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I saw someone asking this EE.SE questionthis 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 suggestedsomeone 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?

Thanks!

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?

Thanks!

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?

Thanks!

2 Added links to the question & answer from the previous thread mentioned by the OP. Added tag "level-shifting" as that is the topic. Put the earlier answer included by the OP, into a block quote, to make it clear which part is the quote (and added missing character from word).

I saw someone asking this questionthis 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 suggestedsomeone 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 stat to turn the transistor on.

[...] 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?

Thanks!

I saw someone asking this 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 stat 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?

Thanks!

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?

Thanks!

1

Changing a 3.3V signal to a 5V signal using a BJT transistor

I saw someone asking this 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 stat 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?

Thanks!