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

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

Steve

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

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.

How do you know it is a PWM fan ??

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.

So neither do we.

.... Phil

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

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

Cost it is obviously all crap.

Why the secrecy ?

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

You must be some sort of anal nut bag.

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

I have a few questions. You put what you say was 2ma to the control pin and the fan spun up.

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.

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

reciate it.

Mikek,

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:

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

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... ... ...

beled "basics". You remind me of why I stopped using Usenet back in the da y

le jusr want others to do well in electronics.

ut that.

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

His type is pervasive. They are lacking something in either themselves or t heir lives and they fill the hole with, "Oh yeah, I'm smarter than you!"

Thanks again to all who have helped.

Steve

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

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?

That was my thinking. Thanks for confirming.

Steve

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

linear relationship, as you say. Mikek

ppreciate it.

and moving a lot of air.

ng (lower voltage, about 0.6V),

owever, if I disconnected the

applying the 200 Ohm setting

ching to 2000 Ohms.

really okay with full speed.

complished by probing the motor leads, but I'm game.

o come on involves high temp readings on the DC-DC converter,

Mikek,

I'm trying to understand PWM, and your questions are a bit confusing. Base d on what I've read about how computer PWM fans work, they "pulse" a curren t (around 3-5V and

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.

.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.

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.

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.

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.

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.

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

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 ( +12V360 ohms120 ohmsGND). 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.