Why use transistors when a uln2003 can drive a stepper by itself?

Thread Starter

Younes Thabet

Joined Jan 9, 2019
53
Hello all,

In the picture attached below, you can see that this circuit uses a transistors with the uln2003 to drive a 5-Lead unipolar stepper motor, but isn't the ULN2003 capable of driving the stepper without the extra transistors and diodes!? as it can output a current of 500 mA per driver and the 24byj48 stepper motor requires only 40mA per phase! and it already has the diodes to protect from inductive spiking..

can anyone explain what's the purpose of using the transistors with the ULN2003?

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Thanks,
 

ericgibbs

Joined Jan 29, 2010
12,884
hi YT,
A ULN2003 cannot source current, it is open collector output, so it can only sink upto 500mA.
The circuit uses the ULN as a level shifter from 5V to 12V , in order to control the Darlington Bases

E
Corrected MOS error > BJT
AAA 1127 08.13.gif
 

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Thread Starter

Younes Thabet

Joined Jan 9, 2019
53
hi YT,
A ULN2003 cannot source current, it is open collector output, so it can only sink upto 500mA.
The circuit uses the ULN as a level shifter from 5V to 12V , in order to control the MOSFET Gates

E
View attachment 229435
Got it,
in this case can I remove the ULN2003 and keep only the transistors (PDTB123YT) and drive them directly from mcu pins. is this gonna cause problems with mcu?

Thanks again,
 

AlbertHall

Joined Jun 4, 2014
11,305
The mcu outputs can only get up to 5V and so the transistors will never turn off. Also the current from the transistor bias resistors will flow through the protection diodes in the mcu. Depending on the mcu, this may cause latch-up and damage the mcu. Also this current may disrupt the operation of the mcu.
 

sagor

Joined Mar 10, 2019
439
While a ULN type can sink current, the specs for it have to read carefully. Total dissipation is a limiting factor, you cannot drive all lines 100% at the same time at full pin rating. Same goes for the UDN series (they source current). Most are rated at about 400 to 500mA per pin, but only if one pin is active. If using 2 pins, current per pin drops and so on until all 7 (or 8) pins are on, each is limited to about 100-120mA each.
So, application and design dictate how you can use the ULN or UDN series of buffer chips. I use them to drive relays, and if more currecnt is needed I either pair up a set of inputs and outputs for double the current, or use external transistors/mosfets to drive even higher currents.
 

shortbus

Joined Sep 30, 2009
8,558
The motor is a 5-lead unipolar stepper. The fifth lead is not shown in the diagram. Eric has added the connection to GND on the fifth lead.
And that would be wrong. The fifth lead is a voltage supply, not ground. So the chip in the original post would/should make the motor rotate, it just wouldn't be possible to have any load of the shaft.
 

MrChips

Joined Oct 2, 2009
23,505
And that would be wrong. The fifth lead is a voltage supply, not ground. So the chip in the original post would/should make the motor rotate, it just wouldn't be possible to have any load of the shaft.
The power transistors shown in the original schematic are PNP on the high side.
 

AnalogKid

Joined Aug 1, 2013
9,248
isn't the ULN2003 capable of driving the stepper without the extra transistors and diodes!?
Yes - IF -

IF you can connect the motor's common lead to the +12 V instead of GND

AND IF

the peak current at any time through any transistor does not exceed 500 mA

AND IF the total current through all transistors does not exceed 1 A at any time.

If all of that is true, then the four external transistors and four suppression diodes can be eliminated. Note that 2003 pin 16 must be connected to the same supply as the motor.

ak
 

shortbus

Joined Sep 30, 2009
8,558
The power transistors shown in the original schematic are PNP on the high side.
That may be so in the original schematic, but that wasn't his question. It was,
but isn't the ULN2003 capable of driving the stepper without the extra transistors and diodes?
And then someone else said the ULN2003 couldn't source current, only sink current. The way a 5 lead stepper is USUALLY wired is with the fifth lead sourcing the power and the other 4 sinking it.
1612390260788.png
 
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