How do you solve this 2 BJT circuit ?

Thread Starter

DarthVolta

Joined Jan 27, 2015
456
I'm trying to work through an audio amp circuit, and I found a BJT current source driving 3 resistors with another BJT across them, like to limit current.

Since Vee is -10V, Ve is known and hence Vb, but I can't get Vc right. I'm not doing anything different than normal.

I'm getting Vc=2.10V and Ib=114nA from this, imagine its a matrix eq for Vc and Ib (dam I hate these font's where I looks like 1 so much). The transistor in the real circuit is not fully on. I tried solving with Vbe=0.6V and get Vc=1.26V and Ib=1.34uA

2bjt eq.PNG




2bjt.PNG


2bjt 2.PNG
 

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ci139

Joined Jul 11, 2016
1,470
Q10 works as CV(/zener) for it's Vb & Ve against the Vc of the Q9 - so the more the Q9 conducts the higher the voltage drop on R17 & R7 (with the proportion about 1:50) . . . the type of "active zener" is not too constant over the entire current range (the fet version vould be better but also a lot more sensitive , considering the width of it's linear region also keeping in mind it's non-linearity over the full width of it's linear region)

if the R16 , R17 were swapped the ↑above↑ would apply better — with the shown values the Q10 acts more as the stabistor or the zener at it's breakdown region . . . e.g. — as a dynamic resistance with negative voltage dependency component . . . well - maybe i'm stupid . . . at least the transistor acts as described → http://tinyurl.com/taa6tdt
 
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Thread Starter

DarthVolta

Joined Jan 27, 2015
456
If I did the 3 node eqn's right for Vc Vb Ve , am I supposed to get a real answer? Sometimes IDK if I'm making the mistake, or if the model is wrong, or the calculator/matrix solver

So does a diode Id eqn work for whats above Vb ?
 

Thread Starter

DarthVolta

Joined Jan 27, 2015
456
Ok and so I'm not doing school or building something, so the actual value's not important, but using KVL/KCL and the model in the 1st pic, is that solvable, or is there's something else about the BJT sharing current with a resistor I'm missing ? I don't know linear algebra well enough, but I reduced it to 2 eqn's and 2 unknown's,...so it's solvable right ?

Yeah and I need to learn better transistor models too thanks ci139
 
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Thread Starter

DarthVolta

Joined Jan 27, 2015
456
I'm worried physics and math are collapsing, which usually means I'm at the limit's of my knowledge or making a mistake, which is it ?

When I try to search about DC biasing of BJT circuits, pretty much all I get is for 1 transistor, on google or utube. I have a good 1st/2nd yr university EE textbook on the way, and another on semi-conductors and circuits. I hope the explain it (assuming my math was right in the 1st place...I own my own check mark pen, and no one to ask in the real world)
 

ci139

Joined Jul 11, 2016
1,470
there is also the "self heating effect of the transistor" due the current through it . . .

+ the 2N3904 (→PDF) and 2N3906 (→PDF) do not have a linear ß throughout the Ic range (unlike the 2SC1815 2SA1015)

i usually don't do any ... (complete) calculations unless there is no any other options left . . . (↑ is the very cause ↑ - you need to adjust it to your real circuit by testing it out anyway) . . . and i don't have enough time and/or patience and/or need

you can replace your dynamic collector load with some 2 terminal ((math.-cal)model )transfer-function . . . "solve" for PNP -- shouldn't be that hard for DC assumed by your circuit . . . there's some formulas
 
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Thread Starter

DarthVolta

Joined Jan 27, 2015
456
Hang on tho what about just the model in the 1st pic, with the simple BJT model, my answer still doesn't match LTspice so I'll have to try again, to re-check myself (the equations), but just the circuit with those elements, I still waiting for someone to say I'm missing something about that.
 

Jony130

Joined Feb 17, 2009
5,170
The equation for the Vbe multiplier output Vo will look like this:

untitled.PNG


Vo = Vbe*(1+ R1/R2 * β/(β +1) )+ Ic*R1/β

In your case we have:

Vo = 0.7V * (1 + 2kΩ/1kΩ * 300/301) + 1.638mA*2kΩ/300 = 2.11V

Ve4 = 1.638mA * 5kΩ - 10V = -1.81V
Vc3 = Ve4 + Vo = -1.81V + 2.11V = 0.3V

Is this what you want?
 

Thread Starter

DarthVolta

Joined Jan 27, 2015
456
Right so I made a mistake brute forcing it then

I missed getting the current across Vbe was easy too.
 
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ci139

Joined Jul 11, 2016
1,470
? the first figure both the Ic3 & B4 must be passive -e.g.- the current limiters (if they are NOT) ← as much as i am aware the B4 must be arbitrary current source (for you to be able to type in the formula) . . . the correct way would be using a resistor instead (the B4) where on the resistance field you write R=Rce() and define it as (a resistance value in) spice directive .func Rce(){ V( Vc3 , Ve4 ) / ( uramp( I(Vbe2) ) + ( 1 – u( I( Vbe2 ) ) ) * 1E-8 ) * beta() } , where the beta() is either .func beta(){Const} or beta(){function accounting the ***T.amb + ***etc.} a definition of the least requires including the BV sources that "integrate" required data for ***

PS! the Rce() does not account the case where V(Vc3,Ve4) is less than saturated voltage drop (? 380mV) on the CE junction of the 2N3904 -- it requires some lab to define more correct Rce for -- first improvment is Rce(){ if( u( V( Vc3 , Ve4 ) – .38 ) , V( Vc3 , Ve4 ) / ( uramp( I(Vbe2) ) + ( 1 – u( I( Vbe2 ) ) ) * 1E-8 ) * beta() , fxcp( V ( Vc3 , Ve4 ) ) ) } where fxcp() is .func fxcp(v){ ( ( 1 – uramp(v) / .38 )**5 + uramp(v) ) / ( uramp( I(Vbe2) ) + ( 1 – u( I( Vbe2 ) ) ) * 1E-8 ) * beta() } ←← as you study this sentence and the text above -- then it's clearly visible that there is a high simplification to the model for your actual circuit !!! . . . also such "math" may dirve the Spice Engine nuts . . . + the formula fxcp(v) used by the improved formula has a bug . . . as the Rce should go to infinity for both Vce→0 and Ib→0 independently (for the case when B and C are fed from different voltage sources . . . and we don't want to introduce the BC forward diode to the model) . . . not easy to grasp it out of the thin air for desired 100MΩ OFF resistance ← the least is required to make things easier for the solver

(Fixed some errors in the previous v. of #11)
 
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