Using a 12V 2A power supply direct and with 3V voltage divider?

"S Keith" <stkeith11@gmail.com> wrote in message 
news:c7b48a48-89c0-4b50-8918-27a2794fd93a@googlegroups.com...
> I meant that since the VR puts out 0.3W, I must meet that input plus a 
> little for inefficiencies.  With 12V/2.1A PS, that shouldn't be an issue.
>
> It's probably been lost in the mess, but I'm trying to simulate what 
> amounts to a PWM signal, so I need to cut the current. Reading up on 
> computer fan PWM control suggests 3.3V/5-8mA max if this fan is analogous.
>

I am guessing that you do not understand the voltage current relationship.

The load wihch in your case is 3 volts and 2 ma (assuming it is not 
changing) is the equal of 1500 ohms.  YOu do not need to limit the current 
as that is automatically done if you only supply 3 volts.  By simple math, 
if you have 3 volts, then the only current that can flow is 2 ma.  Also by 
the same math if 2 ma is flowing only 3 volts will cause that.

Just plug in 1500 ohms for the 3 V, 2 ma load in the formulars you are 
using.

That is why I said all you need is one resistor if the load is a constant 3 
V, 2 ma.  You are starting with 12 V so 12-3= 9.  YOu must drop 9 V at 3 ma. 
So 9/.002 = 4500 ohms.
 

On Thursday, February 26, 2015 at 11:59:32 AM UTC-7, Ralph Mowery wrote:

> I am guessing that you do not understand the voltage current relationship.
> 
> The load wihch in your case is 3 volts and 2 ma (assuming it is not 
> changing) is the equal of 1500 ohms.  YOu do not need to limit the current 
> as that is automatically done if you only supply 3 volts.  By simple math, 
> if you have 3 volts, then the only current that can flow is 2 ma.  Also by 
> the same math if 2 ma is flowing only 3 volts will cause that.
> 
> Just plug in 1500 ohms for the 3 V, 2 ma load in the formulars you are 
> using.
> 
> That is why I said all you need is one resistor if the load is a constant 3 
> V, 2 ma.  You are starting with 12 V so 12-3= 9.  YOu must drop 9 V at 3 ma. 
> So 9/.002 = 4500 ohms.

I understand that much, but I got hung up on the concept of a voltage divider.  Thanks for the clarification.

Steve

"S Keith" <stkeith11@gmail.com> wrote in message 
news:7bcdfd31-6458-47d4-a4db-4ec8aaee807e@googlegroups.com...
> >>
>> Just plug in 1500 ohms for the 3 V, 2 ma load in the formulars you are
>> using.
>>
>> That is why I said all you need is one resistor if the load is a constant 
>> 3
>> V, 2 ma.  You are starting with 12 V so 12-3= 9.  YOu must drop 9 V at 3 
>> ma.
>> So 9/.002 = 4500 ohms.
>
> I understand that much, but I got hung up on the concept of a voltage 
> divider.  Thanks for the clarification.
>
>
The voltage devider only works if there is no load.  As you add a load, the 
voltage across the series resistor will drop as the current is made larger. 
To account for that, you have to know the equivilent resistance value of the 
load.  Which in this case is 1500 ohms. So you have 2 resistors in series 
and in parallel with one of them which you are calling R2 you have the 
equivilent of 1500 ohms to account for.  That is why if you use the values 
you started with you will only have half the voltage or 1.5 volts instead of 
3 volts.

It is easy to get confused on the voltage deviders if you do not allow for 
the ammount of current that the load on across the R2 has.  If you put  a 
meter that has about 10 meg of input impedance like many of the digital 
meters have, you see 3 volts, but with the load only about 1.5 volts.  The 
digital meter has a high enough impedance to make the very slight error 
unnoticable.  Sort of like putting a single brick on a pickup truck.  You 
know it is there, but the truck is so heavy you just can not measure it with 
most practical scales.

S Keith wrote:

> >  
> > > Thanks John G. While my ignorance is vast, I know V=IR. I need to supply 3V, 2mA
> > 
> > ** You are being far to literal. 
> > 
> >  3V at 2mA MEANS with a voltage supply of 3V, the LOAD draws 2mA. 
> > 
> > 
> > > I have a 12V PWM controlled fan.
> > > I don't know anything other than it takes 12V ~1.6A
> > 
> > 
> > ** That is a very big PWM fan. 
> > 
> >   High time you supplied a link to the thing. 
> > 
> > 
> > > to run it and a 3V, 2mA signal to activate it.
> > 
> > 
> > ** PWM fans normally require a PWM signal to control them.
> > 
> >    You are contradicting yourself not making sense.

> 
> If I could link it for you, I would know more about it.  It's the cooling fan for a Honda Civic Hybrid battery pack. 


** Figured it was something weird and automotive. 

   How do you know it is a PWM fan ?? 


> I "discovered" that I can activate the fan with a multimeter that puts out a 3V, 2mA current. 


** OK - so we FINALLY  get to know where your 3V, 2mA nonsense came from.

   The spec sheet for a DMM. 
 
   And it does not mean what you assume. 


I wish to duplicate this to drive the fan as I thought it would be a simple solution compared to figuring out how to supply a PWM signal.

** You do not know what is needed. 

   So neither do we. 


....  Phil 

On Fri, 27 Feb 2015 08:58:38 +1000, Ralph Mowery  
<rmowery28146@earthlink.net> wrote:

>
> "S Keith" <stkeith11@gmail.com> wrote in message
> news:7bcdfd31-6458-47d4-a4db-4ec8aaee807e@googlegroups.com...
>> >>
>>> Just plug in 1500 ohms for the 3 V, 2 ma load in the formulars you are
>>> using.
>>>
>>> That is why I said all you need is one resistor if the load is a  
>>> constant
>>> 3
>>> V, 2 ma.  You are starting with 12 V so 12-3= 9.  YOu must drop 9 V at  
>>> 3
>>> ma.
>>> So 9/.002 = 4500 ohms.
>>
>> I understand that much, but I got hung up on the concept of a voltage
>> divider.  Thanks for the clarification.
>>
>>
> The voltage devider only works if there is no load.  As you add a load,  
> the
> voltage across the series resistor will drop as the current is made  
> larger.
> To account for that, you have to know the equivilent resistance value of  
> the
> load.  Which in this case is 1500 ohms. So you have 2 resistors in series
> and in parallel with one of them which you are calling R2 you have the
> equivilent of 1500 ohms to account for.  That is why if you use the  
> values
> you started with you will only have half the voltage or 1.5 volts  
> instead of
> 3 volts.
>
> It is easy to get confused on the voltage deviders if you do not allow  
> for
> the ammount of current that the load on across the R2 has.  If you put  a
> meter that has about 10 meg of input impedance like many of the digital
> meters have, you see 3 volts, but with the load only about 1.5 volts.   
> The
> digital meter has a high enough impedance to make the very slight error
> unnoticable.  Sort of like putting a single brick on a pickup truck.  You
> know it is there, but the truck is so heavy you just can not measure it  
> with
> most practical scales.
>
>

the voltage divider works providing the load is small relative to the  
current trough the divider

On Thursday, February 26, 2015 at 4:28:35 PM UTC-7, Phil Allison wrote:
> S Keith wrote:
> 
> > >  
> > > > Thanks John G. While my ignorance is vast, I know V=IR. I need to supply 3V, 2mA
> > > 
> > > ** You are being far to literal. 
> > > 
> > >  3V at 2mA MEANS with a voltage supply of 3V, the LOAD draws 2mA. 
> > > 
> > > 
> > > > I have a 12V PWM controlled fan.
> > > > I don't know anything other than it takes 12V ~1.6A
> > > 
> > > 
> > > ** That is a very big PWM fan. 
> > > 
> > >   High time you supplied a link to the thing. 
> > > 
> > > 
> > > > to run it and a 3V, 2mA signal to activate it.
> > > 
> > > 
> > > ** PWM fans normally require a PWM signal to control them.
> > > 
> > >    You are contradicting yourself not making sense.
> 
> > 
> > If I could link it for you, I would know more about it.  It's the cooling fan for a Honda Civic Hybrid battery pack. 
> 
> 
> ** Figured it was something weird and automotive. 
> 
>    How do you know it is a PWM fan ?? 
> 
> 
> > I "discovered" that I can activate the fan with a multimeter that puts out a 3V, 2mA current. 
> 
> 
> ** OK - so we FINALLY  get to know where your 3V, 2mA nonsense came from.
> 
>    The spec sheet for a DMM. 
>  
>    And it does not mean what you assume. 
> 
> 
> I wish to duplicate this to drive the fan as I thought it would be a simple solution compared to figuring out how to supply a PWM signal.
> 
> ** You do not know what is needed. 
> 
>    So neither do we. 
> 
> 
> ....  Phil

Phil,

Since you're clearly not willing to read what I write as evidenced by your response (I said nothing of any DMM spec sheet), I'm going to disregard any of your future responses.

I'm not sure what you're getting from your participation in a group labeled "basics".  You remind me of why I stopped using Usenet back in the day (started in the late 80s).

Simply put, you're an arrogant asshole. I've seen it time and time again in dozens of groups covering as many topics. Take comfort in your false sense of superiority.  I'm far kinder to those ignorant of my areas of expertise.

Steve

To everybody else, thank you for your patience and input.  I really appreciate it.

Steve

Shithead Keith wrote:

> > 
> > ** You do not know what is needed. 
> > 
> >   So neither do we. 
> > 
> 
> Since you're clearly not willing to read what I write 


** No, I am simply not willing to BELIEVE anything what you write. 

  Cost it is obviously all crap. 


> as evidenced by your response


** No, they all show I read what you posted very carefully. 


> (I said nothing of any DMM spec sheet)

** Looked just like you were quoting from one though.

   Why the secrecy ?

   Why refuse to post where those two numbers came from ?? 

   You must be some sort of anal nut bag.


> I'm going to disregard any of your future responses.


** No you won't. 


> I'm not sure what you're getting from your participation in a group > labeled "basics".

** To find an shoot down trolling nut cases like you - buddy. 


 > > Simply put, you're an arrogant asshole.

** Fraid the REAL arrogant ASSHOLE is YOU  !!
   
I've seen fuckwits like YOU, time and time again in dozens of groups covering  many topics. 

Instead of posting all the facts ( so others would know what they actually have and what they are really doing ) smartarse anal nut cases like YOU hide all the facts.

They foolishly imagine doing this is the way to control the discussion and make it go their way. 

Got news for you pal - it don't work like that.

No poster can control the discussion on such a public forum. 

However, that is just what TROLLs all try to do. 

Consider yourself thoroughly outed. 

Then FOAD. 



...  Phil 

On 2/26/2015 7:22 PM, S Keith wrote:
> To everybody else, thank you for your patience and input.  I really appreciate it.
>
> Steve
>

  I have a few questions.
You put what you say was 2ma to the control pin and the fan spun up.
1)Did it spin up to full speed?
2)Do you want full speed?
3)Do you want it variable?



1) If it wasn't full speed, the 4500 ohms resistor might need to be 
lower in value, but I wouldn't jump right in and raise it to 12 volts.
But you could lower it 10% and see if the motor speeds up.

2) If it was full speed, and you want variable, put a 10K variable 
resistor in series with the 4500 ohm resistor, then you can adjust the 
10k variable resistor to vary the speed.

3) My opinion is you don't need a regulator, and a single series 
resistor will do. If you want it variable you still want a fixed 
resistor in series so if you turn the variable resistor all the way up 
you don't put 12 volts on the control pin.
  Although we still don't know the range the control input needs.

Can you get a meter on the actual motor leads?
If yes, we can have some real fun!

                               Mikek

On Thursday, February 26, 2015 at 7:50:22 PM UTC-7, amdx wrote:
> On 2/26/2015 7:22 PM, S Keith wrote:
> > To everybody else, thank you for your patience and input.  I really app=
reciate it.
> >
> > Steve
> >
>=20
>   I have a few questions.
> You put what you say was 2ma to the control pin and the fan spun up.
> 1)Did it spin up to full speed?
> 2)Do you want full speed?
> 3)Do you want it variable?
>=20
>=20
>=20
> 1) If it wasn't full speed, the 4500 ohms resistor might need to be=20
> lower in value, but I wouldn't jump right in and raise it to 12 volts.
> But you could lower it 10% and see if the motor speeds up.
>=20
> 2) If it was full speed, and you want variable, put a 10K variable=20
> resistor in series with the 4500 ohm resistor, then you can adjust the=20
> 10k variable resistor to vary the speed.
>=20
> 3) My opinion is you don't need a regulator, and a single series=20
> resistor will do. If you want it variable you still want a fixed=20
> resistor in series so if you turn the variable resistor all the way up=20
> you don't put 12 volts on the control pin.
>   Although we still don't know the range the control input needs.
>=20
> Can you get a meter on the actual motor leads?
> If yes, we can have some real fun!
>=20
>                                Mikek

Mikek,

1) I can't say that it was full speed, but it was certainly high speed and =
moving a lot of air.  FWIW, when I changed the setting of the multimeter to=
 the 2000 Ohm setting (lower voltage, about 0.6V), the speed slowed to abou=
t half based on noise and perceived air flow.  However, if I disconnected t=
he DMM and reconnected on the 2000 Ohm setting, the fan would not turn.  Re=
applying the 200 Ohm setting would fire the fan back up and I could repeat =
the speed decrease by switching to 2000 Ohms.
2) I'm good with the speed attained.
3) If it can be done simply as you describe, it would be nice, but I'm real=
ly okay with full speed.

I plan to experiment a bit this weekend.  I'm not sure what would be accomp=
lished by probing the motor leads, but I'm game.  I can't probe the fan as-=
installed as the criteria for getting the fan to come on involves high temp=
 readings on the DC-DC converter, BCM or battery pack.  However, I have a s=
pare that I can play with if you have suggestions.  it's one of these:

http://hybridrevolt.com/catalog/images/DSC02766.JPG

The whole purpose of this exercise is to be able to run the fan while I'm g=
rid charging at ~195V/350mA.  It doesn't generate a lot of heat, but the se=
aled battery compartment allows it to accumulate.  Full speed is overkill, =
but it beats a cooked pack.

Thanks,

Steve