# Beginner's question about a NOT inverter

#### Light Yagami

Joined Sep 15, 2021
3
Hello,

This is my first post here. I'm trying to learn electronics on my own via different sites

At the moment, I'm trying to understand how a NOT inverter works. I perfectly understood what it was used for, but something escapes me on this diagram: https://www.dummies.com/programming...-how-to-create-a-transistor-not-gate-circuit/

If I understand correctly, if the input (base) of the transistor is HIGH, then the current flows through the transistor from the emitter to the collector, without passing through the output. And if it is at LOW, then the current goes directly to the output without passing through the transistor (normal).

What I don't understand is why, when the input is at HIGH, the current flow doesn't split between the path through the transistor and the path to the output, but only "flows" through the transistor. Indeed, since the output and the transistor are both connected to VCC, shouldn't the current take both paths?

Thanks for any explanation

Last edited:

#### Ya’akov

Joined Jan 27, 2019
8,158
Welcome to AAC.

#### Light Yagami

Joined Sep 15, 2021
3
Oups, sorry. I've just fixed it. Thanks.

#### Ya’akov

Joined Jan 27, 2019
8,158
HIGH and LOW logic levels are thresholds. A HIGH will be above a certainly voltage and a LOW below one.

In that circuit, when the transistor turns on it creates a voltage divider, and you are not wrong that the current will be split between the two paths, but, the voltage present at the output will be very low when the transistor is on and so be considered logically LOW.

It doesn’t have to be 0V, but even in a circuit like that it can approach it.

#### Ya’akov

Joined Jan 27, 2019
8,158
An addition: imagine if another gate of similar construction followed this one. HIGH (TRUE) would require a voltage high enough to turn on the transistor connected to the output, anything less would be LOW (FALSE).

In this way you can see that LOW doesn’t have to be 0V and HIGH the Vcc.

#### Light Yagami

Joined Sep 15, 2021
3
Thank you very much for this perfectly clear explanation. I had assumed that HIGH == 5V and LOW == 0V and that these values were fixed. But if they are thresholds, then I understand better.

Everything is clear, thank you very much

#### Audioguru again

Joined Oct 21, 2019
6,153
The threshold voltage for a red LED is about 2.0V. If it gets less than about 1.8V then it is turned off.
The transistor's collector becomes 0.1V to 0.2V when it is turned on then the LED does not produce light.

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#### MrChips

Joined Oct 2, 2009
29,207

Remove the LED from the circuit for now since that is confusing you.

When the switch is not pressed (floating input) the NPN transitor is OFF and 6V appears on the output.
When the switch is pressed (HI input to the base of the transistor), the transistor is conducting and about 0.2V appears on the output.

#### dl324

Joined Mar 30, 2015
15,818
Welcome to AAC!
I'm trying to learn electronics on my own via different sites
Just be aware that there's all kinds of garbage on the internet. I've found that a large number of so called "experts" don't know what they're talking about.
What I don't understand is why, when the input is at HIGH, the current flow doesn't split between the path through the transistor and the path to the output, but only "flows" through the transistor. Indeed, since the output and the transistor are both connected to VCC, shouldn't the current take both paths?
You should start with the basics. There are 4 modes of operation for a BJT (Bipolar Junction Transistor). You're using two of them: cutoff and saturation.

In cutoff mode (or just off), the BE and CB junctions are unbiased or reverse biased. This is the condition you have when SW1 is opened. For now, you can ignore leakage currents.

When the switch is closed, current will flow into the base. That current will be:
$$I = \frac{V}{R2} = \frac{6V-0.7V}{1k\Omega} = 5.3mA$$

With a general purpose 2N transistor, such as 2N3904, that will support a collector current of up to 53mA. That data is in the datasheet.

From On Semiconductor (now just onsemi):

You can see that for Ic=50mA, Ib=5mA. This is the beta the manufacturer used for saturation mode operation. We use this rule of thumb for all silicon transistors unless we have reason to use some other number. For Instance, it will be lower for power transistors and higher for something like BC547. The datasheets will give the correct beta to use.

I used Vbe=0.7V in the base current calculation because that's what I was taught in school.

When the transistor is saturated, Vce will be a maximum of 0.3V. Since the anode of the LED is connected to the transistor collector, the voltage across the LED will be a maximum of 0.3V.

If you look at the IV (current vs voltage) graph for an LED, you'll see that most LEDs won't emit light (photons) at that voltage:

These are standard brightness LEDs. If you extrapolate the curve, if it intersects Vf = 0.3V, it will be at a very very low current and you wouldn't expect to see any light emitted.