TIP120 + 555 DC Motor Speed Control = Hot Battery

Thread Starter

yegglet

Joined Feb 4, 2023
5
Hi Everyone!

I'm using a 9V to power this circuit meant to control the speed of a 6-9V DC motor. The only difference is that I'm using a 50k pot instead of a 100k pot. This is meant for a handheld device so ideally I'd like the battery to last a while, the hot battery indicates me to that this circuit is drawing a lot of current. I would appreciate any suggestions for modifying this circuit to be more energy efficient! I was also thinking about maybe replacing the 9V with two LiPo batteries in series because of improved current output and rechargability, does that seem like a reasonable idea? Thanks in advance!

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ElectricSpidey

Joined Dec 2, 2017
2,758
The pot is making very little difference...it's the motor that is drawing the most current.

You can get rid of that pull down resistor it's not needed. (Darlingtons already have internal pull downs)

Change LED1 to a super bright and increase R4.

You may be able to raise R3 depending on the current needed by the motor.
 
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Thread Starter

yegglet

Joined Feb 4, 2023
5
The pot is making very little difference...it's the motor that is drawing the most current.

You can get rid of that pull down resistor it's not needed. (Darlingtons already have internal pull downs)

Change LED1 to a super bright and increase R4.

You may be able to raise R3 depending on the current needed by the motor.
Thanks, I'll try replacing R3 with a 5k resistor thanks!
 

panic mode

Joined Oct 10, 2011
2,715
i agree about using mosfet with low Rdson.

using Darlington here is not helping...

increasing R3 will lower the base current. in original circuit that current is barely enough to turn the transistor on. this is switching mode (PWM) and transistor should be fully on or fully off. 555 output is about 2V less than supply. so 12V-2V-Vbe divided by 1k was resulting in under 10mA of base current. hot battery suggests significant current and you are not going to make notable savings by reducing down 10mA current. also Darlington has huge Vce, resulting in considerable losses (even when fully on). at lower voltage such as 6V this gets even worse, the 2V lost on 555 and 1.4V lost on Vbe means R3 only gets about 2.6V. using 5k for R3 value means base current will be just 0.5mA. This means Vce will be huge 2-3V for 2A motor current. That is comparable to voltage across motor. You probably do not want half of battery voltage (and energy) lost on the transistor. some drop is unavoidable but loss of 0.1V is much better than loss of 3V.


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BobTPH

Joined Jun 5, 2013
8,813
Changing the darlington to a MOSFET will cause more current draw because more voltage will be supplied to the motor.

Measure the current draw of the motor connected directly to the battery. The problem is likely that the motor requires too much current for that battery. Either that, or you have a short somewhere.
 

ElectricSpidey

Joined Dec 2, 2017
2,758
True a MOSFET will allow more current to the motor but the TS is using PWM so that could be compensated there.

Also using a low power 555 will save a little power. (if not already being used)

But in the end changing the circuit will only help in the margins.
 

ElectricSpidey

Joined Dec 2, 2017
2,758
I didn't calculate the frequency of your circuit, but another thing to consider in reducing switching losses would be to use the lowest frequency possible while still getting acceptable low motor performance.

This is especially important if selecting a low Rdson MOSFET because it requires shuffling a large amount of charge in and out of that large gate.
 

crutschow

Joined Mar 14, 2008
34,285
I agree that the only change to significantly increase the efficiency of the circuit is to use a low Rds(on) N-MOSFET in place of the Darlington.
That should reduce the circuit power loss by at least 10%.

For the MOSFET, eliminate R5 and reduce R3 to about 20 ohms.

If you reduced the pot to 50kΩ, then increase C2 to about 200nF.

If you measure the motor current draw, you can then determine the approximate battery life, from its ampere-hour (Ah) rating.
 

Audioguru again

Joined Oct 21, 2019
6,674
The current for a brand new motor is very low when it is not doing anything. The current for an old motor might be much more when it needs lubrication and it is driving a heavy car or a large propeller.
You never told us what the motor is driving.
 

MaxHeadRoom

Joined Jul 18, 2013
28,619
Also a good idea is to get in the habit of return the pot to zero before the next turn-on time.
Mimicking the typical treadmill boards that do this automatically. although theirs is for safety reasons. ;)
 

Audioguru again

Joined Oct 21, 2019
6,674
One spec on the internet shows that the motor draws a high current when loaded or stalled. That would cook the little 9V battery quickly.
My radio controlled airplanes shut off the motor to avoid a fire when it becomes stalled.
 

Thread Starter

yegglet

Joined Feb 4, 2023
5
The current for a brand new motor is very low when it is not doing anything. The current for an old motor might be much more when it needs lubrication and it is driving a heavy car or a large propeller.
You never told us what the motor is driving.
It's driving a 1oz 45mm diameter acrylic disc with magnets in it. The field created by the spinning magnets vibrates steel guitar strings when in close proximity. I think the drag of the magnets to the strings is introducing a lot of friction too.
 

Thread Starter

yegglet

Joined Feb 4, 2023
5
I agree that the only change to significantly increase the efficiency of the circuit is to use a low Rds(on) N-MOSFET in place of the Darlington.
That should reduce the circuit power loss by at least 10%.

For the MOSFET, eliminate R5 and reduce R3 to about 20 ohms.

If you reduced the pot to 50kΩ, then increase C2 to about 200nF.

If you measure the motor current draw, you can then determine the approximate battery life, from its ampere-hour (Ah) rating.
Wow this is great advice! Thanks so much! Do you think a 30n06 MOSFET would suffice? That's all I have at the moment.
 

LowQCab

Joined Nov 6, 2012
4,029
Everybody missed the fact that there are NO Bypass-Capacitors in this Circuit.

This can cause all sorts of wacky, and unpredictable behaviors.

A ~470uF, plus a 100nF Ceramic-Capacitor,
attached as closely as possible to the 555 should be the #1-priority.

Bi-Polar-555s are notorious for acting badly with no Bypass-Caps.
Read the Spec-Sheet and believe it.
.
.
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