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

[John Fields]

On Thu, 26 Feb 2015 06:39:34 -0800 (PST), S Keith
<stkeith11@gmail.com> wrote:

On Wednesday, February 25, 2015 at 10:15:00 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.


... Phil

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. It's a "squirrel cage" blower type fan. It blows air into the trunk thereby creating a vacuum in the sealed case, which then draws cabin air in through the battery pack and the related electronics.

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

I do not know the components involved or their limitations. I only know the "experimental" input and outcome.

---
Steve, forget the regulator for the time being.

How did you use your multimeter to arrive at 3V 2mA?

 

[Phil Allison]

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

 

[Ralph Mowery]

"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.

 

[S Keith]

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

 

[S Keith]

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

Steve

 

[Phil Allison]

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

 

[Ralph Mowery]

"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.

 

[David Eather]

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

 

[S Keith]

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 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

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 about half based on noise and perceived air flow. However, if I disconnected the DMM and reconnected on the 2000 Ohm setting, the fan would not turn. Reapplying 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 really okay with full speed.

I plan to experiment a bit this weekend. I'm not sure what would be accomplished 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 spare 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 grid charging at ~195V/350mA. It doesn't generate a lot of heat, but the sealed battery compartment allows it to accumulate. Full speed is overkill, but it beats a cooked pack.

Thanks,

Steve

 

[amdx]

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

 

"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). "

He is an arrogant asshole but he knows WTF he is talking about. Some people jusr want others to do well in electronics.

Indeed, why is he here ? He certainly needs little help if any. Think about that.

Well I mean help in electronics. Psyche on the other hand... ... ...

 

[S Keith]

On Friday, February 27, 2015 at 6:34:04 AM UTC-7, jurb...@gmail.com wrote:

"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). "

He is an arrogant asshole but he knows WTF he is talking about. Some people jusr want others to do well in electronics.

Indeed, why is he here ? He certainly needs little help if any. Think about that.

Well I mean help in electronics. Psyche on the other hand... ... ...

Indeed. I'm not contesting his expertise. I'm questioning his motivation. Given the expertise, I see only two real motivations in this situation, 1) he wants to contribute his knowledge, or 2) he wants to demonstrate his superiority. The bulk of the data thus far supports #2.

His type is pervasive. They are lacking something in either themselves or their lives and they fill the hole with, "Oh yeah, I'm smarter than you!"

Thanks again to all who have helped.

Steve

 

[S Keith]

On Friday, February 27, 2015 at 9:34:16 AM UTC-7, amdx wrote:

The point was, if you can measure the motor voltage, then you could
increase the control pin until you get 12 volts on the motor. You
wouldn't want to raise it any more without knowing the circuit.
This assumes it is PWM.

More potential ignorance... Since it's PWM, the fan voltage is always 12V, it's just pulsed. Will a meter read an "average" voltage, e.g., if it's at 75% duty cycle, I will read 9V assuming a linear relationship between duty cycle and voltage?

On the 200 ohm setting you had enough current to over come the torque
required to start the motor, when you switched 2000 ohm it had enough
current to keep it rotating. Trying to start at the 2000 ohm setting
there was not enough current to produce the torque required to start the
motor.

That was my thinking. Thanks for confirming.

Steve

 

[S Keith]

On Friday, February 27, 2015 at 9:34:16 AM UTC-7, amdx wrote:

On 2/26/2015 11:43 PM, S Keith wrote:
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 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

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 about half based on noise and perceived air flow. However, if I disconnected the

DMM and reconnected on the 2000 Ohm setting, the fan would not turn. Reapplying 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 really okay with full speed.

I plan to experiment a bit this weekend. I'm not sure what would be accomplished 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,

The point was, if you can measure the motor voltage, then you could
increase the control pin until you get 12 volts on the motor. You
wouldn't want to raise it any more without knowing the circuit.
This assumes it is PWM.

On the 200 ohm setting you had enough current to over come the torque
required to start the motor, when you switched 2000 ohm it had enough
current to keep it rotating. Trying to start at the 2000 ohm setting
there was not enough current to produce the torque required to start the
motor.

Mikek





---
This email has been checked for viruses by Avast antivirus software.
http://www.avast.com

Mikek,

I'm trying to understand PWM, and your questions are a bit confusing. Based on what I've read about how computer PWM fans work, they "pulse" a current (around 3-5V and <5ma). When that pulse is active, it applies 12V to the fan motor. When there is no pulse, there is 0V at the motor. The voltage of the PWM signal doesn't vary as part of the fan speed control, just the duration of the pulse within the cycle.

The pulse occurs at a nominal frequency of 25kHz. For the sake of argument, if the pulse width duration is 10ms, and the pulse is active for 7 of those 10 ms, the fan is at 70% duty cycle (analogous to 70% speed).

I believe what I have done is inadvertently applied a 100% duty cycle PWM signal with 3V and 2mA, so I suspect the fan is running at 100%. I will try to confirm this over the weekend. Where my logic fails is at the second setting of much lower voltage and current. If my understanding of PWM is accurate, I would expect it to shut off if the current is too weak rather than decrease speed.

Am I totally confused?

Thanks,

Steve

 

[whit3rd]

On Wednesday, February 25, 2015 at 6:59:28 PM UTC-8, S Keith wrote:

> I have a 12V PWM controlled fan. I don't know anything other than it takes 12V ~1.6A to run it and a 3V, 2mA signal to activate it. I'm trying to accomplish this. If I supply a higher current, I am concerned I might damage something.

If the fan is on a fused automotive circuit, and you're bypassing that power source, you might want to
insert a similar fuse in your 'alternate' power connection. As for the 'signal to activate',
that sounds like it might be TTL logic (and that means we cannot be sure that
it sinks 2 mA; it could as easily SOURCE 2 mA, the polarity DOES matter).

If the fan runs when the signal is at +3V, and does NOT run when the signal pin is left open,
and does NOT run when the signal pin is grounded, then probably that signal
pin has a pulldown resistor and runs on 3V logic. The "PWM" feature could be
implemented by putting a variable duty cycle onto the signal pin, i.e. constant 3V
drive makes the motor run full speed.

A two-resistor voltage divider, output 3V and impedance about 100 ohms, would be
suitable to arrange this ( +12V<--->360 ohms<--->120 ohms<--->GND).
If the 12V source is not regulated, some provision to prevent overvoltage
events on the logic signal pin is recommended. Automotive +12 power is allowed
to have occasional 60V spikes... a fan doesn't usually care, but logic inputs will.

 

[amdx]

On 2/26/2015 11:43 PM, S Keith wrote:

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 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

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 about half based on noise and perceived air flow. However, if I disconnected the

DMM and reconnected on the 2000 Ohm setting, the fan would not turn. Reapplying 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 really okay with full speed.

I plan to experiment a bit this weekend. I'm not sure what would be accomplished 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,

The point was, if you can measure the motor voltage, then you could
increase the control pin until you get 12 volts on the motor. You
wouldn't want to raise it any more without knowing the circuit.
This assumes it is PWM.

On the 200 ohm setting you had enough current to over come the torque
required to start the motor, when you switched 2000 ohm it had enough
current to keep it rotating. Trying to start at the 2000 ohm setting
there was not enough current to produce the torque required to start the
motor.

Mikek





---
This email has been checked for viruses by Avast antivirus software.
http://www.avast.com

 

[S Keith]

On Friday, February 27, 2015 at 1:08:23 PM UTC-7, Ralph Mowery wrote:

I don't know the setup of your motor, but I was thinking that the supply is
the 12 volts and the lead with the 3 volts on it is the speed determing
wire. With a constant voltage on it,it is running full speed as it is a
'pulse' of 100 % . If you fed that wire some pulses then the speed would
change. Just a guess.

That is exactly what I was thinking.

When meters are fed pulses they can show many differant voltages depending
on the meter. Most will try and average them. Some will try to show a
true RMS value. Often depends if an analog or digital meter.

These are NOT precision instruments; however, the half dozen I have all correlate nicely with each other and other hardware. I'll find out what it does this weekend.

Thanks,

Steve

 

[S Keith]

On Friday, February 27, 2015 at 1:14:10 PM UTC-7, Tim Wescott wrote:

On Thu, 26 Feb 2015 17:19:18 -0800, S Keith wrote:

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, an essential USENET skill (any-"NET" for that matter) is to just
ignore the trolls. Phil oscillates between being a valuable resource on
audio circuits and an incensed troll. No amount of talk will get him down
from the trees when he's chosen to go there, so you need to just ignore
him. Or bait him on purpose, if it makes you feel better. But trust me
-- many people have tried talking sense to him, and all it does is get him
worked up.

He's:

--> very trustworthy on audio stuff

--> less reliable on regulatory and "that'll burn your house down" stuff

--> not at all reliable when you can tell that there's spittle hitting
the computer screen.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Tim,

Agreed. I have been away from usenet for the better part of 7-8 years, and I lost that skillset. My biggest regret is that plonking with Google Groups isn't an option.

Lesson learned!

Thanks,

Steve

 

[S Keith]

On Friday, February 27, 2015 at 1:23:41 PM UTC-7, Tim Wescott wrote:

As has been expressed to you before, nothing can, always and
simultaneously, put out a given voltage at a given current, because the
current at any given voltage (or voltage at any given current) depends on
the load, not the source.

I see contradiction in your statement. The VR suggested early in the thread puts out 3V/100mA (max).

Would it be more accurately stated that I wish to produce a 3V, 2mA max power supply?

So in the context that you quote it, your "3V, 2mA output" is completely
meaningless. Something else is happening, and I suspect that you want to
know what that is.

Perhaps, but I definitely want to know what it is.

I do not know the components involved or their limitations. I only know
the "experimental" input and outcome.

From elsewhere I know that you learned the "3V, 2mA" figure from a meter
data sheet. Here are some observations:

1: That's a data sheet. Data sheets lie on rare occasions, commonly
obfuscate, and are often not written to be easy to understand for
the general public. So you don't want to trust your reading of one
unless it's been a close reading, and you're firm in your knowledge
of the subject.

2: You've been assured that it's physically impossible by more than
one person, the majority of whom have been calm and helpful about
it.

3: A common practice, when one runs into questions on a data sheet, is
to MEASURE. Get another meter (or two), and measure the voltage
on the "3V, 2mA" input, and measure its current, too, if that floats
your boat.

Phil is the source of that assertion. I explicitly stated that I measured the output of one DMM with another DMM. The DMM that spun the motor MEASURED 3V, 2mA

4: Just because it works with 3V (or whatever) on the control input
doesn't mean that it won't suffer long-term damage. Your best bet
would be to measure what's going on at that input when the thing is
in a car, working. If you can get access to a car (even a junker)
and MEASURE, you'll know way better. It could be that the input is
supposed to be 12V, or 5V, or something, and that 3V just barely
turns it on, and leaves it sputtering on and off. It could be that
3V is exactly right -- you don't know.

Agreed. As I mentioned before, the fan only runs when temperature hits certain significant values. I am unable to create that environment reliably to get the fan to come on AND measure it. When Phoenix temps hit 110°F+, then I can probably do exactly that.

The values I have measured are consistent with 12V computer fan PWM specifications. I realize that ultimately means nothing, but I choose to use it as supporting data.

Thank you for your help.

Steve

 

[amdx]

On 2/27/2015 12:41 PM, S Keith wrote:

On Friday, February 27, 2015 at 9:34:16 AM UTC-7, amdx wrote:

The point was, if you can measure the motor voltage, then you could
increase the control pin until you get 12 volts on the motor. You
wouldn't want to raise it any more without knowing the circuit.
This assumes it is PWM.


More potential ignorance... Since it's PWM, the fan voltage is always 12V, it's just pulsed.

Will a meter read an "average" voltage, e.g., if it's at 75% duty cycle,

I will read 9V assuming a linear relationship between duty cycle and voltage?

Yes it will, possibly somewhat meter dependent, but as I understand it,

linear relationship, as you say.
Mikek



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