# What is a load line - intuitively

Started by April 30, 2017
```http://pix.toile-libre.org/?img=1493527288.png

That's a typical graph of Vce vs Ic for a bunch of Ib as provided by the
manufacturer. Is my understanding correct:

We create a loadline so that we can operate the transistor's collector ckt
in a linear fashion for a varying base input.

Basically there are an infinite set of Vce vs Ic curves for each and every
base current value possible - obviously manufacturer's can't plot all that
so they give us certain typical curves.

Q represents the DC operating point for some base current but when an input
signal is fed for amplification, the base current changes, and we basically
move to a different Vce vs Ic curve (the new Q point on this curve
represents the output for the new base current flowing as a result of
changed Ib due to signal).

Because the load line is linear, for every change in Ib we get a linear
change in Ic; if the load line was somehow a sinusoid we'd get a sinusoidal
amplification action for Ib?
```
```veek wrote:

> http://pix.toile-libre.org/?img=1493527288.png
>
> That's a typical graph of Vce vs Ic for a bunch of Ib as provided by the
> manufacturer. Is my understanding correct:
>
> We create a loadline so that we can operate the transistor's collector ckt
> in a linear fashion for a varying base input.
>
> Basically there are an infinite set of Vce vs Ic curves for each and every
> base current value possible - obviously manufacturer's can't plot all that
> so they give us certain typical curves.
>
> Q represents the DC operating point for some base current but when an
> input signal is fed for amplification, the base current changes, and we
> basically move to a different Vce vs Ic curve (the new Q point on this
> curve represents the output for the new base current flowing as a result
> of changed Ib due to signal).
>
> Because the load line is linear, for every change in Ib we get a linear
> change in Ic; if the load line was somehow a sinusoid we'd get a
> sinusoidal amplification action for Ib?
Also, would it be fair to say:
Once the transistor is amplifying a signal.. the Q point would wander/walk
all along the load-line and that the current Q-point position represents the
Vce*Ic output for corresponding Ib?
```
```On 4/30/2017 12:57 AM, veek wrote:
> veek wrote:
>
>> http://pix.toile-libre.org/?img=1493527288.png
>>
>> That's a typical graph of Vce vs Ic for a bunch of Ib as provided by the
>> manufacturer. Is my understanding correct:
>>
>> We create a loadline so that we can operate the transistor's collector ckt
>> in a linear fashion for a varying base input.
>>
>> Basically there are an infinite set of Vce vs Ic curves for each and every
>> base current value possible - obviously manufacturer's can't plot all that
>> so they give us certain typical curves.
>>
>> Q represents the DC operating point for some base current but when an
>> input signal is fed for amplification, the base current changes, and we
>> basically move to a different Vce vs Ic curve (the new Q point on this
>> curve represents the output for the new base current flowing as a result
>> of changed Ib due to signal).
>>
>> Because the load line is linear, for every change in Ib we get a linear
>> change in Ic; if the load line was somehow a sinusoid we'd get a
>> sinusoidal amplification action for Ib?
> Also, would it be fair to say:
> Once the transistor is amplifying a signal.. the Q point would wander/walk
> all along the load-line and that the current Q-point position represents the
> Vce*Ic output for corresponding Ib?

I'm not sure of all the things you wrote, but the load line is the line
drawn *on* the Ic vs Vce curves to show the possible operating points.
With a constant power supply voltage the voltage across the load
resistor will be the difference of power supply minus Vc, so the load
line is drawn from zero current and max Vcc to zero voltage and max Ic.

Where the load line intersects the transistor curves for a given base
current is where the transistor will operate.  I would not say the
operation of the transistor is necessarily linear.  To see that clearly
you need to plot Ic vs Ib for a given load.  In general it won't be
completely linear.

--

Rick C
```
```rickman wrote:

> On 4/30/2017 12:57 AM, veek wrote:
>> veek wrote:
>>
>>> http://pix.toile-libre.org/?img=1493527288.png
>>>
>>> That's a typical graph of Vce vs Ic for a bunch of Ib as provided by the
>>> manufacturer. Is my understanding correct:
>>>
>>> We create a loadline so that we can operate the transistor's collector
>>> ckt in a linear fashion for a varying base input.
>>>
>>> Basically there are an infinite set of Vce vs Ic curves for each and
>>> every base current value possible - obviously manufacturer's can't plot
>>> all that so they give us certain typical curves.
>>>
>>> Q represents the DC operating point for some base current but when an
>>> input signal is fed for amplification, the base current changes, and we
>>> basically move to a different Vce vs Ic curve (the new Q point on this
>>> curve represents the output for the new base current flowing as a result
>>> of changed Ib due to signal).
>>>
>>> Because the load line is linear, for every change in Ib we get a linear
>>> change in Ic; if the load line was somehow a sinusoid we'd get a
>>> sinusoidal amplification action for Ib?
>> Also, would it be fair to say:
>> Once the transistor is amplifying a signal.. the Q point would
>> wander/walk all along the load-line and that the current Q-point position
>> represents the Vce*Ic output for corresponding Ib?
>
> I'm not sure of all the things you wrote, but the load line is the line
> drawn *on* the Ic vs Vce curves to show the possible operating points.
> With a constant power supply voltage the voltage across the load
> resistor will be the difference of power supply minus Vc, so the load
> line is drawn from zero current and max Vcc to zero voltage and max Ic.

(anyway yeah - that's the math procedure for computing loadline - plug and
chug thing]

> Where the load line intersects the transistor curves for a given base
> current is where the transistor will operate.  I would not say the
> operation of the transistor is necessarily linear.  To see that clearly
> you need to plot Ic vs Ib for a given load.  In general it won't be
> completely linear.
>
Ah.. so the transistor operates in a range between Qh and Ql on the load
line? For a sine wave, first half it would operate between Q and Ql? The
operating point would shift from Q to Ql and then back to Q for the zero
crossover?
```
```veek wrote:

> http://pix.toile-libre.org/?img=1493527288.png
>
> That's a typical graph of Vce vs Ic for a bunch of Ib as provided by the
> manufacturer. Is my understanding correct:
>
> We create a loadline so that we can operate the transistor's collector ckt
> in a linear fashion for a varying base input.
>
> Basically there are an infinite set of Vce vs Ic curves for each and every
> base current value possible - obviously manufacturer's can't plot all that
> so they give us certain typical curves.
>
> Q represents the DC operating point for some base current but when an
> input signal is fed for amplification, the base current changes, and we
> basically move to a different Vce vs Ic curve (the new Q point on this
> curve represents the output for the new base current flowing as a result
> of changed Ib due to signal).
>
> Because the load line is linear, for every change in Ib we get a linear
> change in Ic; if the load line was somehow a sinusoid we'd get a
> sinusoidal amplification action for Ib?

https://qph.ec.quoracdn.net/main-qimg-752573d7d6bc8c20319db701ddc539d3
In this graph/image:
1. Can the transistor work at A? No - right? The curved part represents a
non-linear part where Vce is not large enough to draw all electrons from the
emitter - you'd be leaving behind part of th signal (so to speak) for some
particular Ib?

2. Note the Q Point (active REGION). The transistor is active across all Ib

As in: the Q point (that point on graph) is where the transistor stays when
it has no input signal because of DC biasing. When there is an input signal
the Q point moves along the load-line between A and B driven by the input
signal (assuming the signal is large enough to drive the Q point that far) -
correct?

If it goes all the way to the curved region near A - the Ic value will no
longer be proportional to Ib like it was near say where Ic=50 cuts the load
line?
```
```On 4/30/2017 3:23 AM, veek wrote:
> rickman wrote:
>
>> On 4/30/2017 12:57 AM, veek wrote:
>>> veek wrote:
>>>
>>>> http://pix.toile-libre.org/?img=1493527288.png
>>>>
>>>> That's a typical graph of Vce vs Ic for a bunch of Ib as provided by the
>>>> manufacturer. Is my understanding correct:
>>>>
>>>> We create a loadline so that we can operate the transistor's collector
>>>> ckt in a linear fashion for a varying base input.
>>>>
>>>> Basically there are an infinite set of Vce vs Ic curves for each and
>>>> every base current value possible - obviously manufacturer's can't plot
>>>> all that so they give us certain typical curves.
>>>>
>>>> Q represents the DC operating point for some base current but when an
>>>> input signal is fed for amplification, the base current changes, and we
>>>> basically move to a different Vce vs Ic curve (the new Q point on this
>>>> curve represents the output for the new base current flowing as a result
>>>> of changed Ib due to signal).
>>>>
>>>> Because the load line is linear, for every change in Ib we get a linear
>>>> change in Ic; if the load line was somehow a sinusoid we'd get a
>>>> sinusoidal amplification action for Ib?
>>> Also, would it be fair to say:
>>> Once the transistor is amplifying a signal.. the Q point would
>>> wander/walk all along the load-line and that the current Q-point position
>>> represents the Vce*Ic output for corresponding Ib?
>>
>> I'm not sure of all the things you wrote, but the load line is the line
>> drawn *on* the Ic vs Vce curves to show the possible operating points.
>> With a constant power supply voltage the voltage across the load
>> resistor will be the difference of power supply minus Vc, so the load
>> line is drawn from zero current and max Vcc to zero voltage and max Ic.
>
> (anyway yeah - that's the math procedure for computing loadline - plug and
> chug thing]

Yes, Rc (c for collector) is the load resistor in this case.

>> Where the load line intersects the transistor curves for a given base
>> current is where the transistor will operate.  I would not say the
>> operation of the transistor is necessarily linear.  To see that clearly
>> you need to plot Ic vs Ib for a given load.  In general it won't be
>> completely linear.
>>
> Ah.. so the transistor operates in a range between Qh and Ql on the load
> line? For a sine wave, first half it would operate between Q and Ql? The
> operating point would shift from Q to Ql and then back to Q for the zero
> crossover?

I was looking at the schematic for the other post with 100 kohm and 10
kohm resistors.  This one with 500 kohm base and 3 kohm collector
resistors is *not* in saturation.

I don't remember the Q point thing, at least the name.  But the idea is
that the bias puts the static point somewhere which you are calling Q
and the AC signal creates the movement on the load line between Ql and
Qh.  So I think you get it.

--

Rick C
```
```On 4/30/2017 6:36 AM, veek wrote:
> veek wrote:
>
>> http://pix.toile-libre.org/?img=1493527288.png
>>
>> That's a typical graph of Vce vs Ic for a bunch of Ib as provided by the
>> manufacturer. Is my understanding correct:
>>
>> We create a loadline so that we can operate the transistor's collector ckt
>> in a linear fashion for a varying base input.
>>
>> Basically there are an infinite set of Vce vs Ic curves for each and every
>> base current value possible - obviously manufacturer's can't plot all that
>> so they give us certain typical curves.
>>
>> Q represents the DC operating point for some base current but when an
>> input signal is fed for amplification, the base current changes, and we
>> basically move to a different Vce vs Ic curve (the new Q point on this
>> curve represents the output for the new base current flowing as a result
>> of changed Ib due to signal).
>>
>> Because the load line is linear, for every change in Ib we get a linear
>> change in Ic; if the load line was somehow a sinusoid we'd get a
>> sinusoidal amplification action for Ib?
>
> https://qph.ec.quoracdn.net/main-qimg-752573d7d6bc8c20319db701ddc539d3
> In this graph/image:
> 1. Can the transistor work at A? No - right? The curved part represents a
> non-linear part where Vce is not large enough to draw all electrons from the
> emitter - you'd be leaving behind part of th signal (so to speak) for some
> particular Ib?

I think you are referring to "working" as amplifying a signal with
linearity.  No, biasing the transistor to point A will not let it swing
both ways so an input signal of any amplitude will be clipped.

But amplifying an AC signal is not the only reason to use a transistor.
There are switching uses where bias at point A would be a good idea.

> 2. Note the Q Point (active REGION). The transistor is active across all Ib

Not sure what this implies, "active across all Ib"?  There are limits to
the Ib range you can use no matter where you bias it.  But yes, this
bias point will give you a good range of operation.

> As in: the Q point (that point on graph) is where the transistor stays when
> it has no input signal because of DC biasing. When there is an input signal
> the Q point moves along the load-line between A and B driven by the input
> signal (assuming the signal is large enough to drive the Q point that far) -
> correct?

Yes, when you apply an input signal the collector voltage moves across

> If it goes all the way to the curved region near A - the Ic value will no
> longer be proportional to Ib like it was near say where Ic=50 cuts the load
> line?

Yes, it only has to get close and the non-linearity gets bad.

--

Rick C
```
```rickman wrote:

> On 4/30/2017 3:23 AM, veek wrote:
>> rickman wrote:
>>
>>> On 4/30/2017 12:57 AM, veek wrote:
>>>> veek wrote:
>>>>
>>>>> http://pix.toile-libre.org/?img=1493527288.png
>>>>>
>>>>> That's a typical graph of Vce vs Ic for a bunch of Ib as provided by
>>>>> the manufacturer. Is my understanding correct:
>>>>>
>>>>> We create a loadline so that we can operate the transistor's collector
>>>>> ckt in a linear fashion for a varying base input.
>>>>>
>>>>> Basically there are an infinite set of Vce vs Ic curves for each and
>>>>> every base current value possible - obviously manufacturer's can't
>>>>> plot all that so they give us certain typical curves.
>>>>>
>>>>> Q represents the DC operating point for some base current but when an
>>>>> input signal is fed for amplification, the base current changes, and
>>>>> we basically move to a different Vce vs Ic curve (the new Q point on
>>>>> this curve represents the output for the new base current flowing as a
>>>>> result of changed Ib due to signal).
>>>>>
>>>>> Because the load line is linear, for every change in Ib we get a
>>>>> linear change in Ic; if the load line was somehow a sinusoid we'd get
>>>>> a sinusoidal amplification action for Ib?
>>>> Also, would it be fair to say:
>>>> Once the transistor is amplifying a signal.. the Q point would
>>>> wander/walk all along the load-line and that the current Q-point
>>>> position represents the Vce*Ic output for corresponding Ib?
>>>
>>> I'm not sure of all the things you wrote, but the load line is the line
>>> drawn *on* the Ic vs Vce curves to show the possible operating points.
>>> With a constant power supply voltage the voltage across the load
>>> resistor will be the difference of power supply minus Vc, so the load
>>> line is drawn from zero current and max Vcc to zero voltage and max Ic.
>>
>> (anyway yeah - that's the math procedure for computing loadline - plug
>> and chug thing]
>
> Yes, Rc (c for collector) is the load resistor in this case.
>
>
>>> Where the load line intersects the transistor curves for a given base
>>> current is where the transistor will operate.  I would not say the
>>> operation of the transistor is necessarily linear.  To see that clearly
>>> you need to plot Ic vs Ib for a given load.  In general it won't be
>>> completely linear.
>>>
>> Ah.. so the transistor operates in a range between Qh and Ql on the load
>> line? For a sine wave, first half it would operate between Q and Ql? The
>> operating point would shift from Q to Ql and then back to Q for the zero
>> crossover?
>
> I was looking at the schematic for the other post with 100 kohm and 10
> kohm resistors.  This one with 500 kohm base and 3 kohm collector
> resistors is *not* in saturation.
>
> I don't remember the Q point thing, at least the name.  But the idea is
> that the bias puts the static point somewhere which you are calling Q
> and the AC signal creates the movement on the load line between Ql and
> Qh.  So I think you get it.
>
cool - thanks!
```
```On Sun, 30 Apr 2017 10:24:17 +0530, veek <vek.m1234@gmail.com> wrote:

>http://pix.toile-libre.org/?img=1493527288.png
>
>That's a typical graph of Vce vs Ic for a bunch of Ib as provided by the
>manufacturer. Is my understanding correct:
>
>We create a loadline so that we can operate the transistor's collector ckt
>in a linear fashion for a varying base input.
>
>Basically there are an infinite set of Vce vs Ic curves for each and every
>base current value possible - obviously manufacturer's can't plot all that
>so they give us certain typical curves.
>
>Q represents the DC operating point for some base current but when an input
>signal is fed for amplification, the base current changes, and we basically
>move to a different Vce vs Ic curve (the new Q point on this curve
>represents the output for the new base current flowing as a result of
>changed Ib due to signal).
>
>Because the load line is linear, for every change in Ib we get a linear
>change in Ic; if the load line was somehow a sinusoid we'd get a sinusoidal
>amplification action for Ib?

If you're designing a linear transistor amplifier, you usually don't
current will be constant at some value, and pick the collector
resistor accordingly, typically to park midway between VCC and
transistor saturation.

analysis was one easy way to define the operating point.

And sure, if the collector load is nonlinear, the gain will be
nonlinear.

--

John Larkin         Highland Technology, Inc

lunatic fringe electronics

```
```On Sun, 30 Apr 2017 16:06:13 +0530, veek <vek.m1234@gmail.com> wrote:

>veek wrote:
>
>> http://pix.toile-libre.org/?img=1493527288.png
>>
>> That's a typical graph of Vce vs Ic for a bunch of Ib as provided by the
>> manufacturer. Is my understanding correct:
>>
>> We create a loadline so that we can operate the transistor's collector ckt
>> in a linear fashion for a varying base input.
>>
>> Basically there are an infinite set of Vce vs Ic curves for each and every
>> base current value possible - obviously manufacturer's can't plot all that
>> so they give us certain typical curves.
>>
>> Q represents the DC operating point for some base current but when an
>> input signal is fed for amplification, the base current changes, and we
>> basically move to a different Vce vs Ic curve (the new Q point on this
>> curve represents the output for the new base current flowing as a result
>> of changed Ib due to signal).
>>
>> Because the load line is linear, for every change in Ib we get a linear
>> change in Ic; if the load line was somehow a sinusoid we'd get a
>> sinusoidal amplification action for Ib?
>
>https://qph.ec.quoracdn.net/main-qimg-752573d7d6bc8c20319db701ddc539d3
>In this graph/image:
>1. Can the transistor work at A? No - right? The curved part represents a
>non-linear part where Vce is not large enough to draw all electrons from the
>emitter - you'd be leaving behind part of th signal (so to speak) for some
>particular Ib?
>
>2. Note the Q Point (active REGION). The transistor is active across all Ib
>
>As in: the Q point (that point on graph) is where the transistor stays when
>it has no input signal because of DC biasing. When there is an input signal
>the Q point moves along the load-line between A and B driven by the input
>signal (assuming the signal is large enough to drive the Q point that far) -
>correct?
>
>If it goes all the way to the curved region near A - the Ic value will no
>longer be proportional to Ib like it was near say where Ic=50 cuts the load
>line?

Two small points:

Beta is poorly controlled, so using Ib to set Ic is hazardous. One
generally designs a circuit to force some desired Ic indepentent of
beta.

"Q-point" is, I think, an amateur audio term. We talk about bias
current and collector voltage. Or more generally, maybe bias point.

--

John Larkin         Highland Technology, Inc

lunatic fringe electronics

```