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Problem with darlington circuit

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aliraza54

New Member
darlington.png
Hi all,
I have ORCAD Capture 16.6. In the software i implemented a darlington circuit taken from the example of an electronics book. The picture of the circuit is attached.
The problem is that when i run the simulation i get double values of all the currents and voltages in the circuit.
Can anyone help me in this regard.
 

MikeMl

Well-Known Member
Most Helpful Member
Here is the circuit:

D15.png

Since the NPNs are both operated in their non-saturated condition, the voltages and currents in the circuit are very dependent on the β of the respective transistors. I show various node voltages V(b) and V(e), and the resistor currents I(R1) and I(R2).

This is not a practical circuit because there is no pull-down on the base of Q2. Note how long it takes the transistors to turn off. Also note that the base of Q2 is "floating" due to reverse collector-base leakage...
 

Tony Stewart

Well-Known Member
Most Helpful Member
Maybe you should change V+ to 9V.

rCE=Rb/(hFE1*hFE2). and Rb=R12 for hFe=100
rCE=3.3e6/1e4=330 Ohm

Using Ve as a voltage divider, Ve= R13/(rCE+R13)*(V+ - 2Vbe)

for 5uA=Ib, Vbe=0.55V

thus Ve=390/(330+390)*(18-1.1)=9.15V

if hFE=200, rCE= 330/4 and Ve = 13V.
 
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aliraza54

New Member
Hi tony,
I have changed the v+ to 9v but still dont get the values you mentioned. Even i have changed the hFE to 200 to 100 and still not getting the values you have mentined.
 

aliraza54

New Member
I am also facing a problem that when i change the value of hFE the output voltage does not change it remains the same no matter what value of hFE i set. Kindly help me in this is it the problem of ORCAD Pspice or problem in my circuit
darlington.png
 

Les Jones

Well-Known Member
Most Helpful Member
Can you post the values of voltage and current that you expect and the values you get from your initial simulation. Also what hfe values did you use in your initial test. I agree with the way Tony has done the calculation. (Although initially I forgot to take into account the two vbe drops.) This information may help us understand what the problem is.

Les.
 

JimB

Super Moderator
Most Helpful Member
To re-iterate and emphasise what MikeML said:

This is not a practical circuit because there is no pull-down on the base of Q2.
All the best books on how transistors work will start off with a (non-darlington) circuit similar to this and then proceed to tell you how bad it is.

The main problems are:
1 the gain variability of the transistors from one unit to the next.
2 the variation of gain and leakage current with temperature.

Building a darlington circuit like this just compounds the problems.
Maybe your odd calculation results are proof of this.

JimB
 

MikeMl

Well-Known Member
Most Helpful Member
I redid the simulation of post #2 to show the effect of changing the Bf (Beta forward) parameter in the 2n3904 transistor model that comes with LTSpice. I created a custom .model statement where I replaced Bf=300 with Bf={beta} so that I can use parameter substitution to vary {beta}. Note that I set both instances Q1 and Q2 to reference the modified model instead of the default 2n3904 model. I use the .step command to change {beta} from 50 to 300 in steps of 50 during multiple, overlaid simulation plots.

D15b.png

In the plot, the green trace is when both betas are 50. The progression from green to gray is 50, 100, 150, ... , 300. Note the great variability in the output voltage at V(e).

In the real world, it is unlikely that both 2n3904s would have exactly the same beta, so to be more rigorous in the simulation, we should really be doing an analysis that steps both betas independently, but I will leave that to the student...
 

Flyback

Well-Known Member
does your simulator (orcad) give more than 18V/390R in the 390R resistor?.....if it does, then it is a faulty simulator.
 

Colin

Active Member
The first thing we must know is the actual application.
Just stating this circuit is absolutely pointless.
 

Tony Stewart

Well-Known Member
Most Helpful Member
I am also facing a problem that when i change the value of hFE the output voltage does not change it remains the same no matter what value of hFE i set. Kindly help me in this is it the problem of ORCAD Pspice or problem in my circuit
View attachment 103312
The main problems I see with your query is , no details. No node voltages anywhere. no effort to follow my calculations, nada, no response to other people's help.
so. my friend, what's up?
 

aliraza54

New Member
Can you post the values of voltage and current that you expect and the values you get from your initial simulation. Also what hfe values did you use in your initial test. I agree with the way Tony has done the calculation. (Although initially I forgot to take into account the two vbe drops.) This information may help us understand what the problem is.

Les.
Hello Les Jones,
The values that i expect to come are:
Ib (base curremt)= 2.55uA, Ie (emitter current) = 20.48mA= Ic (collector current), Ve (emitter voltage)= 8.06 V, Vb (base voltage)= 9.65V and Vc (collector voltage)= 18V.
hFE is 443.3. The transistor model i am using in simulation is Q2N3904.
While the values am getting are:
Ib (base curremt)= 2.585uA, Ie (emitter current) = 42.54mA= Ic (collector current), Ve (emitter voltage)= 16V, Vb (base voltage)= 18V and Vc (collector voltage)= 18V.
hFE is 443.3.
 

aliraza54

New Member
The main problems I see with your query is , no details. No node voltages anywhere. no effort to follow my calculations, nada, no response to other people's help.
so. my friend, what's up?
Hello Tony,
The values that i expect to come are:
Ib (base curremt)= 2.55uA, Ie (emitter current) = 20.48mA= Ic (collector current), Ve (emitter voltage)= 8.06 V, Vb (base voltage)= 9.65V and Vc (collector voltage)= 18V.
hFE is 443.3. The transistor model i am using in simulation is Q2N3904.
While the values am getting are:
Ib (base curremt)= 2.585uA, Ie (emitter current) = 42.54mA= Ic (collector current), Ve (emitter voltage)= 16V, Vb (base voltage)= 18V and Vc (collector voltage)= 18V.
hFE is 443.3.
 

aliraza54

New Member
The main problems I see with your query is , no details. No node voltages anywhere. no effort to follow my calculations, nada, no response to other people's help.
so. my friend, what's up?
Dear Tony,
I have tried to follow your calculations but the problem is same that i am not getting the calculated voltages when i run the simulation.
 

Colin

Active Member
You are going about the problem entirely incorrectly.
You have to look at the arrangement and see the 3M3 is pulling the pair up.
To say there needs to be a resistor on the base of the lower transistor is absolutely absurd.
You have to start with an assumption.
Assume there is 1v across the 3M3.
This will produce 3.3uA into the base of the upper transistor.
The gain of the transistor is stated as 100 to 300.
Let us assume 200.
The emitter current will be 0.6mA
This is the current into the base of the lower transistor.
If the gain of the lower transistor is 200, the emitter current will be 120mA.
120mA though 390R result in more than the supply.
So, we have the situation that the current through the 3M3 is less than 3uA and both transistors will be turned ON.
The base voltage will be less than 1v lower than the supply, the emitter will be 0.6v lower and the emitter of the lower transistor will be 0.6v lower again.
That is all you can assume.
 

Les Jones

Well-Known Member
Most Helpful Member
The result I get is close to the simulation. This is how I went about it. We know the base current of the top right hand transistor is the emitter current of the bottom right transistor divided by 443.3 x 443.3 (The product of the assumed hfe value) 443.3x 443.3 = 196514.89. We can now think of the circuit in a different way by thinking of the combination of the 3.3M resistor and the darlington as a resistor and two diodes. (The top of the resistor connected to +18 v and the bottom connected to the anode of two diodes in series. The cathode of the lower diode connected to the top of the 390 ohm resistor. The value of this virtual resistor will be 3300000/196514.89 = 16.79 ohms We can now do a rough calculation of the current through the chain of resistors (Ignoring the voltage drop across the diodes.) so the current through the resistors will be 18/(390 + 16.79) = 18/406.79 = 44.2 mA So if we look at the data sheet for the transistor we see from the graphs the for a collector current of about 40 mA the base emitter voltage is about 0.75 volts. We now do the same for the top left transistor. We know it's collector current will be 44/443.3 mA = about 0.1 mA So looking at the graph again we see that for this collector current the base emitter voltage will be about 0.4 volts. Now we calculate the the current. The voltage across the two resistors will be 18 - (0.75 + 0.4) = 16.85
We now calculate the current 16.85/406.79 = 41.4 mA so the voltage at the top of the 390 ohm resistor will be 390 x 0.0414 = 16.146 volts
 

Colin

Active Member
Les Jones

Where do you get all the above RUBBISH from????

I have never read such GARBAGE.

Start to learn about how circuits work before launching into such stupidity.
How can a 390 ohm resistor become 16.85 ohms ??

It's irrelevant that you got the approximate value correct. It is the way you have gone about it that is totally incorrect.
The diode drops are the only values that come into the equation.
 

Les Jones

Well-Known Member
Most Helpful Member
Colin,
It is the combination of the 3.3 meg and the two transistors that behave like a 16.79 ohm resistor. This is because the voltage drop across this virtual resistor will be the same as the voltage drop across the 3.3 meg resistor but the current through it will be the product of the two hfe values times the current that would be passing through the 3.3 meg resistor.

Les.
 

MikeMl

Well-Known Member
Most Helpful Member
It requires an iterative solution (or solving a set of equations simultaneously) to find all of the node voltages and device currents.

Refer back to post #8. To solve for a consistent set of node voltages and branch currents with pencil and paper, you would have to guess a value of V(b), which determines Ib(Q2) =(18-V(b))/3.3meg, which makes Ib(Q1)=β2*Ib(q2), which makes V(e) = β1*Ib(Q1)*390. Now, we know that V(b) is V(e) + two Vbe drops, so V(b)= V(e)+2*0.65= V(e)+1.3.

You will have to keep iterating around that loop until you find a voltage at V(e) that is consistent with the steps above. I cannot see how you can do this without an iterative solution, or using an assumption as to what the Vbe of the two transistors is and then setting up a solution of two equations in two unknowns (V(b) and V(e)).

That is what the simulator does intrinsically. It effectively iterates the expression(s) above until it reaches a consistent set of branch currents and node voltages, all based on the transistor models. In post #8, I showed how V(e), and by implication V(b), is effected as β1 = β2 ranges from 50 to 300.

Here, I repeat the simulation asking LTSpice to do just the DC (.op) solution as a function of β (the x-axis). I plot V(b), V(e), the difference V(b)-V(e), and the Q2 base current Ib(Q2), all vs β of both transistors.

Les Jones and aliraza54, compare your pencil and paper solutions to these plots:

D15c.png
 
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