Dear helpers 
Below are two diagrams. My question is about the diodes.
I cannot figure out, maybe I am reading the lay-out wrong, why it seems that the diodes are placed differently.
In one diagram the cathode is connected to the collector while in the other diagram the anode is connected to the collector.
Can someone help me with some clarification about why that is?
Regards.
That's the problem with not having a schematic - you can't tell the function of the pins that are connected.
Where is the Gate-Drain-Source? Where is the Base-Collector-Emitter? We can't tell, thus it is hard to help.
Thank you for your reply. I have added schematics with other diagrams. Hope you can help me further.
In both cases, the diodes are designed to protect the transistors, which are driving inductive loads. They try ensure that, when the loads are turned off, the transistors are not subjected to more than one diode-drop (Vf) of reverse polarity. But don't succeed.
BTW, think you'll find that Darlington pair (TIP120) in the upper schematic is upside-down.
johndg:
In both cases, the diodes are there to protect the transistors, which are driving inductive loads. They ensure that, when the loads are turned off, the transistors are not subjected to more than one diode-drop (Vf) of reverse polarity.
thank you, but I still do not understand why the one diode cathode is connected to the collector while the other diagram the anode is connected to the collector.
The TIP120 in the first diagram is drawn wrong.
Emitter must connect to ground, collector to the solenoid.
Breadboard layout seems ok though.
Switches are safer when connected between pin and ground, with internal pull up (no resistor).
So pin is normally HIGH, and LOW when button is pushed.
The diode in the third diagram does nothing useful across the transistor.
A kickback diode should limit collector voltage, and can only do so if the anode is connected to the collector and cathode to the supply (so across the solenoid).
Leo..
Wawa:
The TIP120 in the first diagram is drawn wrong.
Emitter must connect to ground, collector to the solenoid.
Breadboard layout seems ok though.Switches are safer when connected between pin and ground, with internal pull up (no resistor).
So pin is normally HIGH, and LOW when button is pushed.The diode in the third diagram does nothing useful across the transistor.
A kickback diode should limit collector voltage, and can only do so if the anode is connected to the collector and cathode to the supply (so across the solenoid).
Leo..
The third diagram comes from a university page, so they did it wrong? It is from this tutorial: https://itp.nyu.edu/physcomp/labs/motors-and-transistors/using-a-transistor-to-control-high-current-loads-with-an-arduino/
to be clear the anode of the diode should always be connected to the collector and the cathode to the supply (solenoid, motor etc.). So the diode in the tutorial from above is placed wrong.
praiseworthy:
thank you, but I still do not understand why the one diode cathode is connected to the collector while the other diagram the anode is connected to the collector.
The diode in the second schematic is useless.
And yes, the TIP120 in the first schematic is upside-down.
Apparently, they provide the obsolete TIP120s in "starter kits" - presumably as they are cheap. You really should however, be making a point of using modern parts - logic-level FETs such as an IRL540.
praiseworthy:
The third diagram comes from a university page, so they did it wrong?
Correct. You can't believe everything you read. 
Wrong diode placement, wrong symbol for what is a Darlington transistor, and obsolete part suggested for "high current loads". 
praiseworthy:
to be clear the anode of the diode should always be connected to the collector and the cathode to the supply (solenoid, motor etc.). So the diode in the tutorial from above is placed wrong.
Correct.
praiseworthy:
The third diagram comes from a university page, so they did it wrong?
Yes. Very bad tutorial.
If you don't see a way to email the author of a tutorial on a web page, or leave comments, then you should not trust that page.
That unfortunately is lately also true for the official Arduino tutorials.
We don't have the power to correct mistakes anymore.
Leo..
Wawa:
If you don't see a way to email the author of a tutorial on a web page, or leave comments, then you should not trust that page.
Amen to that - an excellent observation. 
No oversight, no credence.
Paul__B:
The diode in the second schematic is useless.And yes, the TIP120 in the first schematic is upside-down.
Apparently, they provide the obsolete TIP120s in "starter kits" - presumably as they are cheap. You really should however, be making a point of using modern parts - logic-level FETs such as an IRL540.
Correct. You can't believe everything you read.
Correct.
Thank you for your answers. WIth the absolute part, I think you mean using a MOSFET such as (IRL540) is "up-to-date".
I do see some emails from the teachers that contributed to this tutorial. I will inform them accordingly.
praiseworthy:
Thank you for your answers. With the absolute part, I think you mean using a MOSFET such as (IRL540) is "up-to-date".
The point is that a Darlington transistor has a very significant voltage drop when "turned on", of the order of one to one-and a half Volts.
This has two consequences - you lose that voltage drop, not so significant for a 12 V or greater supply but quite a significant drop if operating at 5 Volts.
The second is that if you are switching a substantial current - and that is of course why you chose the Darlington - the transistor dissipates significant power - at least 1 Watt per Amp - and needs to be provided with a heatsink,
The logic-level power FETs have an effective resistance rather than a voltage drop, and that resistance is quite low, specified as no more than 0.11 Ω for the IRL540 at 4 V on the gate, somewhat less with higher gate drive. This means that it can pass several Amps before it dissipates a Watt or two and you require a heatsink. In comparison, there is no argument for deliberately using the TIP120 and it is essentially obsolete.
IIRC Darlingtons are faster than MOSFETs unless you have a fancy MOSFET driver. Maybe sometimes this can make them better for applications using PWM?
