I saw someone asking this EE.SE question in a thread over 2 years old and could not really understand some things about the answer. He had a 3.3V input signal that he wanted to change into 5V signal.

This is a circuit someone suggested:

Here is a comment he made about the circuit:

[...] the transistor is configured as an emitter follower and the voltage on the emitter is base voltage minus about 0.6V. If the emitter got higher it would turn off the transistor thus preventing the voltage rising much above about 3V. Think of base and emitter and what differential voltage they must be at to sta[r]t to turn the transistor on.

What I don't understand is:

  1. What is Vb when there is 0V in the emitter? I know that Vbe = Vb - Ve, and that Vb is supposed to be 0.6V, but why? There is the 3.3V supply in the base, doesn't it contribute anything? Vb is determined only and only by Ve?
  2. Pretty much the same question but about Ve. If there is a voltage Vb that is set by the 3.3V supply and the R1 resistor, there is supposed to be a Ve according to the equation in (1). But if Ve is set by the 0-3.3V input - isn't there some kind of a clash?
  3. Why is the transistor off when the input is 3.3V (in the emitter)? According to the equation in (1), Vb is supposed to be Vb = Vbe + Ve = 0.6+3.3 = 3.9V. That means, the base has '1' (high), which means the transistor should be on, no? I assume the 3.3V supply is limiting Vb to 3.3V, but I'm asking anyway.
  4. Any reason why the resistors have these values?



2 Answers 2


I have re-drawn your schematic, because I like to see more "positive voltage" supplies at the top, "less-positive" voltage supplies lower down, and ground at the bottom.
The case for a logic low input (0v) is on the left, the case for logic high input (+3.3v) is on the right.
The logic source that provides the logic low (on the left) will have to work at it, because it must sink current from both the 3.3v supply, and from the 5v supply. The transistor's emitter is pulled to zero volts by that logic source. The transistor's base has no choice but to follow, and is about 0.65v higher. Considerable base current flows (about 1.2 mA). The current ensures that the transistor is heavily saturated. It is so saturated that collector voltage actually falls below the base voltage, and provides a logic low at the output. Collector current is 0.72 mA. Many saturated transistor circuits allow collector current to be ten or twenty times larger than base current, but not in this case - no current gain is required of the transistor (actually less than one). Not shown is the current that might flow from the 5v logic output. That current must flow through the transistor, and the input driver must sink this too.


simulate this circuit – Schematic created using CircuitLab On the right show the case where the logic input is in its "high state", up at 3.3v. In this case base and emitter are kept at the same voltage. The base-emitter junction of the transistor has no voltage drop as before, so no base current can flow through the 2.2K resistor. The transistor is OFF. Since it is off, no collector current flows. The collector resistor is left on its own to pull the output up to +5v.

Answer to your question (1): Base is at 0.65v. The 3.3v must supply current through the 2.2k resistor, since the transistor base is being dragged down by the logic source providing "logic 0 input".
Answer to your question (2): Yes, a clash. That "logic 0 input" is working hard to pull transistor's emitter down to 0v.
Answer to your question (3): Run Kirchoff's voltage law around the base-emitter loop of the schematic (right). Everything's at 3.3v. There can be no voltage between base and emitter. That forces the transistor "off".
Answer to your question (4):Depends on what you're driving. This circuit has wide latitude in choosing these values. It won't be fast, because their values are high. If you lower them, the poor 3.3v logic source must work even harder. If the 5v logic output must drive a significant load, that 3.3v logic source must sink its current too.

  • \$\begingroup\$ Thank you for a very detailed answer! Just want to make sure about question (3). You said that the base and emitter have the same voltage in this case. I would understand that completely if it weren't for R6. I tried to make sense of your answer, and if I understand correctly, the reason that Vb is also 3.3V despite R6 is because there is a short between one end of the R6 resistor and the emitter, and that current prefers to flow through a short, therefore causing no voltage drop on the resistor? \$\endgroup\$
    – nettek
    Commented Oct 1, 2016 at 20:58
  • \$\begingroup\$ Also can you please explain what you mean by "sink"? I'm not a native English speaker so I'm not sure what that means. \$\endgroup\$
    – nettek
    Commented Oct 1, 2016 at 20:59
  • 1
    \$\begingroup\$ We use terms "sink", and "source" to describe current flow. @Eran sink as in water running down a drain. source as in water flowing down from a faucet. For example, the logic input must "sink" current to ground to maintain a logic 0. It must neither sink nor source current to maintain logic 1. If left unconnected, this logic translator defaults to logic 1. \$\endgroup\$
    – glen_geek
    Commented Oct 1, 2016 at 21:18
  • 1
    \$\begingroup\$ Both ends of R6 (righthand schematic) are at 3.3v Base is at 3.3v. Emitter is at 3.3v. With all nodes in this loop are at 3.3v, no current can flow. \$\endgroup\$
    – glen_geek
    Commented Oct 1, 2016 at 21:20
  • 1
    \$\begingroup\$ @Eran, When we look at that input side, consider that base-emitter is a diode junction. No current can flow until its cathode (emitter) is dragged more than 0.6v below anode (base). Current starts to flow when "logic_in" starts on its way from logic"1" to logic"0". The diode becomes forward-biased, and the base is dragged down by the emitter (lagging by a bit more than 0.6v). \$\endgroup\$
    – glen_geek
    Commented Oct 2, 2016 at 13:42

It might be instructive for you to do some experiments with an LED, a resistor, a battery, and a multimeter...

An LED is a diode. Similar to the base-emitter junction of your transistor. When a diode is turned on it has a constant voltage drop across it. (Not exactly, but we often assume that it does for analysis and such.)

So let's talk about an LED that has a forward voltage drop of 1.8V. If you apply less than 1.8V across it, it will not turn on. If you apply more than 1.8V across it, it will turn on and maintain 1.8V across it.

Now, you can't do this with just the LED. Because if you do the LED will allow so much current to pass through that it will heat up and burn itself out. This is the purpose of a current limiting resistor.

So now think of an LED in series with a resistor, say 1K ohms. Again, if we apply less than 1.8V across it, the LED will not turn on. And, again, if we apply more than 1.8V across it, the LED will turn on and the resistor will limit the current. The difference between the battery voltage and the LED voltage drop is the voltage across the resistor. From that you can calculate how much current is flowing.

This applies to the transistor as well. Your typical NPN bipolar transistor will need about 0.65V across the base-emitter junction (base being more positive) in order to turn on and allow collector current to flow. If you apply less than 0.65V the transistor will not turn on. If you apply more than 0.65V the transistor will turn on and the difference between the voltage drop (0.65V) and the applied voltage will be the voltage that the current limiting resistor will drop.


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