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Take a potentiometer as a voltage divider?

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Jimmylee

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Good day, all
I have a small fan setup and I wanted to add a potentiometer to use as a variable resistor. I've come up with this design which uses a logarithmic potentiometer in place of a series resistor.
Initially, the red line did not exist, I was only going to use one terminal and the wiper, but while trying to come up with an appropriate resistance size I could only find information about a voltage divider.
potentiometer in place of a series resistor.png

(I'm sorry, there is a typo on the diode label)
I was wondering if this is correct,it doesn't look right to me, but I'm relying on my terrible electronics knowledge from elementary school, which is known to be less than optimal at times.

Anyone have ideas of it? Many thanks.
 

No way will that work unless your fan draws very tiny current.
Most likely scenario is R1 will burn out.
What you have to consider is that the fan probably presents a resistance of only a few Ohms so whether you use the potentiometer or variable resistor it will not let enough current pass through unless it is adjust almost at the end of its travel. Under those circumstances, the short length of track carrying current in R1 will have to dissipate quite a lot of heat.

A far better solution which is also energy efficient is to omit R1 and wire the fan directly between the transistor and +5V then drive GPIO18 with a PWM signal to control the average motor current.

Brian.
 

Thanks for helping.
The fan is rated for 5v, 0.15A 0.75W
I just reread Potentiometer Voltage Divider Basic (may be i don't understand well here mentioned). However, for a bit more info, I already do use PWM to control the fan speed, which in turn is regulated by the temperature of the SOC, it works quite well.
This idea was to add an additional option of powering the fan with 5v or about 3v that I didn't fully consider the values of some things.
When using 5v, I was going to enable the software control, and when using 3v, I would disable it.
 

Very simply
Your Fan have resitance 5V/0,15 = 33 Ohm if you add in series resistor 1-10kOhm, current over fan on 1kOhm resitor will be
5/1033= 4,8mA (much less because you have more losses on the transistor)
If you insist on a potentiometer, must use any as 50ohm potenciometr
5/(33+50)= 60mA 2V on Fan 3V on potenciometr on 50Ohm.
if you want 3-5V on the fan use 50ohm potentiometer and paralel fixed resistor 39 Ohm (R1 resistance will be 0 -22 Ohm).
Any as

PT.jpg

A
 
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But note that the start-up current is likely to be much higher than the running current. You may have to set the potentiometer to a lower resistance to start the motor then increase the resistance to slow it to the desired speed. Simple PWM control would work better at all supply voltages.

Brian.
 

A 33Ω load on a 1kΩ potentiometer isn't going to work very well, because the pot will be burn out.
So I have the possible solutions from my online search to adjust:
1) Use an emitter follower stage between the pot and the fan to reduce the load on the pot.
2) Use PWM to control the motor speed.
 

Hi,

Yes, makes sense.
The PWM solution is the better in my eyes, but needs a higher oart count.

I guess you won't continously adjust the pot, rather find a suitable value and you are done.
If so you may use suitable fixed "speed reducers".
Thus you may try fixed resistors in the range 3Ohms (rather fast) ... 300 ohms (rather slow)
Or better in terms of voltage stability (torque stability, RPM stability): series zener diodes.

The more sophisticated solution is a "fan control IC".

Klaus
 

Simple solution: Calculate. and repeat with new guess values.

By the way, logarithmic potentiometers are used for audio output circuits. Reason? Human ear sensitivity is logarithmic in nature. You will need a linear potentiometer.

Next you will need to estimate the power rating for the potentiometer. Your 5V motor fan will consume about 1W power - say about 200mA. You need to see the motor label.

The fan will stall below 20mA (guess value)- but most low power fans running on 5V are BLDC and they are notoriously difficult to speed control. Regular brushed DC motor fans are noisy and unreliable but they respond to voltage well as speed control.

Because of the transistor switch, you fan will never see 5V (about 4.5V). The estimated resistance for the fan will be around 25 ohms. For a current of 20mA, you will need a resistance of 250 Ohms. That means your series resistors will be about 200 Ohms. So you need a potentiometer for this value.

A 200 Ohm potentiometer carrying about 20mA shall have a power rating of 4W; double this as a safety measure (input and output impedances of potentiometers are really messy to calculate). So you will need a 5-10W rating 200 Ohm potentiometer.

If you want to get fancy, use a DC motor control IC, they can do almost everything except making coffee! And they are cheap and widely available. But they will not work with the modern instrument cabinet fans that run at 5-12V and have no brushes (BLDC).
 

Yep...before going further, doing some calculations are necessary, because there are undoubtedly wrong assumptions here about the resistance of the fan at different voltages.
 

Capture.PNG

Maybe others can correct me here. But why not use your transistor as a variable current source instead of static? Use your voltage divider to drive your base. Sorry about the bad drawing. Perhaps place the potentiometer as the top resistor and pick a lower resistor to ensure your transistor never turns off. Plus, no risk of burning your pot.

All those PWM suggestions are good, but this seems to really simple, perhaps I am missing something.
 
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If you want speed control then use a NPN emitter follower
and run the base off the potentiometer wiper, but be sure
to add some series resistance so that "full on" won't be able
to burn out the pot with (5V-Vbe)/0ohms worth of current.

If you want torque control then a current source/sink
makes sense. You can get some control with a current
source as fan load will follow speed, but it'll not be a
very linear relation.
 

View attachment 169155


All those PWM suggestions are good, but this seems to really simple, perhaps I am missing something.
You are really missing something. I cannot figure out how turning the potentiometer can affect the base current.

Transistors are current devices; FETs are voltage dependent devices.

Using a potentiometer as a voltage divider is a good idea. Bit they cannot source much current. So you should not use a power transistor driven by a potentiometer.
 

Transistors are V controlled devices -

1619783001809.png


Regards, Dana.
 

Hi,

I agree that a BJT is voltage driven.

But if you drive it with a constant voltage you get thermal drift problems.
Since V_BE drifts with about -2mV/K you need to modify the drive voltage with the same -2mV/K.

*****
If one "wrongly ?" treats the BJT as current driven ... one reduces/avoids this problem.

*****

It´s good to know the details, the benefits and drawbacks, so one can use the one or the other method.

Klaus
 

There is a continual back and forth about whether a BJT is voltage or current controlled on these forums.

The pedantic types insist that the solid-state physics shows that it is voltage-controlled current-source, which is technically true, and that can be useful for small signal analysis of a circuit using the transconductance gain of the device.

But for DC bias and large signal/switching design it works better to consider the BJT as a black-box current-controlled current-source with a low input-impedance (as a forward biased diode), using the current-gain (Beta or hFE) for the calculations.
Trying to use the voltage-control equations for these calculations will be difficult and generally give poor results.

A MOSFET is an example of a device that is always considered a voltage-controlled current-source, with a very high DC input impedance.
 

    Hawaslsh

    Points: 2
    Helpful Answer Positive Rating
But for DC bias and large signal/switching design it works better to consider the BJT as a black-box current-controlled current-source with a low input-impedance (as a forward biased diode), using the current-gain (Beta or hFE) for the calculations.
Capture.PNG

That is a great way to explain it, thanks! I definitely applied the small signal model to my previous thought. Not sure if even useful to the OP, but for my edification and correction to my previous circuit: would a simple current mirror do the trick? Doesn't solve the problem of the current going through the POT, but easier than filtering and buffering a PWM signal IMHO.
I spend my entire carrier stuck in a cleanroom making devices, its about time I learn to use them :D
 

Hi,

I agree that a BJT is voltage driven.

But if you drive it with a constant voltage you get thermal drift problems.
Since V_BE drifts with about -2mV/K you need to modify the drive voltage with the same -2mV/K.

*****
If one "wrongly ?" treats the BJT as current driven ... one reduces/avoids this problem.

*****

It´s good to know the details, the benefits and drawbacks, so one can use the one or the other method.

Klaus
Of course then we have to deal with beta dependence on T which gives us grief. Driving
with a current , unless we compensate for this, is akin to your point about Vbe being a
f(T).

1619800119176.png


Regards, Dana.
 

Hi,

I agree with you.

May I ask for three V_BE values? (2N3903, V_CE = 1.0V, 25°C)
1) V_B for I_C = 20mA
2) V_B for I_C = 10mA
3) V_B for I_C = 7mA

Thank you

Klaus
 

Both Vbe and current-gain change with temperature.
My point is that it's easier to use a current model than a voltage model for the large signal or bias analysis of a BJT.
 

*****
If one "wrongly ?" treats the BJT as current driven ... one reduces/avoids this problem.

*****

It´s good to know the details, the benefits and drawbacks, so one can use the one or the other method.

Klaus
If you look at the device, the current carriers injected at the base is the single most important controlling factor of the collector current.

That the base emitter junction acts like a typical diode is incidental.

The most important parameter for a bipolar junction transistor is the current gain and that is seen in the hybrid model.

The voltage applied to the base is important because that controls the base current. The base current in turn controls the collector current.

A small signal equivalent circuit model of a transistor clearly shows what I am trying to say.

Glad to see that there is no confusion about a FET being a voltage controlled device.
 

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