The circuit is simple to analyze numerically.

The voltage drop over R2 is Q2's Vbe. So the current going through R2 / LED / Q1 is 0.65/33=20ma;

The base of Q1 (=Q2's collector) sits one Vbe above R2 = 0.65 + 0.65 = 1.3v;

That means the current going through R1 / Q2 = (10 - 1.3) / 10k = 1ma.

Dahdah!

You can introduce other factors in but 10% of the work (=above) gets you within 90% of the correct answers.

That's the difference between an engineer (10% input for 90% of the output) and an engineering professor (90% input for 10% of the output).

Did you read posts #13 and #14? Those are from an engineering professor.

 

I made a lot of errors in my equations. And they where complicated.
And defining quiescent values doesn't necessarily mean one understands the circuit operation.
After doing this maths, I can see, thanks to hindsight and other posts, that there are easier and more intuitive ways.
This is my first attempt at a 'real' analysis, and I am happy that I solved all the variables, perhaps not 100%.
I would much prefer my equations be debated if they are wrong, and any errors brought to the front. After all, I am only human.
I will say though, that having defined all the currents and voltages, the circuit makes more sense.
For instance, I know I3 is dependent on beta squared. Which makes sense.
Keep up the discussion.
Perhaps someone else can draw the diagram and solve the equations more coherently, this would be good.
Give it a go.

 

Perhaps someone else can draw the diagram and solve the equations more coherently, this would be good.

There are only a few interesting equations.

\(\small R2 = \frac{V_{BE2}} {I_{LED}}\)

\(\small R1 <= \frac{V1 - (V_{BE1} + V_{BE2})} {I_{LED}/({\beta_{1} + 1})}\)

\(\small V1 >= V_{BE2} + n * V_{f} + 1\) Note: 1V added to keep Q1 out of saturation. Making V1 significantly larger than necessary increases dissipation in Q1.

 

Many times an accurate description doesn't help understand the problem.

That's why we have models: by ignoring those factors that are not as important.

 

Here's another go. Hopefully its right, and a bit easier to follow.

Here's another circuit that will do the job.

In order to find out how well each circuit maintains a constant current through the diodes, we need to see how this current changes with beta and Vbe, which are dependent on temperature.
Replace Vbe and β with Vbe + ΔVbe and β + Δβ, where ΔVbe and Δβ are small changes.
What is ΔI(diode).

 

Hopefully its right

Ic1=56ma

That pretty much dashed any hope of it being right,

Here's another circuit that will do the job.

When the supply voltage changes, what does the current through the load?

This is one of those cases where you may be able to lay out the (wrong, in this case) equations. If you don't understand intuitively how the circuit works, it likely will lead you down to the wrong path.

Having the right intuition is far more important than having math.

 

In order to find out how well each circuit maintains a constant current through the diodes, we need to see how this current changes with beta and Vbe, which are dependent on temperature.

You’re overthinking this circuit. The human eye response to light is logarithmic and changes of less than 50% are generally unnoticeable and not objectionable.

LED luminous intensity does vary with temperature, but a 20C increase from room temp (i.e. 110F) will result in an intensity change of around 20%.

Intensity also changes with current, but a 25% change in current (from 20mA) will result in an intensity change of around +/- 10%.


Transistor beta changes with temperature, but a well designed circuit minimizes the effect of absolute or varying beta. Making R1 smaller reduces the effect of beta variation; at a cost of increased power dissipation.

The effect of temperature on Vbe will have a small effect on intensity; on the order of a percent per 10C.

 

Hopefully its right, and a bit easier to follow.

I'm really sorry but you wrong again. Ic1 cannot be equal to 56mA.
And for this circuit we have
Ic1 = (β*(R2*Vcc + R1*Vbe*β - R2*2*Vbe))/(R1*R2 (1 + β + β*β)) = (400(0.033kΩ*20V + 10kΩ*0.7V*400 - 0.033kΩ*1.4)/(10kΩ*0.033kΩ *(401 + 400*400) ) = 21.1637mA
Ib1 = (R2*Vcc + R1*Vbe*β - R2*2*Vbe))/(R1*R2 (1 + β + β*β)) = (0.033*20 + 10*0.7*400 - 0.033*1.4)/(10*0.033 *(401 + 400*400)) = 0.052909mA = 52.909μA
Ic2 = ( (R2 * (Vcc - 2*Vbe)*(1 + β) - R1*Vbe)*β)/(R1*R2 (1 + β + β*β)) = (0.033*(20 - 1.4)*401 - 10*0.7)*400)/(10*0.033 *(401 + 400*400) ) = 1.8070mA
Ib2 = ( (R2 * (Vcc - 2*Vbe)*(1 + β) - R1*Vbe))/(R1*R2 (1 + β + β*β)) = (0.033*(20 - 1.4)*401 - 10*0.7)/(10*0.033 *(401 + 400*400) ) = 0.0045177mA = 4.5177μA

 

Ic1 = (β*(R2*Vcc + R1*Vbe*β - R2*2*Vbe))/(R1*R2 (1 + β + β*β)) = (400(0.033kΩ*20V + 10kΩ*0.7V*400 - 0.033kΩ*1.4)/(10kΩ*0.033kΩ *(401 + 400*400) ) = 21.1637mA

Or 650mv/33 = 20ma.

Ic2 =

( (R2 * (Vcc - 2*Vbe)*(1 + β) - R1*Vbe)*β)/(R1*R2 (1 + β + β*β)) = (0.033*(20 - 1.4)*401 - 10*0.7)*400)/(10*0.033 *(401 + 400*400) ) = 1.8070mA

or (20v-1.3v)/10k = 2ma.

Much simpler calculations that gets 90% of the job done with 10% of the effort.

 

I'm really sorry but you wrong again.

Yeap. No amount of math can help if you don't understand how the circuit works.

 

Or 650mv/33 = 20ma.



or (20v-1.3v)/10k = 2ma.

Much simpler calculations that gets 90% of the job done with 10% of the effort.

So? No one has said that you can (or should) use more detailed techniques than are warranted. The issue is knowing what is and is not warranted and which 10% of the job can be left undone.

 

It's right this time. I messed up the simultaneous equations. I'v got one foot in the mud.

An opto-coupler used in a critical feedback path of a switch mode power supply, for instance, would have to have its intensity precisely controlled.
And there are other factors influencing this.
How Vs is regulated, how resistance, beta, Vbe and Vd change with temperature, and so on.
And this is purely a quiescent analysis. Even if I understood this, what about the dynamic analysis. Would I understand this.
The thing is, with so many currents and voltages happening in so many places, it can only be understood with mathematics.

A typical dynamic analysis would be finding how Ic changes with changing β:
Say β is the value of the current gain at 25°C, and Ic(β) is the value of Ic at this beta.
Say Δ.β is the change in β due to a change in temperature, and that Δ.Ic is a change in Ic due to Δ.β.

Then Δ.Ic = Ic(β + Δ.β) - Ic(β)

 

The problem with your approach is that you are trying to use both higher-order effects and zero-order models at the same time. You fix Vbe at 0.7 V (one sig fig) yet want to get into the weeds with taking the base currents into account which requires calculations good to five sig figs.

 

You are right. But there's nothing wrong with finding a value disappearing due to rounding, if only to show it is insignificant.
My way of doing it covers all angles, and I tend to re-invent the wheel, but I know I don't miss anything. It is important to have feedback though, because I couldn't figure this **** out on my own. Now that the equations are written and correct, they are written for good. They can be simplified but it doesn't work the other way around.

Back to the (concise and complete) analysis of the dynamic operation of the circuit. (I still don't know if it's a practical circuit, or something invented by a university student to test his maths and circuit analysis prowess).

ΔIc1 = Ic1(β1 + Δβ1) - Ic1(β1)
Ic1 = β1.Ib1
ΔIc1 = (β1 + Δβ1).Ib1 - β1.Ib1
= Δβ1.Ib1
So Ic1 is directly dependent on Ib1, but only if Ib1 is unchanging.
But Ib1 is dependent on Vbe1, Ib2 and β2 etc.
Isn't maths fun

 

**** !
So Ic1 is directly dependent on Ib1, but only if Ib1 is unchanging. =>
So Ic1 is directly dependent on β1, but only if Ib1 is unchanging.

 

It's right this time.

I am afraid not. This is the 4th/5th time you have tried so solve it, and you still couldn't get it right. Just shows you how important it is to understand what you are doing before doing it.

Δ.Ic = Ic(β + Δ.β) - Ic(β)

It just doesn't work that way.

My way of doing it covers all angles, ...
Back to the (concise and complete) analysis

It certainly doesn't cover all angles or isn't complete: if it were, it wouldn't be a model anymore. It would be the reality.

Models are useful only because they are incomplete and are wrong in that sense. They allow you to isolate the effect you are trying to understand without being confused by other effects that you don't care as much.

Take the impact of temperature for example. Its most significant mechanism would be through Q2's Vbe changes - my simplified approach shows that clearly. Q2 is a small signal transistor running at low current levels and dissipating very little heat. So its temperature changes are mostly due to ambient temperature changes. At -2mv/C, and a change of 15C (from 25C to 40C) would cause a change of 30mv, or 1ma on the leds. Check out the luminosity chart of any led and you will be hard pressed to find that kind of changes material, or even noticeable to naked eyes.

The same with beta changes: the biggest impact beta changes can have are from Q2, where the likely changes in beta are from Q1. If you box the issues through some reasonably assumptions, you will conclude that its effect is 2ndary, at the most.

As to a closed form solution for beta changes, it is going to be polynomial and you will not be able to solve for it analytically.

 

**** !

Yeah, no kidding.

 

An opto-coupler used in a critical feedback path of a switch mode power supply, for instance, would have to have its intensity precisely controlled.

Can you post a schematic that uses an optocoupler in this fashion? Optocouplers are more typically when high electrical noise is present or galvanic isolation is required.

Linear operation of optocouplers requires biasing the light emitter at a relatively high current to avoid non-linearities that are more significant at low bias currents; but high bias currents contribute to CTR degradation.

 

Can you post a schematic that uses an optocoupler in this fashion?

Almost every real world off-line SMPS used optocoupler and TL431 in the feedback loop
https://www.fairchildsemi.com/application-notes/an/an-4137.pdf

 

Here's the feed back circuit with optocoupler.

Here's the KA431 shunt regulator driving the LED.


Showing how critical it is to precisely control the LED intensity.