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I'm trying to understand the following circuit diagram:

http://pinout-circuits-images.dz863.com/6/L293D-circuit-8.png

Specifically, the four triangles within the L293D component. So I understand there are two transistors in there, connected in series from 24v to ground, but what is connected to the bases of the transistors, the lines just end...

From what the circuit should do, I guess that the line coming in to the triangles from the 'and' gates (at the top left of the triangles) should some how enable the whole triangle, and that the line coming in from 1A\2A\3A\4A should toggle 5v between the two transistors. Is that right? And how do I deduce that from the schematic?

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The triangle is a symbol for a buffer, driver or amplifier, whatever you call it.

The transistors drawn inside the triangle are there to let you know what is the type of your output. Thanks to this drawing you can see that the outputs are switched via BJTs between the power supply rails using some kind of Push-Pull topology.

Knowledge about the type of output is useful. If those were power MOSFETs, you would expect them to exhibit different behavior than BJTs. A BJT will have a voltage drop whereas a MOSFET will have a resistance between drain and source.

The drawing doesn't provide the exact schematic for the amp/buffer and it doesn't have to. The exact schematic would be a lot more complicated and this image is there to clarify things, not complicate them.

So, in short, the triangle symbolizes your driver/buffer and the transistors are there to let you know that the output was made using bipolar transistors.

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  • \$\begingroup\$ So is it possible to tell by this diagram what the circuit does? BTW this diagram is missing some text, you can see it here in full. Thanks. \$\endgroup\$ – pseudoDust Oct 12 '12 at 12:02
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    \$\begingroup\$ Yes. You can expect the output to have a fixed voltage drop relative to the VCC. This little simulation helps understanding what's happening with a BJT push pull follower. Whereas this is the MOSFET version of the same circuit. \$\endgroup\$ – Christoph B Oct 12 '12 at 16:23
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That is just a notation, not by any means a schematic. The "triangles" themselves symbolise high current drivers, the two "transistors" drawn there are just to symbolise what the circuit is doing. There more components "inside the triangle", that for sake of brevity are not displayed.

This is similar to pseudocode and real source code in a given programming language. This is a "pseudo-schematic" of the chip.

BTW: Why do you want to get to know what's inside the IC? Would you like to get higher current output, because the chip itself could not drive the motor you want to use?

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    \$\begingroup\$ Well, I just want to understand exactly what the IC does and how it does it. I'm trying to teach myself electronics, and I've always found that I learn best by taking on a project while making sure I understand everything I do. My project is a CNC plotter, I'm using L293D to drive a stepper motor. I'm planning on running about half an amp at 24v through this IC, I think it should be able to handle it... Thanks for your reply, sorry I can't vote up yet, and I can only accept one answer. \$\endgroup\$ – pseudoDust Oct 12 '12 at 11:55
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    \$\begingroup\$ Thanks for the thorough explanation, now I understand, and respect your motivation. I hope you find a more detailed schematic of the chip, though I have doubts that there is one with all the values included. \$\endgroup\$ – ppeterka Oct 12 '12 at 12:31
  • \$\begingroup\$ I'm not sure I could handle a full detail schematic of the chip in any case, but I've had penny drop, I just can't shake thinking like a programmer, I think in terms of input and output, linear instead of circular. My mistake was that I didn't realize that a logical 0/low and ground are the same thing, it seems to stupid now... but that's how it always is when learning something new... \$\endgroup\$ – pseudoDust Oct 12 '12 at 12:53
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    \$\begingroup\$ Ah yes 'domain specific knowledge' - at the border of two adjacent domains... Hope you have a lot more pennies drop, and get rich in the end :) \$\endgroup\$ – ppeterka Oct 12 '12 at 13:20
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A triangle with one input and one output usually represents a buffer: it copies its input to its output, usually at a greater signal strength, usually cleaning up the input signal to logic 0 or 1.

Those transistors are probably exactly what the output stage looks like: a PNP transistor connected to the power line and an NPN connected to ground, only one of which is on at once.

A triangle with two inputs and one output usually represents an op-amp.

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  • \$\begingroup\$ Thanks for your reply. Aren't both of the transistors NPN? \$\endgroup\$ – pseudoDust Oct 12 '12 at 11:57
  • \$\begingroup\$ Yes, you're right, that was what I expected rather than what was actually shown.. \$\endgroup\$ – pjc50 Oct 23 '12 at 8:41
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The simplified schematic of the buffer inputs "just ending" is supposed to be intuitive to indicate the decoupling of noise from the load to the input as well as source supply.

When you drive high currents with logic you risk getting supply noise back into your logic and this degrades the reliability of your signals which can cause glitches or false triggering. This is especially true of any reactive loads and motor loads with commutation noise.

The beauty of this design is that the logic and drivers use separate supplies. Although they share common ground potentials, careful design (beefy ground tracks and/or ground plane and/or isolated ground distribution can mitigate noise ground-shift false triggering from EMF kickback diode shunt current or other surges. By the way these drivers also have ESD protection and diode clamps built-in for inductive loads. Designers are wise to use common mode chokes on stepper motor loads to reduce EMI.

The complementary driver is often called a "half-H" or "half bridge" switch. When the load is connected between two such half-H drivers with one side inverted, you can get twice the load swing useful with bi-directional motor control with PWM controlled by the enable signal then you have a full-bridge of full-H driver configuration. Dead-time control to prevent shoot-thru or shorting of the supply to ground via the complementary outputs becomes a critical design factor in high current inductive loads.

TI describes it as follows:

The L293 and L293D are quadruple high-current half-H drivers. The L293 is designed to provide bidirectional drive currents of up to 1 A at voltages from 4.5 V to 36 V. The L293D is designed to provide bidirectional drive currents of up to 600-mA at voltages from 4.5 V to 36 V. Both devices are designed to drive inductive loads such as relays, solenoids, dc and bipolar stepping motors, as well as other high-current/high-voltage loads in positive-supply applications.

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  • \$\begingroup\$ Thanks for the thorough explanation, I'm afraid that most of it is way over my head, I am but a mere programmer, and only started with electronics about a week ago in a "learn by doing" approach, I think I'll need to burn a couple dozen components before I'll be able to fully appreciate your post. \$\endgroup\$ – pseudoDust Oct 12 '12 at 12:36
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    \$\begingroup\$ No problem. Just think of it as an electronic switch with 4 devices inside, each a SPDT with the output being a Pole and the double "Throw" internally connected to V+ and ground which is separate from the logic supply but shares a common ground. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Oct 12 '12 at 12:46
  • \$\begingroup\$ Yes, I got that much from your post, it's just the EMF kickback and EMI that I don't understand yet, I mean, I can see (with difficulty) the benefit of the diodes in the above circuit for example, but I don't really comprehend the dynamics of electricity, I'm very far from being able to design a circuit, or even to appreciate the beauty of a design, but I'll get there... \$\endgroup\$ – pseudoDust Oct 12 '12 at 13:08
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    \$\begingroup\$ get some little torroids to wrap each motor wire set to reduce the spike noise when each winding is turned off. The diodes absorb the potential high voltage by clamping to supply or ground, which results in a current spike that may couple to nearby wires or circuits. digikey.com/product-search/en/filters/… then use one for each motor wire set with as many turns as you can wrap or use clamshell type like on your VGA cable or DC charger. 50 cents ~ $1 each. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Oct 12 '12 at 14:32
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    \$\begingroup\$ then silicone the ferrite as it is very brittle if it does not come with plastic wrap. EMI noise couples into high impedance logic and sensors and can be heard with an AM radio between channels. the long wires should be limited to 1 or 2 meters act as antenna spewing noise up the rise time of the switches f=1/t approximately. Also consider shielding sensitive sensors and long wires if you get erratic results. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Oct 12 '12 at 14:42

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