Is there a reason that you have ch1 and ch2 overlapping each other? Please separate the baseline traces so they're not overlapping each other. It's also good practice to assign Ch1 to the Input and Ch2 to the Output.
Chris
Chris
Thats is because the transistor will not be fully turned on. will probably be within the "active region" if i that is the correct term in english. Am i wrong ?The output won't swing below ~ 3V. If R2 is reduced further to 47Ω the output probably won't swing below 4.5V.
I look forword to that sir Chris. I have to let you know though that i will be away for work for the next 6 days without internet acces. So please take your time.Constantine, I've done some simulations since my last post. I believe that later today I can clearly explain the scope readings you're experiencing. Right now I'm long overdue for a haircut.
This can't be. The base of the transistor will not rise much above ~0.6 V ... 0.7 V. At that voltage the base-emitter diode becomes conducting and the resulting base emitter current will drop the rest of the input voltage across the series resistor from signal source to base. Your scope signal shows ~ 5 V signal level, thus you surely are measuring at the signal source or the left side of the series resistor, not at the base of the transistor.Channel 1 = input signal (at base of transistor)
Your signal source is very probably far from an ideal voltage source. It will have some internal resistance.when connected to the base of the transistor, the input signal drops about 0.4Volt
When you operate an NPN transistor with collector to ground, it will seem to work regardless of the mixed up connections. This is called reverse mode of operation.Picture 2 = with collector to gnd
Picture 3 = with emitter to gnd.
Harald, that's an issue that I never questioned him about because I assumed he was scoping the base signal at the input end of the base resistor, as you surmised.This can't be. The base of the transistor will not rise much above ~0.6 V ... 0.7 V. At that voltage the base-emitter diode becomes conducting and the resulting base emitter current will drop the rest of the input voltage across the series resistor from signal source to base. Your scope signal shows ~ 5 V signal level, thus you surely are measuring at the signal source or the left side of the series resistor, not at the base of the transistor.
Yes thats right. What i ment was that this is the signal i am feeding to base through the resistor.hus you surely are measuring at the signal source or the left side of the series resistor, not at the base of the transistor.
When you operate an NPN transistor with collector to ground, it will seem to work regardless of the mixed up connections. This is called reverse mode of operation.
However, the current gain is greatly reduced. An easy explanation you may read up here.
How can that be explained ? maybe when i tie the emitter to vcc and collector to gnd, the transistor works in "reverse-active" mode ?
like explained here ? https://learn.sparkfun.com/tutorials/transistors
A simple repeat option"Repeat" option