SimonnomiS
- Apr 5, 2023
- 1
- Joined
- Apr 5, 2023
- Messages
- 1
You don't say how you determined the value of the base resistor, but it is correct. The circuit should work as intended.I hope some one can verify this basic circuit. I will use an ESP32 O/P to switch the transistor.The load is 25mA View attachment 58638
As noted frequently on these sites, that's not good design practice when using the transistor as a switch.calculate the current flowing in the base resistor and multiply it by the current-gain (beta) of the device.
You only took the first part of my post - when added to the rest of it, i.e. saturation, there's a distinct difference. My post was on the basis that this seems like a homework question and that 'helpful advice' is preferred over a simple 'answer'.As noted frequently on these sites, that's not good design practice when using the transistor as a switch.
A forced beta of 10 is the usual value used.
Okay.You only took the first part of my post - when added to the rest of it, i.e. saturation, there's a distinct difference.
Thank you sir. I will try it.The rule of thumb to force the transistor into saturation is Ib = Ic / 10.
Its a term called forced beta in the industry. Generally speaking when
we use a transistor as a switch we want the Vce to go as low as possible,
so that its power dissipation is kept low and the switch acts as close as
possible as a short, like relay contacts typically do.
So the base R, Rb =~ ( Voh - Vbe ) / Ib =~ ( Voh - Vbe ) / [ Ic / 10 |
Voh comes from micro sdatasheet, when an outpuit is turned on to high
logic level, and given at a specific current. Some datasheets show a typical
graph of Voh versus load current.
Typical curves that apply (2N3904 as example) :
View attachment 58640
As you can see in the graph, 10 mA case, we get a low Vce when we force 1 mA into
the base. Any more base current has very little egffect, eg. no more than Ic / 10 is
needed in most designs.
Note some transistors have even lower Vce when collector and emitter are reversed,
but takes more base current then Ic / 10 to get that increased performance.
Regards, Dana.
Most Arduino, if not all, have this aspect ......input default.I have programmed PICs where I found it was necessary to change an input/output pin from its initially programmed output status back to being an input pin, and then change back again to output.
The rule of thumb to force the transistor into saturation is Ib = Ic / 10.
Do you mean Ic/Ib=10
Typical curves that apply
Silly! Don't use plots, that's just typical behavior.
Look at the limits!!
Maximum , Minimum
normal criteria is for a saturated transistor, so that's why I stated it
but what's this hfe? Oh... That's for an unsaturated transistor used as an amplifier they call it gain
I will stop repeating what everyone else's just said.@Delta Prime : How do you think this is going to help the op?
Thanks for that link.Most Arduino, if not all, have this aspect ......input default.
According to below, same for pic on reset......
https://microchip.my.site.com/s/article/What-is-the-default-state-of-pins-on-a-PIC-microcontroller#:~:text=What is the default state of pins on a PIC microcontroller?,-Answer&text=The state of the pins,pin has an analog function.
Yes sir it is. Ain't is not word so don't say ain't.
I'll turn it down keep the shenanigans to a minimum thank you for your patience "all of you."I do realize I used to have an ego problem when I was arrogant, but no longer... cuz now I'm perfect.I will stop repeating what everyone else's just said.