Hi:
I am trying to understand the current flows in the following circuit that I ran in LT Spice.





I need to understand how you would determine the current split at both point A and point B if you did not have LT Spice
to do it for you. I keep coming up against conflicting information depending on the source. It makes my head hurt.

Thanks,

David

 

This circuit cannot be algebraically solved. We need to know the BJT's Is current.
For example if I assumed that the Is1 = Is2 = 1E-14A and β1 = β2 = ∞ and Vbe = 0.65V at 1mA.

We can find Ic2 this way

Ic1 ≈ (12V - 0.65V)/2.8kΩ = 4.05mA and solve for the new Vbe1 value

Vbe1 = Vt * ln(Ic1/Is1) = 26mV*(4.05mA/1E-14A) = 0.694V

So the new Ic1 current is

Ic1 ≈ (12V - 0.694V)/2.8kΩ = 4.04mA. Not a big difference so I stop the iteration process.

Now the Q2 current:

VR2 = Vbe1 - Vbe2 = Vt*ln(Ic1/Ic2)

Therefore

Ic2 = VR2/R2 = Vt/R2*ln(Ic1/Ic2)

Again we are forced to use the iteration techniques.

I start the iteration process by assuming that Ic2 = 3mA

Ic2 = Vt/R2*ln(Ic1/Ic2)

Ic2 = 26mV/10Ω*ln(4.04mA/3mA) = 0.774mA (1)

And the new value will be ( (3 + 0.774)/2 = 1.887mA )

2.6mA *ln(4.04mA/1.887mA) = 1.979mA (2)

(1.887 + 1.979)/2 = 1.933mA

2.6mA *ln(4.04mA/1.933mA) = 1.916mA (3)

(1.933 + 1.916)/2 = 1.924mA

2.6mA *ln(4.04mA/1.924mA) = 1.928mA (4)

(1.924 + 1.928)/2 = 1.926mA

2.6mA *ln(4.04mA/1.926mA) = 1.926mA (5)

So we are done the answer is Ic2 = 1.926mA

Of Course, we would include the β effect into the equation and solve it again by using iteration. And use the spreadsheet to do the iteration for us.

https://en.wikipedia.org/wiki/Widlar_current_source#Finding_the_current_with_given_resistor_values

But this calculations not include the Early effect.

But no one does this type of calculation when designing the audio amplifier mainly due to transistor mismatches (differences in Is current, β, Vce, temperature e.t.c), junction temperature has also a big effect. And normally we set the output stage bias current after we build the amplifier and wait until the temperature to stabilizes. We then correct the setting and wait again and so on.

 

Hi:
I am trying to understand the current flows in the following circuit that I ran in LT Spice.

View attachment 168277



I need to understand how you would determine the current split at both point A and point B if you did not have LT Spice
to do it for you. I keep coming up against conflicting information depending on the source. It makes my head hurt.

Thanks,

David

Thanks Jony130

I guess this process works for small-signal only. If I look at a datasheet for say
the 2N3904, how would I determine the range in which this process would be
valid? If I use current values much higher than Io = 2 mA and Iref =4 mA, then
the sim results don't agree with my calculations. I think people used to use an
IV curve but datasheets don't give them.

David

 

I guess this process works for small-signal only.

No, it works all the time.

If I use current values much higher than Io = 2 mA and Iref =4 mA, then
the sim results don't agree with my calculations.

Can you show it to me?

 

Here is my spreadsheet with Io = 2 mA and Iref = 4 mA.

 

I where is the problem then?

 

I where is the problem then?

If I choose Io = 10 mA and Iref = 25 mA, R1 s/b 451Ω and RE = 2.23KΩ per the
spreadsheet. When run in LT Spice, Io = 73μA and Iref = 25mA.

 

Something is not right with your spreadsheet. Because Io*Re = 10mA *2.23kΩ = 22.3V What ??

Let me check your calculations

Vbe1 = Vt * ln(Ic1/Is1) = 25.85mV *ln(25mA/1E-14A) ≈ 0.738V

R1 = (12V - 0.738V)/25mA ≈ 450Ω

and

RE = ( 25.85mV *ln(25mA/10mA))/10mA ≈ 2.3Ω

 

Something is not right with your spreadsheet. Because Io*Re = 10mA *2.23kΩ = 22.3V What ??

Let me check your calculations

Vbe1 = Vt * ln(Ic1/Is1) = 25.85mV *ln(25mA/1E-14A) ≈ 0.738V

R1 = (12V - 0.738V)/25mA ≈ 450Ω

and

RE = ( 25.85mV *ln(25mA/10mA))/10mA ≈ 2.3Ω

 

My spreadsheet did give me RE = 2.3Ω, I just entered the wrong value into the sim.
Even at that, the sim gives me Iref = 25 mA and Io = 11.83 mA, not 10 mA.

Is the objective here one of trying different values of Iref until you arrive at the
desired Io? And you just use whatever value of Iref is required in order to get
to that desired Io? Although my Iref = 4 mA and Io = 2 mA worked out without
doing the iterative process.

What is puzzling, is that I went into LT Spice left R1 = 450Ω but changed RE =
3.2Ω. Now Iref = 25mA and Io = 10mA as desired. But why would I have to
change Re from the calculated value of 2.3Ω to 3.2Ω?

 

Why I always have to remind you that the transistor is a highly nonlinear device.
And for our hand calculation, we are using a very simplified version of a transistor model ( the ideal transistor). And this model does not include a lot of "nonidealities" that LTspice includes.
For example, LTspice does the calculations for T = 27°C hence the VT ≈ 25.86492mV, the BJT's beta is not constant in LTspice but will vary with temperature, collector current, and Vce voltage.

Also, did you include the beta effect in your calculations?

If we want to do the calculations for the "Ideal" Transistor with this parameters Is = 1E-14; β = 100; and VT ≈ 25.86492mV (27°C)
And Io = 10 mA and Iref = 25 mA.

You should start like this:


IB2 = 10mA/100 = 100μA
Ie2 = IB2 + Ic2 = 10.1mA

Ic1 = (β1/β1 + 1) * (Iref - IB2) = 100/101*(25mA - 0.1mA) = 24.6534653mA

IB1 = Ic1/β1 = 246.534653μA

Vbe1 = 25.86492mV *ln(24.6534653mA/1E-14A) = 0.738012905V ≈ 738.013mV

R1 = (12V - 738.013mV)/25mA ≈ 450.4795Ω

VRE = 25.86492mV*ln(24.6534653mA/10mA) = 23.3387548mV

Re = 23.3387548mV/10.1mA ≈ 2.311Ω

Look at the simulation results for the ideal transistor



Any questions?

 

Why I always have to remind you that the transistor is a highly nonlinear device.
And for our hand calculation, we are using a very simplified version of a transistor model ( the ideal transistor). And this model does not include a lot of "nonidealities" that LTspice includes.
For example, LTspice does the calculations for T = 27°C hence the VT ≈ 25.86492mV, the BJT's beta is not constant in LTspice but will vary with temperature, collector current, and Vce voltage.

Also, did you include the beta effect in your calculations?

If we want to do the calculations for the "Ideal" Transistor with this parameters Is = 1E-14; β = 100; and VT ≈ 25.86492mV (27°C)
And Io = 10 mA and Iref = 25 mA.

You should start like this:


IB2 = 10mA/100 = 100μA
Ie2 = IB2 + Ic2 = 10.1mA

Ic1 = (β1/β1 + 1) * (Iref - IB2) = 100/101*(25mA - 0.1mA) = 24.6534653mA

IB1 = Ic1/β1 = 246.534653μA

Vbe1 = 25.86492mV *ln(24.6534653mA/1E-14A) = 0.738012905V ≈ 738.013mV

R1 = (12V - 738.013mV)/25mA ≈ 450.4795Ω

VRE = 25.86492mV*ln(24.6534653mA/10mA) = 23.3387548mV

Re = 23.3387548mV/10.1mA ≈ 2.311Ω

Look at the simulation results for the ideal transistor

View attachment 168372

Any questions?

Hi Jony:

One question about the calculation for Vbe1.

Vbe1 = 25.86492mV *ln(24.6534653mA/1E-14A) = 0.738012905V ≈ 738.013mV

This result comes to 916.70 mV, not 738.04 mV. LTSpice says 738.013mV. Something
isn't right.

David