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When the base of the transistor is LOW, no current can pass through the collector to the emitter and therefore if the current is 0 the voltage is 0 and all current will pass through the resistor that is connected between collector and Vcc.

When the base is HIGH the transistor is open and current will pass from collector to emitter and the output will be LOW. That I don't get. Why is the output LOW? Can't current flow from Vcc to the output as well to the emitter?

enter image description here

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  • \$\begingroup\$ If transistor is on, and current flows through collector, what is the output voltage? That's an inverter. \$\endgroup\$
    – Justme
    Jan 25 at 10:16
  • \$\begingroup\$ When the transistor is on, the current still passes through the collector resistor, but the equivalent resistance of the transistor between emitter and collector is much lower than the resistance. If A is tied to Vcc, with a transistor beta=100, transistor acts like a resistance with 1/100 the value of the base resistance. \$\endgroup\$
    – TEMLIB
    Jan 25 at 10:29
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    \$\begingroup\$ also an electrical open means no current (open switch = circuit with a gap in it) and closed means current flows (closed switch = circuit with no gaps) \$\endgroup\$
    – user253751
    Jan 25 at 18:13

4 Answers 4

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You've got it backwards.

When the input at A is high, the transistor acts like a short circuit. The output is shorted to ground.

When the transistor shorts to ground, there's 5V across R1. One end is at 5V (from the power supply) while the other end is at 0V (ground.) Given the value of R1 and the source voltage, 5 milliamperes of current will flow through R1 and the transistor.

schematic

simulate this circuit – Schematic created using CircuitLab

When the input at A is low, the transistor acts like an open circuit. No current flows through the transistor, so the resistor pulls the output to the source voltage (Vcc.)

schematic

simulate this circuit

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  • \$\begingroup\$ When the transistor act like a short circuit Output is shorted to ground - what happen to the voltage across R1? and for the current? \$\endgroup\$
    – lon_ly
    Jan 25 at 10:27
  • \$\begingroup\$ @lon_ly: See the changes to my answer. \$\endgroup\$
    – JRE
    Jan 25 at 10:35
  • \$\begingroup\$ @lon_ly All the current that flows through R1 is busy flowing to ground so none of it flows through the output. Same as if you plug a switch where the transistor is and turn the switch on. \$\endgroup\$
    – user253751
    Jan 25 at 18:13
  • \$\begingroup\$ @lon_ly if the above answer still isn't clicking with you, think of it this way: Think of Vcc going across R1, and the transistor shorted to ground as R2 with a 0 ohm voltage. You've just created a voltage divider, and the "NOT A" output is the output of your voltage divider. Since R2 (the transistor)is 0 ohms, the output would be 0. \$\endgroup\$
    – LarryBud
    Jan 26 at 12:34
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The actual problem

How does a transistor inverter work?

Perhaps this is the most important question in analog electronics just because this circuit represents the most basic transistor amplifier. So the question actually is, "How does a transistor amplifier work?"

And since amplification is actually a voltage regulation, the question is really, "How can we control a voltage through another voltage?"

And only last is the question of why and how the circuit inverts the input voltage...

A bit of history

Interestingly, this problem arose a long time ago. Sometime in the middle of the 19th century, people felt the need to reduce the electrical voltage... and they did. To understand the meaning of all this, let us trace the evolution of this idea from the simple electric circuit of the time to the electronic circuit of today.

Building scenario

I will do it in six consecutive steps illustrating them with simple conceptual diagrams. The values ​​of the quantities are approximate for the purpose of this intuitive explanation.

Reducing voltage divider

Imagine the first three steps (schematics) take place in the 19th century.

STEP 1: Connecting a constant voltage source. We have a source of constant voltage Vcc1 and apply this voltage to a true voltage load (voltmeter VM1) which has a very high resistance and therefore does not consume any current. The voltage is "excessively" high; so our task is to reduce it.

schematic

simulate this circuit – Schematic created using CircuitLab

STEP 2: Trying to remove some voltage through resistance. The first idea that comes to mind is to include a resistor R in series with the voltage source Vcc2 to take away part of its voltage. What is our surprise, however, when we see that the voltage after the resistor does not change. The problem is that current does not flow through the resistor... there is no voltage drop across the resistor... and the voltage does not change... as though the resistor is not a resistor.

STEP 3: Causing current to flow. So we have to cause a current to flow in the circuit. For this purpose, we connect a second resistor R2 in series with the first R1 and in parallel with the "load" VM3.

Regulating voltage divider

Now imagine that the next three steps (schematics) take place both in the 19th and 21st centuries.

To change the voltage, we have three options - to change the resistance R1, R2 or both. If we were in the 19th century, we would manually change the resistance by moving the wiper of the so-called "rheostat" (the diagrams on the left below) but in current electronic circuits, e.g. in the OP's inverter, this has to be done automatically using something like a "voltage-controlled resistor" (the diagrams on the right below). A transistor can serve as such a device. It is an "enhancement-mode resistor" which means that when the control voltage increases, its resistance decreases.

STEP 4: Changing R2. So when decreasng R2 (on the left circuit, in the 19th century:-) or increasing Q2's base-emitter voltage (on the right circuit), the output voltage decreases... and the (OP's) circuit is inverting.

schematic

simulate this circuit

STEP 5: Changing R1. And v.v., when decreasng R1 (on the left circuit) or increasing Q1's base-emitter voltage (on the right circuit), the output voltage increases... and we can say that, in this sense, the circuit is non-inverting.

schematic

simulate this circuit

STEP 6: Changing both R1 and R2. Finally, a brilliant idea occurred to us - to simultaneously change R1 and R2 (on the left) and V1 and V2 (on the right) in the opposite direction. The so-called "complementary circuit" without resistors is obtained, which in both states does not consume current.

schematic

simulate this circuit

What is OP' circuit solution?

This is the configuration we discussed in step 4. The regulator (Q2) is connected at the bottom of the "voltage divider" so that as the input voltage increases, its "resistance" decreases and the output voltage decreases accordingly. In the extreme two cases, at a high input voltage the output is low and conversely at a low input voltage the output is high, i.e. the circuit inverts.

The resistance of the transistor is non-linear but this is not essential to understanding the principle.

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if the current is 0 the voltage is 0 and all current will pass through the resistor that is connected between collector and Vcc.

Incorrect. If the current passing through the collector to the emitter is zero, then the collector voltage will be as high as the power supply voltage for any reasonable value of collector resistor.

When the base is HIGH the transistor is open and current will pass from collector to emitter

We don't talk about transistors as hydraulic valves that when open pass fluid. We say activated (or on or shorting) or deactivated (or off or open-circuit).

Also we don't say "when the base is high" because the base isn't driven with logic levels in these circumstances. The point marked A in your circuit can be driven with logic levels but, that isn't the base.

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You said "When the base is HIGH the transistor is open and current will pass from collector to emitter". This is not right.

When the base is HIGH the transistor is CLOSED and current will pass from collector to emitter. So in this case the collector and the emitter are connected the V_c = V_ce. (Depending of the transistor the voltage at the collector will be around few 100mVs, let's say 0.3; 0.7V this value is taken as Low level.

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    \$\begingroup\$ We should always avoid words like "open" and "closed". When a switch is "closed", a path is "open" and allows a current. You see the problem? \$\endgroup\$
    – LvW
    Jan 26 at 16:44

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