Darlington transistor POWER AMPLIFYING

Thread Starter

Dan Aronin

Joined Dec 14, 2018
37
Hi all, I just couldn't find the answer on web so I had to ask you guys.

Based on my understanding, input power is ideally equal to the output power of a circuit unless it is dissipated to heat.

So I was looking at the ULN2003 Darlington controlling a 28byj-48 stepper motor using an Arduino Uno (I know it is NOT recommended due to high current risk).

I couldn't find any measurements of I\O current\voltage of the chip and my question is -

Is there a power amplification produced in order to drive the stepper, or maybe voltage is dropped and current increased?

Edit: Refernce:
Chip and stepper -
https://www.ebay.com/itm/28BYJ-48-2003-Stepper-Motor-Driver-Module-for-Arduino-DC-5V-Stepper-Motor-New/222714306166?_trkparms=ispr=1&hash=item33dace9e76:m:mjJNkHPamvTJId3Eeur0lNA

Controlling with arduino -
https://www.aranacorp.com/en/control-a-stepper-motor-with-arduino/
 
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neonstrobe

Joined May 15, 2009
81
Look at the web again for the datasheet on your choice of stepper motor. From what I found, it has 5V, 50 ohm coils so will need 100mA per coil. The Darlington chip can provide up to 500mA so should work.
As regards efficiency, you are essentially right that 100% efficiency means that an output power is the same as the input.
But with a motor you are getting mechanical energy out rather than electrical. So your input will be determined by your pulse rates and the output by the mechanical power of the motor. BUt stepper motor output power is not usually great, so will be low I suspect. Then there is as you say a part of the energy absorbed as heat.
In a motor electrical energy=> heat + mechanical
 

Thread Starter

Dan Aronin

Joined Dec 14, 2018
37
Look at the web again for the datasheet on your choice of stepper motor. From what I found, it has 5V, 50 ohm coils so will need 100mA per coil. The Darlington chip can provide up to 500mA so should work.
In a motor electrical energy=> heat + mechanical
Great info! Maybe you can suggest how to calculate the input current needed for the application? and is voltage buffered during the amplification?
 

Thread Starter

Dan Aronin

Joined Dec 14, 2018
37
Thanks for answering and dedicating time!!
I know it will work, but I want to understand how much current and voltage it will draw and output
 

Papabravo

Joined Feb 24, 2006
13,559
Just to be clear, the concept of power amplification does not mean that you can get more power out than power in. The power comes from the power supply and once it is consumed it is gone. The current consumption of the uC and the input side of the ULN2003 will be smaller by an order of magnitude than the power consumed by the motor.

I'm not sure what you mean by asking if "the voltage is buffered during the amplification". As I understand it the answer is no. The voltage from the motor comes from a power supply and there is no "buffering" as I understand that term. The concept of buffering is used in multi-stage amplifiers to prevent a later stage from affecting the behavior of a previous stage.
 

Alec_t

Joined Sep 17, 2013
11,163
I want to understand how much current and voltage it will draw and output
The supply voltage for the controller board appears, from the board picture, to be in the 5V-12V range but isn't necessarily the same as the motor supply voltage.
Depending on which mode the stepper motor is operated in, one or two coils at a time may be energised. As stated in post #2, for a 5V motor supply the (static) current drawn is 100mA per energised coil, so 200mA maximum. The current builds up non-instantaneously because of the coil's inductance.
 

Thread Starter

Dan Aronin

Joined Dec 14, 2018
37
The supply voltage for the controller board appears, from the board picture, to be in the 5V-12V range but isn't necessarily the same as the motor supply voltage.
Depending on which mode the stepper motor is operated in, one or two coils at a time may be energised. As stated in post #2, for a 5V motor supply the (static) current drawn is 100mA per energised coil, so 200mA maximum. The current builds up non-instantaneously because of the coil's inductance.
Thank! I guess the current input will be Iout/1000? (Hfe is 1000)
 

Alec_t

Joined Sep 17, 2013
11,163
So if you put 1mA@5v to the ULN2003 you can get 1000mA@5v (Hfe is 1000). Isn't the power output larger than the input?
No. The ULN2003 input current comes from the power supply and is 1mA base current plus 1000mA collector current =1001mA. The output current is just the collector current. So the useful output power is less than the input power. The power difference is manifested as heat.
 

Thread Starter

Dan Aronin

Joined Dec 14, 2018
37
No. The ULN2003 input current comes from the power supply and is 1mA base current plus 1000mA collector current =1001mA. The output current is just the collector current. So the useful output power is less than the input power. The power difference is manifested as heat.
Ok so you mean the actual current that is controlling the coils is coming from the power source so the Darlington acts like a switch?
 

BobTPH

Joined Jun 5, 2013
2,351
Exactly, it is a switch that controls the current going to the motor. There is also some voltage lost in going through the switch, so, although the motor sees all of the current coming from the power supply, it sees a lower voltage and thus not all of the power coming from the supply. The lost power shows up as heat on the chip.

Bob
 

Thread Starter

Dan Aronin

Joined Dec 14, 2018
37
Exactly, it is a switch that controls the current going to the motor. There is also some voltage lost in going through the switch, so, although the motor sees all of the current coming from the power supply, it sees a lower voltage and thus not all of the power coming from the supply. The lost power shows up as heat on the chip.

Bob
Thanks for clarifying!
So to sum up this thread, the Uln2003 acts in this case as a switch that conrols current flow from power supply to the coils.
 
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