# Concept clarification

#### Shimul07

Joined Oct 24, 2023
13
I was studying BJT voltage divider bias and its characteristics. But it has come to my notice that there is a 180 degree phase shift between input and output signals.My question is-Can anyone explain why is there a phase shift?

I haven't got any answer from textbooks.The schematic :

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#### Papabravo

Joined Feb 24, 2006
21,228
The BJT is a current controlled current device. It works like this:
1. An increase in voltage at the base due to the AC signal input will cause an increase in the base current.
2. The increase in the base current will cause an increase in the collector current.
3. The increase in the collector current will cause the voltage across the collector resistor to increase.
4. More voltage drop across the collector resistor means less voltage with respect to ground
Hence there is a voltage inversion in the Common Emitter configuration.

A similar argument works for a negative going AC signal at the base. Have you ever drawn a load line on a set of characteristic curves? That exercise will answer many questions.

#### crutschow

Joined Mar 14, 2008
34,470
Below is a simple LTspice simulation to show what's happening:
As the base current increases and decreases around its bias point (green trace) the collector and collector resistor current also increase and decrease (yellow trace), but by much more (due to the model transistor current gain of about 100).
This causes the collector voltage (red trace) to decrease and increase due to the voltage drop across the resistor from the current.

So you can see that the collector voltage is out-of-phase with the base current and the base voltage (blue trace).

The base voltage looks distorted due to its diode log relation to the base current.
That's why negative feedback is needed (typically an unbypassed emitter resistor) to reduce the distortion in a transistor amp when driven with a voltage.

Make sense?

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#### MrChips

Joined Oct 2, 2009
30,824
The common emitter transistor is configured as a LOW-SIDE switch.
When the base voltage is LOW, the switch is open, the collector is HIGH.
When the base voltage is HIGH, the switch is closed, the collector is LOW.

#### Shimul07

Joined Oct 24, 2023
13
Below is a simple LTspice simulation to show what's happening:
As the base current increases and decreases around its bias point (green trace) the collector and collector resistor current also increase and decrease (yellow trace), but by much more (due to the model transistor current gain of about 100).
This causes the collector voltage (red trace) to decrease and increase due to the voltage drop across the resistor from the current.

So you can see that the collector voltage is out-of-phase with the base current and the base voltage (blue trace).

The base voltage looks distorted due to its diode log relation to the base current.
That's why negative feedback is needed (typically an unbypassed emitter resistor) to reduce the distortion in a transistor amp when driven with a voltage.

Make sense?

View attachment 305839
what is physically happening inside of the circuit?I mean is the current flow theough collector resistor being halted somehow for which output signal voltage lags behind of input?

#### BobTPH

Joined Jun 5, 2013
9,003
Answer this question: If the collector current increases, the output voltage:

a) increases
b) decreases
c) stays the same

#### crutschow

Joined Mar 14, 2008
34,470
what is physically happening inside of the circuit?
You seem to have only a hazy understanding of voltage across a resistor due to current through the resistor (Ohm's law).

Have you not studied how BJ transistors work?
Basically the collector current equals the base current times the transistor current gain.
If you understand Ohm's law, than that explains what is physically happening.

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#### MrAl

Joined Jun 17, 2014
11,496
I was studying BJT voltage divider bias and its characteristics. But it has come to my notice that there is a 180 degree phase shift between input and output signals.My question is-Can anyone explain why is there a phase shift?

I haven't got any answer from textbooks.The schematic :
You mean the 180 degree phase shift?

To understand this, look at how a bipolar transistor works. More basically, as you increase the input current the output current goes up, and when you decrease the input current the output current goes down. The output current works with a resistor which means you see this as a voltage change on the output when you measure the output voltage, but the output resistor is tied to a DC voltage which means as the current in the output decreases there is less voltage drop across the collector resistor, and hence you see the output as the power supply voltage minus the voltage drop, which shows the output voltage as going up.

Thus, the main inversion mechanism is the subtraction of the DC supply voltage minus the voltage drop in the collector resistor.

Now, take that information and figure out the relationship of the output voltage when the input current goes up, and when the input current goes down as above, and since the input current follows the input voltage (being in phase), you will see the answer to why the output voltage gets inverted (180 degrees out of phase).

#### BobaMosfet

Joined Jul 1, 2009
2,113
I was studying BJT voltage divider bias and its characteristics. But it has come to my notice that there is a 180 degree phase shift between input and output signals.My question is-Can anyone explain why is there a phase shift?

I haven't got any answer from textbooks.The schematic :
Title: Understanding Basic Electronics, 1st Ed.
Publisher: The American Radio Relay League
ISBN: 0-87259-398-3

#### WBahn

Joined Mar 31, 2012
30,087
what is physically happening inside of the circuit?I mean is the current flow theough collector resistor being halted somehow for which output signal voltage lags behind of input?
It's not a phase shift, it's a signal inversion.

What happens to the voltage at the output as the voltage at the input changes?

Let's walk it through.

The signal voltage goes from some voltage to some slightly higher voltage.
That results in the base voltage being increased.
That results in more base current.
That results in more collector current.
More collector current results in a larger voltage drop across the collector resistor.
Since the top of the collector resistor is fixed, that means that the voltage at the bottom of the collector resistor (i.e., the output) has to go down.

So the voltage gain of the amplifier is negative.

#### BobTPH

Joined Jun 5, 2013
9,003
I guess no i else values the Socratic method. I believe he knows enough to work it out if giuided. That is what I am trying to do.

#### MrAl

Joined Jun 17, 2014
11,496
It's not a phase shift, it's a signal inversion.
Hi,

I am not sure if I can agree with the implied semantics.
The phrase "phase shift" is just a more general way to describe a phase difference. For example, if two signals are in phase we can say that there is a zero degree phase shift between the two. If the phase of one is 10 degrees, we can say one is shifted by 10 degrees, and if one is 20 degrees, shifted by 20 degrees, and if 90 degrees then 90 degrees out of phase, and if 170 degrees then we can say 170 degrees out of phase, so I do not see a problem with calling it 180 degrees out of phase from the input, or it has a phase shift of 180 degrees.
If the frequency went up it could actually go to 179 degrees for example.
Thus, I do not think we have to concentrate entirely and solely on the reason behind the phase shift.
I do have to agree though that we should mention that the circuit is in fact called an inverter or inverting amplifier, etc.

#### WBahn

Joined Mar 31, 2012
30,087
Hi,

I am not sure if I can agree with the implied semantics.
The phrase "phase shift" is just a more general way to describe a phase difference. For example, if two signals are in phase we can say that there is a zero degree phase shift between the two. If the phase of one is 10 degrees, we can say one is shifted by 10 degrees, and if one is 20 degrees, shifted by 20 degrees, and if 90 degrees then 90 degrees out of phase, and if 170 degrees then we can say 170 degrees out of phase, so I do not see a problem with calling it 180 degrees out of phase from the input, or it has a phase shift of 180 degrees.
If the frequency went up it could actually go to 179 degrees for example.
Thus, I do not think we have to concentrate entirely and solely on the reason behind the phase shift.
I do have to agree though that we should mention that the circuit is in fact called an inverter or inverting amplifier, etc.
A phase shift of a signal is equivalent to a delay in that signal. An inversion is very different.

The post that I was responding to very clearly indicates that the TS is trying to understand the underlying mechanism for what is being observed in terms of the signal being delayed somehow. That is not what is happening. The signal is being inverted.

Showing that these are NOT the same is quite trivial. Put in a pulse train of short, positive pulses separated by several times the pulsewidth. Do you see a series of short, positive pulses that are delayed by half a period of the overall waveform? No? So much for the output waveform being the input with a 180° phase shift. What do you see? You see a series of short, negative pulses that are aligned with the input pulses with very little delay, which is consistent with a straight and simple signal inversion.

#### BobTPH

Joined Jun 5, 2013
9,003
A 180° phase shift is the same as in inversion for a pure sine wave, or a 50% duty cycle square wave. But not for many other waveforms.

#### MrChips

Joined Oct 2, 2009
30,824
Phase shift and inversion are two different operations.

Phase Shift
y = Asin(ωt +Φ)

Inversion
y = -Asin(ωt)

#### WBahn

Joined Mar 31, 2012
30,087
Phase shift and inversion are two different operations.

Phase Shift
y = Asin(ωt +Φ)

Inversion
y = -Asin(ωt)
Poor example, since, for Φ = 180°, it is true that

Asin(ωt +Φ) = -Asin(ωt)

But for most waveforms, this is not the case.

#### BobTPH

Joined Jun 5, 2013
9,003
They are equal for a sine wave when Φ = π. Also true any waveform containing only odd harmonics. Throw in some even harmonics, and they will be different.

#### MrAl

Joined Jun 17, 2014
11,496
A phase shift of a signal is equivalent to a delay in that signal. An inversion is very different.

The post that I was responding to very clearly indicates that the TS is trying to understand the underlying mechanism for what is being observed in terms of the signal being delayed somehow. That is not what is happening. The signal is being inverted.

Showing that these are NOT the same is quite trivial. Put in a pulse train of short, positive pulses separated by several times the pulsewidth. Do you see a series of short, positive pulses that are delayed by half a period of the overall waveform? No? So much for the output waveform being the input with a 180° phase shift. What do you see? You see a series of short, negative pulses that are aligned with the input pulses with very little delay, which is consistent with a straight and simple signal inversion.
Hi,

But isn't he talking about sine waves?
If I tell you I have a phase shift of 180 degrees in a signal I would think you would know what I mean.

In control theory there's really no difference for example when working in the 's' domain.

Also, I would think that any periodic waveform can be considered to be phase shifted but perhaps not inverted. Thus calling it an "inverter" may not be entirely correct either.

I think what we are talking about here is how the circuit is USED. In your example I think it would be better to call it an inverter. If we have positive pulses going in, we get negative pulses out. With sine waves, as was illustrated in the first post, i think we could call it either an inversion or a phase shift.
Remember we are really talking about the measurements vs the application.
We measure phase shifts, we implement circuits.

But also, we see input and output capacitors. That changes everything. Unless we have some really strange signal, we get a measurable phase shift. Even a train of all positive pulses in various lengths would show a periodic output. It would not really invert the signal with the capacitors as we think of in the pure DC amplifier case.

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#### MrAl

Joined Jun 17, 2014
11,496
They are equal for a sine wave when Φ = π. Also true any waveform containing only odd harmonics. Throw in some even harmonics, and they will be different.
It's interesting that you bright this up because a full wave rectified signal is sometimes confusing. Sometimes the reference of phase is to the original sine, and sometimes to the output half cycle. That means 1/2 cycle is sometimes referred to as 180 degrees and sometimes 360 degrees.

#### MrAl

Joined Jun 17, 2014
11,496
Phase shift and inversion are two different operations.

Phase Shift
y = Asin(ωt +Φ)

Inversion
y = -Asin(ωt)
Hi,

Not sure if I can agree with that, except as a direct application in some application perhaps. For example:
y=A*sin(wt+180)

The real difference I think is that inversion is always 180 degrees, while a pure phase shift can take on any value.
In many cases even a phase shift of 360 degrees + 180 degrees is still an inversion:
y=A*sin(wt+540)
Of course that is not always the case.