Differential Amplifier find currents, and Gain

[The Electrician]

Ok so I am going to put it on here and I am going to upload it later tonight. Ib1=V1-V2/Rb+rpi1, Ib2=V5-V4/Rb+rpi2, Ie1 = V2-V3/ro1, Ie1+Ie2=Ic3, Ie2 = V4-V3/ro2, gm6Vbe6 = V7-V6/ro6

You need to be very careful with your equations. When you type them in text form as you have done here, you need to use parentheses where there could be ambiguity. For example, you have typed:

Ib1=V1-V2/Rb+rpi1, which is the same as:

$$Ib1 =V1 - \frac{V2}{Rb}+ rpi1$$

which is not what you really want. What you really want is:

$$Ib1 =\frac{V1-V2}{Rb+ rpi1}$$

In plain text form you would need to use some parentheses to get a correct representation, like this:

Ib1=(V1-V2)/(Rb+rpi1)

Don't use this for equation 1; I'm going to change the nodes below.

You can use LaTeX on this forum to get the pretty expressions as I've done above. Here's an FAQ that explains how to do it:

https://www.physicsforums.com/showpost.php?p=3977517&postcount=3

Yes I guess so man I don't think my buddies really get this problem either they are just pulling formulas from the book I think I would be better off letting this part go and just going with what Jony and you provided. That being said how would I start on part d ultimatley that was where I was trying to get the total gain?

Parts b and d can both be answered by using the model you have so far. You only need to set up the 7 nodal equations and solve them. Sounds easy, right?

Please use lower case v for the node designators; for small signal analysis, lower case letters are conventionally used for variables.

We will get easier results if two of the nodes are changed slightly. Let's use these node designators:

v1. Top of rpi1
v2. Top of ro1
v3. Top of 80k resistor
v4. Top of ro2
v5. Top of rpi2
v6. Emitter of Q6
v7. Collector of Q6 (your output node)

This will change your equation 1. Here's how we derive it.

The voltage at each node is given by the relevant designator; at node 2, the voltage is v2. Currents leaving a node are considered positive and those entering a node are negative.

There are two paths for current at node 1: the input current in RB1 and the current in rpi1.

1. The current in RB1 is just the voltage across RB1, which is (v1-V1), divided by the resistance RB1: (v1-V1)/RB1. (Here's why you need to use lower case v for the node voltages. v1 is the voltage at the top of rpi1 and V1 is the input voltage). This current is leaving node 1 because we chose to define the voltage as (v1-V1), not (V1-v1); see the explanation at the end of this post for more info about this.

2. The current in rpi1 is the voltage across rpi1, which is (v1-v3), divided by the value of rpi1: (v1-v3)/rpi1

Equation 1 simply sets the sum of these two currents equal to zero:

Eq1: (v1-V1)/RB1 + (v1-v3)/rpi1 = 0



Let's derive equation 2.

We have 3 currents leaving node 2 (If any are actually entering, the algebra will give a negative sign for them; that is taken care of by the order of the subtraction of voltages in the numerator expression for each current).

1. The current in the collector resistor of Q1 (which you should have labeled Rc1, not just Rc) is equal to the voltage across it, which is (v2-0), (remember that V+ is signal ground) divided by the resistance itself: (v2-0)/Rc1 which is just v2/Rc1. That is a current leaving node 2.

2. The current in the resistor ro1 is the voltage across ro1, which is (v2-v3), divided by the resistance of ro1: (v2-v3)/ro1

3. The current supplied by the gm1 source; that current is gm1*vbe1. But, notice that Vbe1 is just (v1-v3), so this current is given by gm1*(v1-v3)

Equation 2 is just the sum of these 3 currents set equal to zero:

Eq2. v2/Rc1 + (v2-v3)/ro1 + gm1*(v1-v3) = 0

You should be able to see how this works. You identify all the paths for current to leave or enter a node; those paths are often just some impedance, although they are sometimes sources such as the gm sources. If the path is an impedance, write an expression for the numerator of a fraction, that expression being the voltage across the impedance in a form like this (va-vb). The va is the voltage at the node being worked on, and vb is the voltage at the other end of the impedance; by ordering the voltages like this: (va-vb) instead of (vb-va) the current is calculated as if leaving the node and is therefore a positive current. The denominator is just the impedance under consideration.

If the path is a controlled source such as your gm sources, write an expression for the current supplied by the source, involving the voltage(s) at some other nodes, where those nodes are the ones whose voltages control the source.

Each current is in the form of a fraction (except possibly for the currents from controlled sources) and all the currents are summed to zero to give you an equation.

Derive the next 5 equations and post them, each on a separate line, not all on one line separated by commas!

You will need to solve 7 simultaneous equations; how will you do that? Do you have access to Matlab, MathCad, or some similar software?

 

[DODGEVIPER13]

[itex]\frac{v1-V1}{R_B1} + \frac{v1-v3}{r_pi1}=0[/itex]

[itex]\frac{v2-0}{R_c1} + \frac{v2-v3}{r_o1} + gm1*(v1-v3)=0[/itex]

$gm1*(v1-v3) + gm2*(v5-v3) + \frac{v3}{80}=0$

[itex]\frac{v4}{R_c2} + \frac{v4-v3}{r_o2} + gm2*(v5-v3)=0[/itex]

[itex]\frac{v5-V2}{R_B2} + \frac{v5-v3}{r_pi2}=0[/itex]

[itex]gm6*(v7-v6) + \frac{v4-v6}{r_pi6} + \frac{v6-v7}{r_o6} + \frac{v6}{R_E}=0[/itex]

[itex]\frac{v7-v6}{r_o6} + \frac{0-v7}{r_c6} + gm6*(v7-v6)=0[/itex]

 

[DODGEVIPER13]

Well I tried LAtex and I don't think it worked but on line 3. Well anyways if you can see it then I believe I may have made mistakes in 6 or 7?

 

[The Electrician]

[itex]\frac{v1-V1}{R_B1} + \frac{v1-v3}{r_pi1}=0[/itex]

[itex]\frac{v2-0}{R_c1} + \frac{v2-v3}{r_o1} + gm1*(v1-v3)=0[/itex]

$gm1*(v1-v3) + gm2*(v5-v3) + \frac{v3}{80}=0$

[itex]\frac{v4}{R_c2} + \frac{v4-v3}{r_o2} + gm2*(v5-v3)=0[/itex]

[itex]\frac{v5-V2}{R_B2} + \frac{v5-v3}{r_pi2}=0[/itex]

[itex]gm6*(v7-v6) + \frac{v4-v6}{r_pi6} + \frac{v6-v7}{r_o6} + \frac{v6}{R_E}=0[/itex]

[itex]\frac{v7-v6}{r_o6} + \frac{0-v7}{r_c6} + gm6*(v7-v6)=0[/itex]

OK, let me see if I can figure out what's wrong with LaTex. Well, I didn't do anything and it works for me.

You have a few errors:

EQ1. $\frac{v1-V1}{RB1} + \frac{v1-v3}{rpi1}=0$

EQ2. $\frac{v2-0}{Rc1} + \frac{v2-v3}{ro1} + gm1*(v1-v3)=0$

In EQ3, the gm sources should have a minus sign because their current flows
INTO node 3, but out of nodes 2 and 4. The resistor from node 3 to
ground should be 80,000 not 80.
EQ3. $-gm1∗(v1−v3)-gm2∗(v5−v3)+\frac{v3}{80000}=0$

In EQ4, you forgot the current leaving that node into rpi6.
EQ4. $\frac{v4}{Rc2} + \frac{v4-v3}{ro2}+\frac{v4-v6}{rpi6} + gm2*(v5-v3)=0$

EQ5. $\frac{v5-V2}{RB2} + \frac{v5-v3}{rpi2}=0$

In EQ6, gm6 is negative and the voltage controlling it is (v4-v6). The term (v4-v6)/rpi6 should
be reversed; remember the first variable in the numerator corresponds to the node you're working on (but not for the gm source).
EQ6. $-gm6*(v4-v6) + \frac{v6-v4}{rpi6} + \frac{v6-v7}{ro6} + \frac{v6}{RE}=0$

In EQ7, the (0-v7) term should be reversed. The controlling voltage for gm6 is (v4-v6), not
(v7-v6). Use capital R for Rc6.
EQ7. $\frac{v7-v6}{ro6} + \frac{v7-0}{Rc6} + gm6*(v4-v6)=0$


You're in the home stretch now. Substitute numeric values for all your parameters and solve the 7 equations.

Here are the values I used:

We want the differential gain, so set V1 to 1/2 and V2 to -1/2; that way the applied differential voltage is 1 volt, and the voltage at node 4 will be the gain you need for part b of your problem. The voltage at node 7, v7, is the gain you need for part d.

Post your results so I can check for errors.

 

[DODGEVIPER13]

Ha well I started doing this by hand it is a nightmare I so far have $V2=898.364V1-790.88$
$V3=2.627V1-1.313$
$V1=0.6172+0.2349V5$

Maybe I should use Matlab but how would I even do that I entered al the variables in but since v1 would be undefined that would confuse me?

 

[DODGEVIPER13]

Oh wait I would need to set up individual functions wouldn't I

 

[The Electrician]

v1,v2,v3,v4,v5,v6,v7 are all unknowns. They are the voltages which exist at the 7 nodes.

There's no way you will be able to solve 7 simultaneous equations by hand without making mistakes (unless you have a great deal of experience doing so)!

Using "solving multiple equations in Matlab", I get a lot of hits, such as:

http://www.ehow.com/how_8718288_solve-simultaneous-equations-using-matlab.html

http://www.mathworks.com/help/symbolic/solve.html

 

[The Electrician]

Don't fail to set all your parameters to their numeric values before you try to execute a multiple equations solver. Do it by including some assignment statements like this before running the solver:

 

[The Electrician]

Just to help you know when you've got it right, I'll tell you the first two voltages:

v1=.362107
v2=-122.569

 

[DODGEVIPER13]

ugggg MATLAB says Undefined function or method 'solve for input arguments of type char

>> V1=0.5;
>> V2=-0.5;
>> RB1=2000;
>> RB2=2000;
>> Rc1=20000;
>> Rc2=20000;
>> gm1=.01904;
>> gm2=gm1;
>> gm6=.0308957;
>> ro1=160000;
>> ro2=160000;
>> rpi1=5252;
>> rpi2=5252;
>> ro6=98605;
>> rpi6=3236.7;
>> RE=5300;
>> Rc6=6000;
>> [v1,v2,v3,v4,v5,v6,v7]=solve('((v1-V1)/(RB1))+((v1-v3)/(rpi1))=0','((v2)/(Rc1))+((v2-v3)/(ro1))+gm1*(v1-v3)=0','-gm1*(v1-v3)-gm2*(v5-v3)+((v3)/(80000)) = 0','(((v4)/(Rc2))+((v4-v3)/(ro2))+((v4-v6)/(rpi6))+gm2*(v5-v3)=0','((v5-V2)/(RB2))+((v5-v3)/(rpi2))=0','-gm6*(v4-v6)+((v6-v4)/(rpi6))+((v6-v7)/(ro6))+((v6)/(RE))=0','((v7-v6)/(ro6))+((v7)/(Rc6))+gm6*(v4-v6)=0')

 

[DODGEVIPER13]

hmmmm it seems solve is undefined in my version of MATLAB it seems will wolfram work?

 

[The Electrician]

A quick look at your Matlab solve statement reveals, for example, unbalanced parentheses in equation 4:

(((v4)/(Rc2))+((v4-v3)/(ro2))+((v4-v6)/(rpi6))+gm2*(v5-v3)=0

You've got parentheses in there you don't need. You could write it like this:

(v4)/(Rc2)+(v4-v3)/(ro2)+(v4-v6)/(rpi6)+gm2*(v5-v3)=0

There may be more, but look over everything VERY carefully. That sort of thing is probably your problem.

 

[The Electrician]

hmmmm it seems solve is undefined in my version of MATLAB it seems will wolfram work?

If you fix the problem I mentioned above and it still won't work (which I don't see why it shouldn't), then you could try another program.

Is Matlab something you found on your school's network?

What do you mean by Wolfram? There is a program called Mathematica made by Wolfram; it might be on your school's network. Then there's Wolfram Alpha, which is a web resource.

 

[DODGEVIPER13]

yah man I just tried it oh well no worries I will have to just go with what I got the test is tomorrow and its late.

 

[DODGEVIPER13]

Hey man thanks to you and all who helped. The way you explained it really helped me understand and I would have had it if I wouldn't have run into issues with matlab.