Electronics > Beginners

Transistor Switching

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tpowell1830:
The observations regarding the lack of 1k resistor is good, also the internal resistance of the battery, however, someone is missing the elephant in the room.. this is not an incandescent bulb. It is an LED.

hamster_nz:

--- Quote from: tpowell1830 on July 08, 2018, 05:10:54 am ---The observations regarding the lack of 1k resistor is good, also the internal resistance of the battery, however, someone is missing the elephant in the room.. this is not an incandescent bulb. It is an LED.

--- End quote ---

Just did a quick test. With a good alkaline 9V battery the short circuit current is about 400mA - giving the battery about 22.5 ohms of internal resistance.

That is enough to burn out an LED, but not enough to burn out a transistor (at best you might be able to get half a watt of heat outside of the battery).

Without the resistor (or the pot dialed to 0 ohm) there will be a few hundred mA flowing through the tranaiator's base, and due to the internal resistance the battery voltage has dropped to about volt or less, causing the LED to go out.

Zero999:

--- Quote from: hamster_nz on July 08, 2018, 07:11:10 am ---
--- Quote from: tpowell1830 on July 08, 2018, 05:10:54 am ---The observations regarding the lack of 1k resistor is good, also the internal resistance of the battery, however, someone is missing the elephant in the room.. this is not an incandescent bulb. It is an LED.

--- End quote ---

Just did a quick test. With a good alkaline 9V battery the short circuit current is about 400mA - giving the battery about 22.5 ohms of internal resistance.

That is enough to burn out an LED, but not enough to burn out a transistor (at best you might be able to get half a watt of heat outside of the battery).

Without the resistor (or the pot dialed to 0 ohm) there will be a few hundred mA flowing through the tranaiator's base, and due to the internal resistance the battery voltage has dropped to about volt or less, causing the LED to go out.

--- End quote ---
The text clearly said LED, so the symbol is just wrong. I suspect it's been copied from another site.

Yes, 400mA is expected from a new alkaline battery, but I think the original poster is probably using a tired zinc battery, which will have a much higher impedance.

MK14:

--- Quote from: hamster_nz on July 08, 2018, 07:11:10 am ---That is enough to burn out an LED,

--- End quote ---

If you examine the pictures, carefully. I can apparently see, a 220 \$\Omega\$ resistor, connected in series with one of the LEDs leads.
I.e. The LED should be fine. Although strictly speaking, the resistor should be higher, if it is intended to run the circuit from 9V, to keep it below 20mA, the LEDs (presumed, as I don't know the exact spec of that LED) max currrent limit.

tpowell1830:
Okay, another hint, how do you dim an LED? The linear region in an LED is tiny, so a pot is difficult to hit that unless there has been careful calculations done. This setup  is very tricky in order to get the LED to dim the full sweep of the pot.

Last hint... hope it helps...

EDIT: Think digital.

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