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Can't figure out this simple problem

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ThE_sPaCeCoWbOy

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I hook up a 2n2700 up to a 9-18VDC motor that can handle 1.98A.

I limited the amps to 200mA and set the Vd = 9V, put a 2.7ohm resistor in series iwth the motor. Turned the transistor on, and it works, but the actual numbers i'm measuring (V, A, R) are not what I calculated and its not the tolerance difference either.

Vd=9
Vmotor = 8 V
Rmotor = 2.4
R = 2.4

I should get Id=208mA

Now when I measure, I get Vmotor = 7.3, Id = 270mA. This is telling me that I have 6.3 ohms of resistance and my meter is telling that this is not what I have. Any clues where this added resistance might be coming from or if i'm doing something wrong?

Thanks,
space
 
If the motor is spinning then it will generate a voltage that opposes the input voltage. It might be this thats throwing off your measurements.
 
ThE_sPaCeCoWbOy said:
I hook up a 2n2700 up to a 9-18VDC motor that can handle 1.98A.

I limited the amps to 200mA and set the Vd = 9V, put a 2.7ohm resistor in series iwth the motor. Turned the transistor on, and it works, but the actual numbers i'm measuring (V, A, R) are not what I calculated and its not the tolerance difference either.

Vd=9
Vmotor = 8 V
Rmotor = 2.4
R = 2.4

I should get Id=208mA

Now when I measure, I get Vmotor = 7.3, Id = 270mA. This is telling me that I have 6.3 ohms of resistance and my meter is telling that this is not what I have. Any clues where this added resistance might be coming from or if i'm doing something wrong?

For a start, a motor isn't a resistor, so you can't use it in calculations as such - it's effective 'resistance' will vary massively with load, and will probably vary considerably with supply voltage.
 
then there is resistance coming from somewhere. By only have a 2.4 ohm resistor, then I would have .416A going through the transistor. I'm only reading .27A which is still higher then I wanted.

Thanks,
space
 
As has been already mentioned, you can't look at a motor as you would at a plain resistor.
A motor is designed to deliver power, Watts = Volts x Amps.
If you have a steady supply voltage then the Amps drawn from it will increase as you demand more power from the motor. If you must use the resistor analogy then look at the motor as a variable resistor whose value depends on the current and hence the load on it.
If there is no load on the motor its 'resistance' will still not equal the value measured at standstill, due to back EMF.
Klaus
 
As has been already mentioned, you can't look at a motor as you would at a plain resistor.
A motor is designed to deliver power, Watts = Volts x Amps.
If you have a steady supply voltage then the Amps drawn from it will increase as you demand more power from the motor. If you must use the resistor analogy then look at the motor as a variable resistor whose value depends on the current and hence the load on it.
If there is no load on the motor its 'resistance' will still not equal the value measured at standstill, due to back EMF.
Klaus
 
I didn't take the motor through my second calculation. Only used the 2.4 resistor I place in series with the motor. I'm still not getting anything close to what I calculated.

Just looking for ideas on what might be wrong, something I'm overlooking, or something that I'm doing wrong.

thanks,
space
 
ThE_sPaCeCoWbOy said:
I didn't take the motor through my second calculation. Only used the 2.4 resistor I place in series with the motor. I'm still not getting anything close to what I calculated.

Just looking for ideas on what might be wrong, something I'm overlooking, or something that I'm doing wrong.

We've already explained that you can't do these calculations with a motor in the circuit!.

If you want to confirm your faith in ohms law, replace the motor by a resistor, and then do the calculations and measurements.

What are you trying to do anyway!.
 
Look Spacecowboy (whats with the 1337 speek for yr name?) a motor is in effect an inductor

It will have some series equiv resistance because it is a real inductor. Thus there will be some reactance (thus some volt drop due to the coils)

You have to remember a motor is a coil, thus an inductor

also

Remember volts = speed
and Amps = torque


--EDIT--

sorry just had my coffee so woken up now.
Also note that as the motor spins it will generate a backEMF, now you are pproviding alot higher voltage from the battery so it wont opose the motor spinning but it will reduce the useful voltage seen by the motor thus the current will dro[
 
We've already explained that you can't do these calculations with a motor in the circuit!.

I don't recall anyone stating that. I know you stated you cant use the motor as a resistor as such in these calculation...but none the less, theres an answer so thanks.

and by the way,

(whats with the 1337 speek for yr name?)

I have no clue what that means.
 
ThE_sPaCeCoWbOy said:
We've already explained that you can't do these calculations with a motor in the circuit!.

I don't recall anyone stating that. I know you stated you can use the motor as a resistor as such in these calculation...but none the less, theres an answer so thanks.

I don't know if that was a typo?, but both myself and Klaus have previously stated that you "can't" consider a motor as a resistor.

and by the way,

(whats with the 1337 speek for yr name?)

I have no clue what that means.

Neither have I?. I presume it refers to your upper and lower case letters?.
 
Can't use the motor as a resistor...edit my previous message.....but noone said you can't use the same equation to solve for it.

anyhow, thanks for the help and also for the 1337 info....i think...

space
 
ThE_sPaCeCoWbOy said:
Can't use the motor as a resistor...edit my previous message.....but noone said you can't use the same equation to solve for it.

anyhow, thanks for the help and also for the 1337 info....i think...

space

What!!!

I said you couldn't in my post that said an machine is an inductor!!!
how can you use Ohm's law with an inductor.

And as to the 1337 stuff im just on a jibe abt people using that style at teh moment to promote themselves

http://www.1337.net/

butsince you idn't know what I was on abt sorry.
 
maybe Nigel was making some joke about how the site didn't make any sense...or something....

"errr?" is a statement I agree with. Whole heartedly.
 
Here is a current limiter that looks about like a 0.7 ohm resistor in series with the motor for currents below 200ma. At 200ma, the voltage across the motor will drop depending on the load on the motor, but the current will be 200ma independent of the load.
For the unused op amps in the package, tie the output to the inverting input, and connect the noninverting input to GND. This will prevent the unused op amps from screwing up the bias on the ones that are used.

A better solution might be a low dropout voltge regulator with current limiting, but this would be more complicated.
 

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