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PN2222A Pin out

I have; pin E to LED's positive pin
- negative pin of the LED is connected to a resistor and then ground.
- If I then leave pin C open and apply power to pin B it will turn the LED on,

It is supposed to not turn the LED on because there is no power to pin C?

This is a problem because;
I am trying to make an AND gate by connecting,
- the E to the C pin of another.

I am supplying power using a Arduino.

What might be my problem?

I have tried using 5v power and 3.3v power to see if I am overpowering it.

EDIT: PN2222A Pin out Here is my circuit. I followed the suggestion and flipped the transistor and the LED still lit up when it should not have. Any other ideas? I also read some other questions and I think that I should not be using PN2222A transistors but instead use a 2n7000 transistor. If that is the case would this work https://www.amazon.com/MCIGICM-2n7000-n-channel-transistor-assortment/dp/B07BKX255D/ref=sr_1_17_sspa?ie=UTF8&qid=1523501912&sr=8-17-spons&keywords=2n7000&psc=1?

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    \$\begingroup\$ Your transistor is working as a diode. Put LED and resistor on collector side. \$\endgroup\$
    – StainlessSteelRat
    Commented Apr 11, 2018 at 20:26
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    \$\begingroup\$ Draw a schematic! We are engineers, not poets, we understand drawings, not poems! \$\endgroup\$
    – Jakub Rakus
    Commented Apr 11, 2018 at 20:26
  • \$\begingroup\$ @JakubRakus Can we not be both.... \$\endgroup\$
    – StainlessSteelRat
    Commented Apr 11, 2018 at 20:27
  • \$\begingroup\$ Duplicate question, Justin. Just in case you do not know how to search here, see top of page. electronics.stackexchange.com/search?q=transistor+and+gate \$\endgroup\$
    – D.A.S.
    Commented Apr 11, 2018 at 21:02

1 Answer 1

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so you are using NPN transistor, as somebody mentioned it in comments you are using it as a diode.

This is the right scheme you should be using

right scheme

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