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

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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.

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    \$\begingroup\$ 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. \$\endgroup\$
    – Transistor
    Mar 14, 2020 at 15:41
  • \$\begingroup\$ 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? \$\endgroup\$
    – The Photon
    Mar 14, 2020 at 15:41
  • \$\begingroup\$ @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? \$\endgroup\$
    – user1999
    Mar 14, 2020 at 15:52
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    \$\begingroup\$ @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. \$\endgroup\$
    – The Photon
    Mar 14, 2020 at 15:56
  • \$\begingroup\$ @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. \$\endgroup\$
    – jonk
    Mar 14, 2020 at 19:08

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is there any sort of remedy to this problem so we get 0.5V and zero outputs instead of 0.5V 0.1V?

All devices have resistance, so you won't ever get exactly zero. Therefore the question is how low do you need to go?

Perhaps you just need a better transistor. The ZXTN25100BFH is a 100V, SOT23, medium power transistor with saturation voltage of 0.02V at 100 mA.

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If this is not low enough then you could use a MOSFET, which is purely resistive in the 'linear' region (equivalent to saturation region in a BJT). To improve on the ZXTN25100 at 100 mA it would need to have RDSon less than 0.02 / 0.1 = 0.2 Ω. A typical logic switching FET like the 2N7000 (RDSon = 1.8 Ω) is not good enough, but an IRLML2502 (typical RDSon = 0.035 Ω) would reduce the voltage at 100 mA to ~3.5 mV.

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A MOSFET is usually the best choice when you want lowest possible voltage drop, but at lower currents a BJT can also work pretty well. One trick that sometimes works is to swap the Collector and Emitter. In this mode the transistor has very low gain, but lower saturation voltage. I tested a BC209 this way with RB = 4k7 and RL = 1k (for 0.5mA Collector current) and it only dropped 2.1 mV.

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