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Transistor DC motor moving, but slowly

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jbchrist

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Hi,

I am studying electronics and am trying to figure out how to use transistors. I have a TIP 31 NPN transistor that I am using to sink the DC motor load (like the image attached). I don't have a resistor from the PIC to the base.

I tested it and it works just fine. When my PIC input goes high it runs, when it goes low it stops just as anticipated. HOWEVER, the motor moves slowly. I don't get it. Can someone help me understand what is going on?

I've read many many threads regarding this topic, but I can't seem to answer my question.

Thanks
 

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Hi,

I am studying electronics and am trying to figure out how to use transistors. I have a TIP 31 NPN transistor that I am using to sink the DC motor load (like the image attached). I don't have a resistor from the PIC to the base.

I tested it and it works just fine. When my PIC input goes high it runs, when it goes low it stops just as anticipated. HOWEVER, the motor moves slowly. I don't get it. Can someone help me understand what is going on?

I've read many many threads regarding this topic, but I can't seem to answer my question.

Thanks

Hi,
You should have a base resistor.
The most likely reason why the motor is running slow is because the base current from the PIC is about 20mA, depending upon the motor current required, the TIP is not saturating.

Whats the voltage on the supply to the motor and also the voltage on the collector when the motor is running.
 
Hi,
You should have a base resistor.
The most likely reason why the motor is running slow is because the base current from the PIC is about 20mA, depending upon the motor current required, the TIP is not saturating.

Whats the voltage on the supply to the motor and also the voltage on the collector when the motor is running.


The voltage on the collector is 12V. I'm not sure what the voltage is on the collector when the motor is running. I THINK the voltage when the motor is running is 8.4V, but I'd have to confirm that.

So from what I gather I'm opening the flow for the current which enables the motor to run, but just not all the way?
 
The voltage on the collector is 12V. I'm not sure what the voltage is on the collector when the motor is running. I THINK the voltage when the motor is running is 8.4V, but I'd have to confirm that.

So from what I gather I'm opening the flow for the current which enables the motor to run, but just not all the way?

hi,
If the voltage on the collector is 8.4V, there's only about 4V across the motor.

Put a 220R in the connection from the PIC pin to the base and change the TIP31 to a npn Darlington transistor TIP120 or TIP 122.
That will increase the current thru the motor.
 
ok, I'll try that tonight, but how did you determine the resistor?


hi,
The TIP darlington will have a Vbe of about 1.4V, the output of the PIC is about 5V, so a 220R will allow a base current of 16mA.
5V-1.4V/220R = 16mA

As the gain of the TIP darlington is about 1000, the current thru the collector will be 1000 * .016 = 16A maximum

If the motor winding is say 6:eek:hm: then the current will be 12V - 1Vsat say, 11V

11V/6==2Amps, so the transistor will be saturated, thats what we want from a switching circuit..OK

What is the resistance of the motor winding.?
 
Last edited:
Use a power MOSFET, like the IRF540.
 
The TIP darlington will have a Vbe of about 1.4V, the output of the PIC is about 5V, so a 220R will allow a base current of 16mA.
5V-1.4V/220R = 16mA

As the gain of the TIP darlington is about 1000, the current thru the collector will be 1000 * .016 = 16A maximum
No.
The output voltage of a PIC is 5V only when it has no load. Its voltage drops with load current. Its output is about 0.8V and about 50mA when it drives the base of a transistor directly. Then the PIC will overheat because its max allowed current is 25mA.
The PIC's output is about 4.2V with a current of about 13mA when it drives the base of the darlington through a 220 ohm resaistor. The darlington will saturate well when its load current is 3.25A or less.

The current gain of a transistor is used when it is linear. A saturated transistor is a switch and is not linear. The max saturation voltage of a transistor is listed when its base current is 1/10th its collector current regardless of its current gain (except the BC548C which has a current gain of up to 800 and its saturation voltage is listed when its base current is 1/20th its collector current).
The saturation voltage loss of a darlington is when its base current is 1/250th its collector current regardless of its current gain.
 
No.
The output voltage of a PIC is 5V only when it has no load. Its voltage drops with load current. Its output is about 0.8V and about 50mA when it drives the base of a transistor directly. Then the PIC will overheat because its max allowed current is 25mA.
The PIC's output is about 4.2V with a current of about 13mA when it drives the base of the darlington through a 220 ohm resaistor. The darlington will saturate well when its load current is 3.25A or less.

The current gain of a transistor is used when it is linear. A saturated transistor is a switch and is not linear. The max saturation voltage of a transistor is listed when its base current is 1/10th its collector current regardless of its current gain (except the BC548C which has a current gain of up to 800 and its saturation voltage is listed when its base current is 1/20th its collector current).
The saturation voltage loss of a darlington is when its base current is 1/250th its collector current regardless of its current gain.

If you take the time and trouble to read my posts carefully, thats what I have said.
Whats the point of rambling on about a BC548C when its not even mentioned by the OP.?

The OP is trying to drive a simple dc motor from the output from a PIC via a medium power transistor.
The way he had it connected is wrong and a way to correct it has been posted, so stop nit picking.:mad:
 
Thanks for the posts everyone! I was able to fix my problem. I'm not exactly sure what I was doing wrong, but I must have been hooking it up wrong or something. What I ended up doing was ripping everything out of my breadboard and start hooking things back up. When I turned it on everything was working just fine! I don't get it. I SWEAR I was hooking it up properly before! Oh well, at least now I have a working example to work from. I made sure to document everything I did so I can reproduce it in the future.

I ended up still using the TIP31 or TIP31A (I can't remember which one). I hooked the 12V to the motor and then the motor to the collector. The emitter was hooked directly to ground. The uC was hooked to the base via a 2K resistor. I made sure that the ground used for the uC was connected to the 12V ground so that I had a common ground.

My code for the uC was simple it just sent a high signal to a pin for a second then Low for a half a second.

Now it works beautifully. I really appreciate your help everyone! Thanks
 
Amazing.
With a 2k base resistor then the base current of the TIP31 is only 2mA. Then the current for the motor is only about 50mA.
If the motor draws more than only 50mA then the TIP31 will not be turned on hard and will get hot. The motor will not run at full speed.
What is the resistance of the motor?
 
12.1 ohms. I just happened to check that last night.
Then the motor draws nearly 1A when it starts and when it works hard. The base current for the TIP31 should be 100ma, not just 2mA.
The motor will run stronger if you replace the 2k base resistor with 220 ohms.
 
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