# transistor base connected to GND [closed]

I made a simple circuit with a transistor, an led and two 100 ohm resistors. Here's a simple diagram:

simulate this circuit – Schematic created using CircuitLab

The problem is as you can see, the base of the transistor is connected to GND, and based on if the button is pressed or not, the led does or does not turn on. Because it should do that while connected to the positive terminal.

• What are you trying to accomplish with this circuit? You can just connect the LED to the switch directly.
– vir
Feb 27 at 19:37
• current wants to flow in the direction of the arrow on your emitter, and you have current wanting to flow in the opposite direction. Feb 27 at 19:47
• @ScottSeidman oh I think I drew the led the wrong way around Feb 27 at 19:49
• No you drew the transistor the wrong way round
– RoyC
Feb 27 at 20:52
• @JustMe It does actually work - not as well as some alternatives, but even if you replaced the transistor by a B-E or B-C diode from the LED to the switch-side resistor, it would still work. So to make it "work", the transistor doesn't even need to be there, a single diode would do. The transistor does provide a bit of gain, even when used in reverse. It does depend a bit on the part number of the transistor of course, but alpha is not zero :) Feb 27 at 21:15

Bipolar transistors work in reverse, at a rather low current gain, but work they do.

Your circuit, basically, works, no matter which way the transistor is inserted, i.e. whether emitter is facing the LED or ground.

simulate this circuit – Schematic created using CircuitLab

Q1b's current gain is orders of magnitude higher than Q1a's current gain, thus R1b can be higher than R1a to get similar LED current. Q1b is the typical way a PNP emitter would be connected.

It will work even better when the PNP transistor is on the supply side of the LED:

simulate this circuit

Again, Q1b would be the usual high-gain configuration. The (a) configuration will waste much more power since the base current has to be rather high with transistor inserted in reverse (collector and emitter swapped relative to normal configuration).

Because it should do that while connected to the positive terminal.

I guess you meant you wanted to use an NPN transistor, then?:

simulate this circuit

As above, Q1b is the usual configuration, while Q1a uses the transistor in reverse, with rather low gain. That is not usually desirable, unless you want a really low saturation voltage.

In all of the above circuits, the performance of the (a) variant will depend quite a bit on the type of the transistor used. A 2N2222 has a poor reverse beta and will not work very well, whereas a 2N3904 is better. Heterojunction transistors, have excellent reverse beta, so a heterojunction RF transistor (careful: they are fairly low voltage!) used as a switch will not care all that much about swapping emitter and collector.

• This circuit will depend on how low Vce can get to light up a red LCD. Feb 27 at 21:25
• @ScottSeidman It's good then that Vce(sat) is very low in reverse gain configuration - even an order of magnitude lower than in forward gain configuration, as long as you can push enough base current before things start to melt :) It would help to have a negative voltage for the base :) It's not as if the asker consciously made that decision, but sometimes beginner's luck does wonders :) All of this is rather moot because with a switch there's no need for the transistor at all, usually, in this simple application. So almost any "circuit" will work, if energy waste is not a figure of merit. Feb 27 at 21:26
• Yes, I did mean to use an NPN transistor, at least in the diagram. I'm pretty sure the transistors I have are NPN, but, the text on the back of the transistor is "BC557" and on the next line it says "B331", when I look both of these numbers, it shows, that they're PNP, while I actually ordered NPN. Feb 28 at 12:03