It's hard to understand why they always want to connect directly to the Arduino

I found a video that uses the diagram below, and notice that it is using a transistor without a base resistor

In the video it shows the assembly, but it also doesn't have the resistor:

FLYBACK DC - DC Converter Theory And Example

¯_(ツ)_/¯

Is there a question?

Yes, there ought to be a base resistor. The Arduino will output 5 volts but the transistor's base-emitter junction will have a forward voltage drop of one diode drop (about 0.7 volts). The large resulting current from the Arduino output will cause it to be much less than 5 volts but this is a poor way of doing things. It would be preferable to not have such a large current from an Arduino output and to drop excess voltage across a resistor.

Of course, an alternative would be to set the pin as an output that is LOW when a low output is desired, and set the pin to INPUT_PULLUP when a high output is desired, but this may not provide sufficient current to fully turn the transistor on.

It's been a hot minute since my last course on transistors, but what about this: With no limiting resistor on the base, the transistor will be put into saturation mode, making Vce ~= 2V. With this, Ie ~= (Vcc - 2) / Rc. This then limits the base current such that Ib = Ie / Hfe. This might be why you can "get away" without using a base resistor and not fry your Arduino. Again, it's dumb not to use either a base or emitter resistor for an NPN and it has been a little since I learned this stuff in college...

Power_Broker:
With no limiting resistor on the base, the transistor will be put into saturation mode, making Vce ~= 2V. With this, Ie ~= (Vcc - 2) / Rc. This then limits the base current such that Ib = Ie / Hfe

Sorry, I didn't understand this explanation.

Sparkfun: Transistors

vaj4088:
Is there a question?

Not really, just a letdown of disappointment.

Power_Broker:
It's been a hot minute since my last course on transistors, but what about this: With no limiting resistor on the base, the transistor will be put into saturation mode, making Vce ~= 2V. With this, Ie ~= (Vcc - 2) / Rc. This then limits the base current such that Ib = Ie / Hfe. This might be why you can "get away" without using a base resistor and not fry your Arduino. Again, it's dumb not to use either a base or emitter resistor for an NPN and it has been a little since I learned this stuff in college...

If I understand your explanation well it is utterly wrong.
In theory the is nothing limiting the base current (I mean "reasonably simplified" theory). In reality there are many features which limit the base current. There is the emitter-base voltage drop which increases with current. Arduino output drivers has some internal resistance. If they are using a soleless breadboard, some other connectors and/or thin wires there may be considerable (a few ohms) resistance between emitter and GND. Depending how they wite it all the current (base, collector and drain current) may go through this resistance causing possibly large voltage drop. Add short pulses from the Arduino output and it easily explains why it survives (at least for limited time of the presentation).
It is surely NOT a good design. OTOH it is good to be aware about this. Here the base current limiting helps protect the pin. Some other day it may cause disappointing results.

Smajdalf:
It is surely NOT a good design. ... Some other day it may cause disappointing results.

Yep!

What this mistake tells you is the person making it has little or no understanding of electronics. It is a coded warning to bin that web page and look for someone else who knows about electronics.

Many beginners think because they can follow a tutorial they know something and want to spread it to the world. Avoid them.

Grumpy_Mike:
What this mistake tells you is the person making it has little or no understanding of electronics. It is a coded warning to bin that web page and look for someone else who knows about electronics.

Amen to that! :astonished:

Power_Broker:
It's been a hot minute since my last course on transistors, but what about this: With no limiting resistor on the base, the transistor will be put into saturation mode, making Vce ~= 2V. With this, Ie ~= (Vcc - 2) / Rc. This then limits the base current such that Ib = Ie / Hfe.

First, Hfe (or beta) is referenced to collector, not emitter, current. So, that equation would be: Ib = Ie / (Hfe + 1). Is say "would" because that relationship only applies when the device is in the Active mode, not when it's saturated.

The 10k collector resistor on the BJT is too large for efficient switching too…

The small transistor issues can be fixed by using a gate driver chip in its place (an inverting one could directly
replace the BJT + 10k resistor, a non-inverting one would need code change).

However there’s no overcurrent protection for the FET/primary part of the circuit either, which is another
thing to bite you hard should the 12V supply be powerful enough - very easy to pop MOSFETs if current limiting
isn’t thought about.

Smajdalf:
If I understand your explanation well it is utterly wrong.

I honestly figured as much, just a random idea I had