Petrucciowns,

You also asked "How is VRload calculated?"

This is the voltage across the 2k load resistor - right?

You know the overall gain is Av(system) = 451.4 according to your notes.

You know the source voltage Vs = 100uV.

So VRload = Av(system)xVs = 451.4 x 100 uV = 0.04514 V or 45.14 mV

Not sure what you are referring to when you asked "What is the current used?"

 

I appreciate both of your help so much, but there is still one thing that I am stuck on. How i the voltage across Rload calculated?

 

I appreciate both of your help so much, but there is still one thing that I am stuck on. How i the voltage across Rload calculated?

Did you see the preceeding post? Did it make sense to you?

 

Oh I'm sorry I must be blind. Thanks for all the help you're awesome, both of you.

 

How is VRload calculated. What is the current used?

I see tnk already answered this question while I was preparing this response, but here it is anyway!

VRload is calculated by multiplying the input voltage (100μV) by the system gain (451.4), giving 45.1 mV.

Regarding your question about numbers being a little off, consider the calculation of Zin.

re1+re11 = 93.9

If you multiply 93.9 * β you get 28170 for Zin(base1), but if you multiply by (1+β) you get 28263.9, a difference in the 3rd decimal place.

Then if you use those two numbers to calculate Zin||R1||R2 you get 11257.19 and 11272.158, respectively. If you later use some number that was rounded off, you accumulate even more error, but as a practical matter for design of circuits, it doesn't matter. You will have more error due to variations in transistor parameters, than will be caused by rounding to 3 or 4 digits.

In textbooks dealing with transistor circuit analysis, you will find graphs showing that the input impedance of a common emitter stage will vary with the collector load resistance. Yet, when we calculate Zin, our calculation only involves R1, R2, and (β+1)*(re1+re11). How then can the input impedance depend on the collector load? It's because we are ignoring a transistor parameter called the "reverse voltage transfer ratio". It's a small number, and it's important in high frequency circuits.

To get very accurate mathematical results (because what we're doing here is using mathematical models of transistor behavior), you must analyze the entire circuit at once, using the nodal method with many digit arithmetic.

Just to show you the effect of ignoring hre, the reverse voltage transfer ratio, the system gain calculated with 12 digit arithmetic using the nodal method on the entire circuit is Av(system) = 447.21755. But if we let hre be .0004, rather than zero, Av(system) = 455.07499. Notice the gain increased, because hre produces positive feedback in common emitter stages.

Without taking into account hre, Zin = 11272.1583, but if we let hre be .0004, then Zin = 11215.9378, essentially the same to 3 digit accuracy.

It was long ago realized that hre can be safely ignored in low frequency transistor circuit design.

 

A great explanation Electrician - thanks from me also.

 

Hi Petrucciowns,

I've had another look at the values you provided in your analysis and have decided you incorrectly calculated the emitter currents.

I think this is how you found the emitter current for stage 1....

1. You took the 10Volt plus 25k / 75k voltage divider to give a base voltage of (25/100)x10 = 2.5 Volts
2. You then found Ie as (2.5-0.7)/1868 = 963.6 uA = 964uA rounded up.

I suggest there is a more correct method.

Proceed as follows ...

1. Convert the same voltage divider to a Thevenin equivalent and find Vth=2.5V and Rth=18.75k
2. The Thevenin equivalent drives the base bias as follows
3. Vb=Vth-Rth*Ib & Vb=Vbe+(1+hfe)*Ib*1868
4. Solve for Ib assuming Vbe=0.7, giving Ib=4.566uA
5. Hence Ie=(1+hfe)*Ib=201*4.566uA=917.77uA

This equates to a 5% error in your value which may not be acceptable.

 

Hi again,

Sorry I used the wrong value for hfe (200 instead of 300).

The values should be

Ib = 3.098 uA
Ie = 932.5 uA

So the error isn't as bad as I though - 3.4%. In any case there is a difference in the method. Maybe Electricain has some thoughts on this as well.

 

My error in using the wrong value of hfe in my last posts raises an interesting point about the stability of the bias method used in relation to any variability in hfe.

If the hfe value changes from 200 to 300 (50%) the emitter current only changes by 1.6% - a great outcome for the designer in terms of predictable circuit behaviour.

 

hi everyone!
i looked out through your solutions and i was amazed ,hope i could add more but i can't find what is lacking!
but since we are dealing witn stages i decided to ask y'all how you will analyse a darlington circuit(two emitter follower in stage) and provide a strict proof of input resistance being greater than in simple emitter followers!
thanx

 

Hi again,

Sorry I used the wrong value for hfe (200 instead of 300).

The values should be

Ib = 3.098 uA
Ie = 932.5 uA

So the error isn't as bad as I though - 3.4%. In any case there is a difference in the method. Maybe Electricain has some thoughts on this as well.

I get exactly those numbers using a different but equivalent method.

For the second stage I get:

Ib2 = 10.09956 uA
Ie2 = 2.030012 mA

These values lead to different values for:

re1 = 26.81Ω
re2 = 12.315Ω

I noticed when this thread started that the Vb's and emitter currents had been calculated with the simple method that ignores the base current effect on the voltage divider, but I decided to go with the values given with the schematic so I could see if my other calculations agreed with the given values.

 

hi everyone!
i looked out through your solutions and i was amazed ,hope i could add more but i can't find what is lacking!
but since we are dealing witn stages i decided to ask y'all how you will analyse a darlington circuit(two emitter follower in stage) and provide a strict proof of input resistance being greater than in simple emitter followers!
thanx

Start a new thread with a schematic and typical values for the transistor parameters. If this is a homework problem, show your attempt at a solution. If it's not homework, post it in the "General Electronics Chat" forum.

 

I noticed when this thread started that the Vb's and emitter currents had been calculated with the simple method that ignores the base current effect on the voltage divider

because the equivalent DC resistance from the emitter resistors is "amplified" by the beta of that transistor. so the 1k resistance among re12 and re1 is effectively 100k or 200k for a beta of 100 or 200, respectively.

paralleling that kind of resistance onto R21 (22k) makes minimal impact.

 

tnk and I have quantified the error. Rather than just assert that it's minimal, we calculated just how much it is.

The error in Ie1 is 3.4% and in Ie2, it's 6.9%.

We provide the actual error, and leave it up to each person to decide for themselves whether it's minimal or not, given whatever their accuracy concerns may be.

 

tnk and I have quantified the error. Rather than just assert that it's minimal, we calculated just how much it is.

The error in Ie1 is 3.4% and in Ie2, it's 6.9%.

Wow, congratulations! you have just wasted some precious time of your own, .

really, you could have figured it out if you just thought it through.

 

I suggest there is a more correct method.

Proceed as follows ...

1. Convert the same voltage divider to a Thevenin equivalent and find Vth=2.5V and Rth=18.75k
2. The Thevenin equivalent drives the base bias as follows
3. Vb=Vth-Rth*Ib & Vb=Vbe+(1+hfe)*Ib*1868
4. Solve for Ib assuming Vbe=0.7, giving Ib=4.566uA
5. Hence Ie=(1+hfe)*Ib=201*4.566uA=917.77uA

This equates to a 5% error in your value which may not be acceptable.

I can give you a quicker approach.

all that transistor does here for DC purposes is to "amplify" the Re resistors and present them as a much larger resistor than they really are, by (1+beta) times.

so, all you do is to parallel (1+beta)*Re to the lower resistor in the divider and then calculate Vb that way.

by my calculation, it yields a Vb of 2.38v, vs. 2.41v for the first stage, calculated by my spreadsheet, for an error rate of 1.4%. Ie is then (2.41-0.7)/1868=.9ma.

the above calculation doesn't factor in the "equivalent" resistance from the b-e junction: 0.7v/0.9ma=778ohm (in terms of emitter current). so to do a 2nd order correction, you add the 778ohm to the Re resistance of 1868, you get the equivalent resistance of (1+200)*(778+1968)=531,846ohm, and Vb=2.4149v, vs. the real thing of 2.4142v, and an error rate of 0.028%.

all that, without killing yourself.

 

the ac input Z calculation is very similar to the dc input Z calculation, with the exception that under high enough frequency, Re1 is shorted and Re2 is shorted for ac purposes.

so your input impedance will just be the divider resistors + the beta amplified Re resistors (for ac purposes), all in parallel.

 

My error in using the wrong value of hfe in my last posts raises an interesting point about the stability of the bias method used in relation to any variability in hfe.

If the hfe value changes from 200 to 300 (50%) the emitter current only changes by 1.6% - a great outcome for the designer in terms of predictable circuit behaviour.

that's only because the emitter current is for the most part determined by Vb. as long as the hFE is substantially large (>50) and the divider network resistance is sufficiently small, the base current will be too small to have a substantial impact on Vb.

that forms the basis to ignore Ib and emitter resistors in the determination of Vb.

now, if you are biasing a low hFE transistor for high current output (2n3055 in a SE power amp), you will see how critical it is to include Re in the calculation of bias point.

it shows why the inclusion of Re in the calculation of Vb serves no practical purposes in this exercise.

 

that's only because the emitter current is for the most part determined by Vb. as long as the hFE is substantially large (>50) and the divider network resistance is sufficiently small, the base current will be too small to have a substantial impact on Vb.

that forms the basis to ignore Ib and emitter resistors in the determination of Vb.

now, if you are biasing a low hFE transistor for high current output (2n3055 in a SE power amp), you will see how critical it is to include Re in the calculation of bias point.

it shows why the inclusion of Re in the calculation of Vb serves no practical purposes in this exercise.

I think the bypassed emitter resistor cannot be ignored in it's role in the transistor amplifier bias stability. It essentially provides the negative feedback to offset the effects of variability in transistor parameters. As I was trying to imply, the aim of the design is to provide some certainty for the designer in the predicting the overall performance - which is a positive outcome.

 

I think the bypassed emitter resistor cannot be ignored in it's role in the transistor amplifier bias stability.

that they do, bypassed by a capacitor or not.

it is to the determination of Vb that those resistors have practically no impact.