too much heat on R0, and in normal mode (blades are running) the battery will be pumped out 3A, while the trimmer would use only 1.5 (and it only needs 1.3 A). Also I suspect that Ud will go up, as Id will increase from 0.5 to 1.5 A. Practically, it will be a good heater, and, probably, an igniter Two batteries will not last 40 minutes.

The voltage regulator like LM338 probably will not work at 2.4 V input.

Of course, one crazy way is to convert DC into AC, then use a transformer to make it low voltage/high current, and then convert it back to DC, but that is definitely a huge overkill.

Darn, this trimmer affair is more complicated than I thought!

Any ideas?

Use two AA rechargeable cells in series for 2.0V to 2.6V, then add a high current diode in series to drop the voltage to 1.2V to 1.9V (it is not regulated).
The diode will get fairly warm.

audioguru said:

Use two AA rechargeable cells in series for 2.0V to 2.6V, then add a high current diode in series to drop the voltage to 1.2V to 1.9V (it is not regulated).
The diode will get fairly warm.

Click to expand...

Hi, audioguru!

Unfortunately this will not do. 1.2V is too little, while 1.9 V is too high.
The task requires 1.5-1.6 V at 2.5 A.
Any more ideas?

How will this diode in series be better than just a resistor in series?

This seems to be to good to be true, but if it is real you could use a 5 volt supply and this guy.

**broken link removed**

GreenBear said:

How will this diode in series be better than just a resistor in series?

Click to expand...

A resistor is linear. If the motor draws 10 times its normal current when it starts then the voltage drop in the resistor is 10 times and the motor might not start (you had that problem with the resistance of your connections).

But a diode has almost a steady voltage when the current chages. The datasheet for a 1N5401 3A diode shows a typical voltage drop of 0.8V at 1.5A. When its current is doubled then its typical voltage drop increases to only 0.85V and at 80A its typical voltage drop is less than 1.6V.

Hi, guys!

ronv said:

This seems to be to good to be true, but if it is real you could use a 5 volt supply and this guy.
**broken link removed**

Click to expand...

Thanks, ronv, for the proposal.
That seems to be a workable solution. But it seems a bit pricy, since I will need to buy a power supply, the regulator, and I expect that there might be some problems with shipping, like high cost, long waiting time and maybe customs (I am from another country).

audioguru said:

A resistor is linear. If the motor draws 10 times its normal current when it starts then the voltage drop in the resistor is 10 times and the motor might not start (you had that problem with the resistance of your connections).

But a diode has almost a steady voltage when the current chages. The datasheet for a 1N5401 3A diode shows a typical voltage drop of 0.8V at 1.5A. When its current is doubled then its typical voltage drop increases to only 0.85V and at 80A its typical voltage drop is less than 1.6V.

Click to expand...

Thanks for the explanation, audioguru.
Your idea may actually work!
But can we make it even better?
A series diode will be wasting 1-2 W of battery power. It would be great if we could make it less by decreasing the current through the diode.
I am now considering these two circuits:

**broken link removed**

What do you think, guys ? Will they work? Which one is better?
I already have a transistor (24 W, 7.5 A, Ube ~ 0.1 V)
I have a difficulty though calculating Ib in the first circuit and I do not know how to calculate R0 in the second.
Can you please help?

Instead of posting your schematics at that PORN site, please attach them HERE with your replies.

Your unregulated power supply voltage is too high which causes a lot of wasted heat. If a resistor or a power transistor wastes the heat it doesn't matter because the waste is the same.

Hi, audioguru!

audioguru said:

Instead of posting your schematics at that PORN site, please attach them HERE with your replies.

Click to expand...

I figured the only way to post a schematics on this forum was to upload it on some external site.
Imagevenue is a very convenient image hosting site, I do not see why you call it a PORN site.
Do you upload porn there? The schematics that I post are family-safe, there are no sexual implications in them at all.
Do you call your computer a "PORN-computer" only because it is possible to use it for PORN?
I read some US senators kept porn on their cell phones, does that mean that my cell phone is a PORN-phone?
I am not trying to promote Imagevenue in any way. I do not mind, of course, to post the schematics directly on this forum, after I figure it out, but how would that matter?

audioguru said:

Your unregulated power supply voltage is too high which causes a lot of wasted heat. If a resistor or a power transistor wastes the heat it doesn't matter because the waste is the same.

Click to expand...

Yeah, that was a crazy idea. I will try to use a single series diode, and will write back about the results. It will need a heat sink, I believe.
Thanks again for the suggestion.