# Remedy for pulling the output to ground in saturation?

Following ubiquitous common emitter configuration receives 0-5V logic input at its base such that the transistor goes into saturation or cut off mode:

So the output taken at collector is either zero or pulled up to 0.5V. Well now not really zero but depending on saturation level could be 100mV a bit less. Because Vce never becomes zero.

My question is: is there any sort of remedy to this problem so we get 0.5V and zero outputs instead of 0.5V 0.1V? Does such a remedy be achieved without using more than one transistor? Using one NPN or PNP in total plus resistors and or diodes.

• How much current do you want to sink? What problem are you trying to solve? Add the information into your question - not in the comments. Mar 14, 2020 at 15:41
• What is the load that will be connected to the output? And why are you restricted to only using BJTs? How close do you need to get to 0 V? Is 10 mV acceptable? 1 mV? Mar 14, 2020 at 15:41
• @ThePhoton It was a question out of curiosity Im not gonna build anything. But I can say the load can be 100Meg. Ground down to 10mV lets say. Do you mean FET has less such residual ground voltage? Mar 14, 2020 at 15:52
• @user1999, A FET is better characterized by an effective resistance in its "ohmic" region. So the less current you're trying to sink, say, 1 mA it's relatively easy to achieve sub millivolt ouput voltage. A BJT should also be able to get closer to 0 V output with low sinking curent, but this isn't usually characterized by the manufacturer so it's harder to design for. Mar 14, 2020 at 15:56
• @user1999 You could just operate a relay with metallic contact points and very, very low resistance. Regarding transistors, I think The Photon has given you "die endgültige Antwort." (definitive answer.) A BJT's collector-emitter pins, in saturation, acts like a voltage source (whose value depends on just how much base current you are willing to expend -- more leading to a lower voltage.) FETs are more like small-valued resistors when "on." So with your 100Meg example, the FET would be a lot better at getting close to zero. A BJT can't take as good advantage at that high of a collector load.
– jonk
Mar 14, 2020 at 19:08