Now that we have IE2, we can go backwards and obtain an equation for VC1.

\(V_{out} - V_{C1} = 0.7, \quad V_{C1} = 2.3V\)

\(I_{C1} = \frac{V_{CC} - V_{C1}}{Rc} = \frac{6.7V}{Rc}\)

Is there another equation we can incoporate Ic and/or Rc so we can establish these two values?

There is something wrong with your equation

 

There is something wrong with your equation

I'm not seeing it, are the arrows in the picture with the voltages labeled beside them suppose to mean that polarity is in that direction i.e. + to -?

The current would be following down into the emitter of T2, and down into the collector of T1, I assume a + to - polarity for these currents.

What did I do wrong?

 

Arrow shows this
\(V_{Rc} = V_{be2}+V_{Re3} = 0.6V+ Ie2*Re3\)

 

Arrow shows this
\(V_{Rc} = V_{be2}+V_{Re3} = 0.6V+ Ie2*Re3\)

Isn't just the following,

\(V_{CC} - V_{C1} = 0.7 + V_{CC} - V_{out}\)

\(V_{C1} = V_{out} - 0.7\)

I do KVL in a clockwise fashion in that loop and it seems to give me the same equation? (In other words inverse the two green arrows on T2)

 

Yes you're right, my mistake.

 

Yes you're right, my mistake.

Okay so is there anyway I can relate Ic and/or Rc to another constraint, say the input resistance?

 

I can relate Ic and/or Rc to another constraint, say the input resistance?

Do you know the hfe (β) of the BJT's or maybe you know Hoe??.

\(R{in} = R_{B1}||R_{B2}||( (\beta+1)*(re1+Re1))\)

So if you want to choose Ic1 you need to know Hfe first.

 

Do you know the hfe (β) of the BJT's or maybe you know Hoe??.

\(R{in} = R_{B1}||R_{B2}||( (\beta+1)*(re1+Re1))\)

So if you want to choose Ic1 you need to know Hfe first.

We are told that,

NPN BJT; ß = 150; VCEmax = 40 V; ICmax = 200mA

and

PNP BJT; ß = 150; VCEmax = -40 V; ICmax = 200mA

So, ß = 150

I'm still confused as to how this allows me to choose an approriate Ic, can you explain?

 

For example form Rin of the BJT himself

\(R{in} = R_{B1}||R_{B2}||R_{inT}\)

\(R{inT} = ( (\beta+1)*(re1+Re1)) = 151 *(re + Re1)\)

And if i choose 40K for RinT, I leave 67K for RB1||RB2
Rin = 40K||67k = 25K

\( (re1+Re1) = \frac{40K\Omega}{151} = 270\Omega \)

Now Rc form the gain

Rc = 50 * 270 = 15KΩ

 

But now if we know Hfe = 150
We see from Rout < 50

\(R{out} = re + \frac{Rc}{\beta +1} || R{e3}\approx re \)

That Rc cannot be larger then 7.5K

 

But now if we know Hfe = 150
We see from Rout < 50

\(R{out} = re + \frac{Rc}{\beta +1} || R{e3}\approx re \)

That Rc cannot be larger then 7.5K

So is it safe to take Rc as 7.5k then?

 

So is it safe to take Rc as 7.5k then?

It's up to you, since β is quite small it is impossible to meet all the requirements.

 

So far I've been able to establish values of Vo, IE2, VC1 RC as follows:

Vcc = 9V

Vo = 3V

IE2 >= 520uA

VC1 = 2.3V

Rc <= 7.5k

Other constraint equations that have been developed are,

(re1 + Re1) = 270Ω

(RB1 // RB2) = 67k

What should I be working on next? Should I start picking values for resistances to meet this requirements?

Also, little re and re1, those resistances aren't actually values I have to allocate, correct? Aren't those resistances due to the T-model of the emitter junction?

What's next?

EDIT:

It's up to you, since β is quite small it is impossible to meet all the requirements.

Also, if we are within 10% of all the requirements they consider it a successfull design.

Other colleagues I've spoken to about this design have iterated a set of equations to determine voltage/current/resistance values on MATLAB, thus find a design that works. What set(s) of equations would I have to iterate through to do such a thing?

 

So far I've been able to establish values of Vo, IE2, VC1 RC as follows:

Vcc = 9V

Vo = 3V

IE2 >= 520uA

VC1 = 2.3V

Rc <= 7.5k

Other constraint equations that have been developed are,

(re1 + Re1) = 270Ω

(RB1 // RB2) = 67k

What should I be working on next? Should I start picking values for resistances to meet this requirements?

Yes you should start picking the values of the components

Also, little re and re1, those resistances aren't actually values I have to allocate, correct? Aren't those resistances due to the T-model of the emitter junction?

re it is small-signal resistance between emitter and base, looking into the emitter.

\( re = \frac{1}{gm} = \frac{26mV}{Ic}\)

Other colleagues I've spoken to about this design have iterated a set of equations to determine voltage/current/resistance values on MATLAB, thus find a design that works. What set(s) of equations would I have to iterate through to do such a thing?

I think that I show you all necessary equations.

 

Yes you should start picking the values of the components

I think that I show you all necessary equations.

Is there anything I should keep in mind while choosing component values? Depending on the value I assign RC, the value of IC will change and thus re will change with it.

So by choosing RC I define IC and as a result IC defines re1 and thus defines RE2.

What about RB1 and RB2, what should I be thinking about when I have to assign them resistance values? We know that their parallel combination must be equivalent to 67K so we can give them each 134K.

Now I need to think about the equations I can use to define RE1 and RE3.

We developed an equation for IE2 such that,

IE2 >= 520uA

From this we should be able to determine RE3 as follows,

\(\frac{V_{cc} - V_{o}}{I_{E2}} = R_{E3}\)

which will vary with the value of IE2.

The only piece I can't figure out is RE2. It might become more obvious once I actually chose some values.

Now the real question is how can I make a smart decision for defining such values?

 

I also attempted a simulation with some values, but I wasn't really able to draw any conclusions,

Here's what I had.

 

If you want 50Ω output impedance, the second transistor emitter current Ie2 cannot be as low as 520uA.

This will set re2 to 50 ohms, but you also have to allow for Rc/(β+1) adding into the impedance.

This suggests you might want to run something more like 1.5mA Ie2, giving re2=17.3Ω, leaving 32.7Ω "left over" to allow for Rc/(β+1).

Frankly, I think you ought to look for transistors with β more than 150. Say β minimum 250: something like these?

NPN: http://www.fairchildsemi.com/ds/2N/2N5210.pdf PNP: http://www.spelektroniikka.fi/kuvat/MMBT5087.pdf

If you had β=250, estimate an possible value for Rc: 32.7*(β+1) gives Rc = 8199Ω.

Your requirement for a gain of 50 suggests (RE1 + re1) = 8.2kΩ/50 = 164Ω.

Next choose T1 collector current (think about collector voltage, but also re1 < 164Ω), and set RE1 to make up the difference.

With β = 250 again, the first transistor input impedance will be about β times 164Ω = 41kΩ which is promising.

The bias chain RB1 and RB2 can be chosen to give Zin = 25kΩ total.

 

The circuit post #36 shows the 2n3906 PNP transistor upside-down. Its emitter is supposed to be positive.

 

I also attempted a simulation with some values, but I wasn't really able to draw any conclusions,

Here's what I had.

Q2 is upside-down. It is PNP, so collector to negative, emitter to positive please.

Your AC voltage gain is also very low because Av≈Rc/(re1+RE1) Why have you got RE1= 1kΩ: maximum gain will be 7.5

Also, your base pot divider voltage may be too high, unless the base current is dragging it down a lot the thing is in danger of saturation.

Edit - Sorry, bracket in wrong place - moved now.

 

In your simulation diagram you have an error. T2 is connect upside down.
As for RB1 and RB2 you should choose Ve equal to 1V ( 1V = Ic1 *(Re1+Re2) ) , so voltage on the base must be equal 1.6V.
And voltage divider formed by (RB1; RB2) must deliver 1.6V.
And i think that it is good to set Vc = 0.5Vcc