Motor acting weird

Thread Starter

Man10

Joined Jul 31, 2018
48
Calculate the base current:
1) Base voltage for a TiP110 is typically 1.8V on its datasheet.
2) The 4.2V battery is almost dead at 3.8V.
3) You have a 1 ohm resistor limiting the base current.
4) Then the base current is (3.8V - 1.8V)/1 ohm= 2A.
5) If the battery is working then you destroyed the base of the TIP110 because its maximum allowed base current is only 50mA.
For a load of 2A then the base current is shown to be usually 2000mA/250= 8mA, then the base resistor is (4.2V - 1.8V)/8mA= 300 ohms.
The saturation voltage loss is 1V to 2.5V so the motor gets 3.2V to 1.7V when the battery is fully charged at 4.2V.
I measured the base current with a multimeter, I measured 430 milliamps base current. And the tip110 transistor will still spin the motor without the arm or any load.
 

Thread Starter

Man10

Joined Jul 31, 2018
48
Why will the motor lift the arm when directly connected to the battery, without the transistor and not lift the arm when motor connected to transistor and transistor connected to battery?
 

LesJones

Joined Jan 8, 2017
3,127
IF the transistor has survived the abuse the voltage between the base and emitter terminals on the transistor should be between 1.2 and 1.4 volts. (This is because the TIP110 is a Darlington transistor which is really two transistors so there is two base emitter junctions in series.) So IF the battery voltage is still 3.8 volts UNDER LOAD there would be about 2.4 volts across the 1 ohm resistor which would give a current through it of 2.4 amps. Even if the transistor was good and being driven with a sensible base current there would be the saturation voltage of 2 volts between collector and emitter so that would be subtracted from the 3.8 volts battery voltage. this only leaves 1.8 volts across the motor. You would be better off using a logic level power mosfet with a low RDS(on) value than the TIP110.

Les.
 
Last edited:

Audioguru again

Joined Oct 21, 2019
2,968
I measured the base current with a multimeter, I measured 430 milliamps base current. And the tip110 transistor will still spin the motor without the arm or any load.
The datasheet for the TIP1210 says its maximum allowed base current is 50mA so you probably destroyed it.

Why will the motor lift the arm when directly connected to the battery, without the transistor and not lift the arm when motor connected to transistor and transistor connected to battery?
The datasheet for the TIP110 says that its maximum saturation voltage loss is 2.5V. Your battery is only 3.8V (with no load?) so it is not charged.
Measure the voltage at the motor when the transistor is used and when the motor cannot lift the arm.
 

Thread Starter

Man10

Joined Jul 31, 2018
48
The datasheet for the TIP1210 says its maximum allowed base current is 50mA so you probably destroyed it.


The datasheet for the TIP110 says that its maximum saturation voltage loss is 2.5V. Your battery is only 3.8V (with no load?) so it is not charged.
Measure the voltage at the motor when the transistor is used and when the motor cannot lift the arm.
Could you post a schematic of how to measure the voltage at the motor?
 

djsfantasi

Joined Apr 11, 2010
7,620
Your multimeter or DMM has two probes marked + and -

Set the multimeter to measure DC volts and choose the 10V scale.

Determine which motor terminal is connected to the + side of the battery. Place the + probe on that terminal. Place the other probe on the other terminal.

Read the voltage and post it here.

Note: Your hand drawn schematic has the battery backwards for the way the transistor is drawn. The - side of the battery should go to the transistor.
 

MrSalts

Joined Apr 2, 2020
237
Have you connected the transistor correctly?

I assume your transistor is dead because of the 1ohm transistor. Get a new transistor and try again.

7E42C1A9-05D3-45E0-A451-2A1C56541696.jpeg
 
Last edited:

MrSalts

Joined Apr 2, 2020
237
Here is a critical value to understand about the TIP110.

It means that the transistor is not very efficient - it is not exactly a good switch for a 4v battery because the voltage across the transistor emitter collector junction is up to 2.5v So you only have 1.5v left to power the motor.

0C2D7E0C-68AF-4BF7-9E35-BF835A09E30E.jpeg
 

MrSalts

Joined Apr 2, 2020
237
The 2.5v emitter to collector voltage is a maximum specification. There is no “typical” value listed but even when a typical value is listed, you can’t ask the store for a “typical” TIP110 - you just get what you get and the manufacturer promises it will be less than 2.5 in the case of a TIP110. It might be as low as 0.75v but it can’t be lower on a Darlington. So likely somewhere from 0.75 to 2.5.

You can measure the value yourself by connecting the motor as shown in post 21 and use your meter to measure from E to C and to confirm your math, measure across the motor connections They should total your battery’s voltage just remember, a but that transistor does have a non-negligible voltage across those terminals
 

MrSalts

Joined Apr 2, 2020
237
This diagram is incorrect. See the attachment to post #27
Yes, thank you. I was trying to emphasize the pi out of the transistor since many people get confused about the difference in assignments between The small packages and the TO-220 packages . It has been corrected.
 

Thread Starter

Man10

Joined Jul 31, 2018
48
The 2.5v emitter to collector voltage is a maximum specification. There is no “typical” value listed but even when a typical value is listed, you can’t ask the store for a “typical” TIP110 - you just get what you get and the manufacturer promises it will be less than 2.5 in the case of a TIP110. It might be as low as 0.75v but it can’t be lower on a Darlington. So likely somewhere from 0.75 to 2.5.

You can measure the value yourself by connecting the motor as shown in post 21 and use your meter to measure from E to C and to confirm your math, measure across the motor connections They should total your battery’s voltage just remember, a but that transistor does have a non-negligible voltage across those terminals
Is it possible to exceed the maximum emitter to collector voltage?
 
Top