am my steps about my question solution about bjt correct?

[circuit975]

Hello there
I am faced with such a question


1) Calculate and determine the value of the resistor RE that will ensure that VC=6 V.

2) Redraw the transistor model of the equivalent circuit for small amplitude AC signals. Determine the voltage gain Av of the circuit.

I solved the question as follows ;

VB = [24K / 24K+82K] x 16V = 3,62 Volt

RB = 24Kx82K / 24k+82k = 18,56K

6 = 16 - ICx5,6k
Ic = 1,78mA

IC = ßxIb = 150xIb
Ib = 11,86 microamper

IE = 11,86x10^-6 x 151 = 1,79mA
re = 26mV / 1,79mA = 14,5
VB - VBE - IBxRB - IExRE = 0
3,62 - 0,7 - [ (11,86 x 10^6) x 18,56K ] - 1,79mA x RE = 0
RE = 402K

Av = -ß x Rc / ßre + (ß+1)RE
Av = -150 x 5600 / 2,175 + 0 (The reason why it is 0 is because the Re resistor and capacitor are in parallel, I know that it is an ineffective element in AC analysis.)

This is my solution, are my solution steps correct? or if there is a point where I made a mistake when converting units, can you warn me, I need to correct my shortcomings and learn the right one. thanks a lot in advance

 

[Diver300]

RE will be less than 5.6 kOhms, as the voltage is less than the collector resistance and the current is slightly more, so you've got something wrong.

As a first estimate, ignore the base current. The emitter voltage will be 3.62 - 0.7 = 2.92 V. The emitter current will be 1.78 mA, the same as the collector current, so the emitter resistor needs to be 2.92 / .00178 = 1.6 kOhm

If you want to be more accurate, you can add in the base current to the emitter current, making it 1.792 mA. Also the base current will reduce the base voltage by 18560 * 0.00001186 = 0.22 V, so the emitter voltage becomes 2.70 V and the emitter resistor is 1.5 kOhm

However, the base-emitter voltage of 0.7 V will depend on temperature, so adjustments of 0.2 V or so probably don't matter on a real circuit.

 

[circuit975]

RE will be less than 5.6 kOhms, as the voltage is less than the collector resistance and the current is slightly more, so you've got something wrong.

As a first estimate, ignore the base current. The emitter voltage will be 3.62 - 0.7 = 2.92 V. The emitter current will be 1.78 mA, the same as the collector current, so the emitter resistor needs to be 2.92 / .00178 = 1.6 kOhm

If you want to be more accurate, you can add in the base current to the emitter current, making it 1.792 mA. Also the base current will reduce the base voltage by 18560 * 0.00001186 = 0.22 V, so the emitter voltage becomes 2.70 V and the emitter resistor is 1.5 kOhm

However, the base-emitter voltage of 0.7 V will depend on temperature, so adjustments of 0.2 V or so probably don't matter on a real circuit.

2,92 / 0,00178 = 1,6k

How right is it to do it directly like this?

Because according to this equation; VB - VBE - IBxRB - IExRE = 0

It is as if we made a transaction without subtracting IBxRB. isn't this wrong?

 

[Diver300]

2,92 / 0,00178 = 1,6k

How right is it to do it directly like this?

Because according to this equation; VB - VBE - IBxRB - IExRE = 0

It is as if we made a transaction without subtracting IBxRB. isn't this wrong?

The first estimate is done in a more simple way by ignoring the base current, so it's somewhat less accurate, but it is more simple and it gets an answer that is close to what will happen in reality. It is always good to have a simple estimate, just as a reality check. It was clear to me that 402 kOhm for RE was very wrong before I did any calculations. The more complications you add, the more possible places to make a huge error. If you have a simple and reliable estimate, it is a good basis and you would expect your more complicated and more accurate estimate to be similar.

In you equation
VB - VBE - IBxRB - IExRE = 0

VB is the base voltage before the base current is taken into account.
VBE is the base-emitter voltage drop
IB is the base current
RB is the effective base resistance
IE is the emitter current
RE is the emitter resistance
Ignoring IB x RB is an inaccuracy of the model, but it's only about 0.22 V in this case so it's the smallest item in the list and only changed RE from 1.6 k to 1.5 k, so it's not a big deal. As I mentioned the model of VBE = 0.7 V is not all that accurate anyhow.

You can have a more accurate representation of the situation, but whatever you do, it will only be a representation as there will always be aspects of the transistor that are not simulated correctly. You have a gain of 150 in the model. In practice that will change with current, temperature and between different examples of the transistor.

Also by estimating the base current as being zero for the simple model, that is modelling the transistor having a much larger (infinite) gain. If there is a large change in conditions, between normal gain and infinite gain, the circuit is probably badly designed.

 

[circuit975]

The first estimate is done in a more simple way by ignoring the base current, so it's somewhat less accurate, but it is more simple and it gets an answer that is close to what will happen in reality. It is always good to have a simple estimate, just as a reality check. It was clear to me that 402 kOhm for RE was very wrong before I did any calculations. The more complications you add, the more possible places to make a huge error. If you have a simple and reliable estimate, it is a good basis and you would expect your more complicated and more accurate estimate to be similar.

In you equation
VB - VBE - IBxRB - IExRE = 0

VB is the base voltage before the base current is taken into account.
VBE is the base-emitter voltage drop
IB is the base current
RB is the effective base resistance
IE is the emitter current
RE is the emitter resistance
Ignoring IB x RB is an inaccuracy of the model, but it's only about 0.22 V in this case so it's the smallest item in the list and only changed RE from 1.6 k to 1.5 k, so it's not a big deal. As I mentioned the model of VBE = 0.7 V is not all that accurate anyhow.

You can have a more accurate representation of the situation, but whatever you do, it will only be a representation as there will always be aspects of the transistor that are not simulated correctly. You have a gain of 150 in the model. In practice that will change with current, temperature and between different examples of the transistor.

Also by estimating the base current as being zero for the simple model, that is modelling the transistor having a much larger (infinite) gain. If there is a large change in conditions, between normal gain and infinite gain, the circuit is probably badly designed.

Yes, RE should not be that high, now I understand exactly what you are saying, thank you.

So, is there a mistake here? Is there a wrong value or unit conversion or something about RE being shorted in the AC state of the circuit or apart from all this is there anything I need to know about the points to be considered when calculating Av or are all my steps here correct? ;

note: RE is short-circuited in AC analysis because it is connected in parallel with the capacitor, hence (ß+1)RE = 0
-ß x Rc / ßre + (ß+1)RE = -150 x 5600 / 2,175 + 0 = -386,206

 

[Diver300]

The equation that you have doesn't work for very low values of RE, so it won't work on AC where you have put in a value of zero for the impedance of the capacitor.

The circuit is stable and the gain is set by the resistor values. That is preferable because the transistor gain varies so much. There is effectively feedback, because if the transistor conducts more, the emitter voltage rises, which reduces the base current.

If the emitter voltage, or the AC component of the voltage, is zero, you need a more complicated model. When the input goes up, there will be a large increase in base current, and it is difficult to predict that. The model of having a fixed base-emitter voltage of 0.7 V is not sophisticated enough to predict the behaviour for low emitter impedance.

 

[circuit975]

The equation that you have doesn't work for very low values of RE, so it won't work on AC where you have put in a value of zero for the impedance of the capacitor.

The circuit is stable and the gain is set by the resistor values. That is preferable because the transistor gain varies so much. There is effectively feedback, because if the transistor conducts more, the emitter voltage rises, which reduces the base current.

If the emitter voltage, or the AC component of the voltage, is zero, you need a more complicated model. When the input goes up, there will be a large increase in base current, and it is difficult to predict that. The model of having a fixed base-emitter voltage of 0.7 V is not sophisticated enough to predict the behaviour for low emitter impedance.

Then the Av value (voltage gain) that I found is -386 and my steps and result when calculating the value are correct?

 

[Diver300]

Then the Av value (voltage gain) that I found is -386 and my steps and result when calculating the value are correct?

It's possible that the voltage gain is as high as that, but the input impedance would be very low, the source impedance would have to be considered.

 

[Tony Stewart]

you said ...
VB - VBE - IBxRB - IExRE = 0
then ... RE = 402K which is the wrong result

Re= (Vb-Vbe)/Ie - Rb/(ß+1). . . . .(1) ok
for level 1 approximations
Hint #1
Since ß >>1 has a high tolerance of 50% ~ 300% replace (ß+1) with ß and let Ie=Ic = 10V/5k6.

#2 if you expect Ic ~ 1mA then use Vbe = 0.6V
Solve Re in (1) becomes (Vb-0.6)*5k6/10V. So you expect Re to be much smaller than Rc.

#3 For reality check on Zc=0, compute 1/(2pifC)= Zc(f), for Ce, let Zc (f) = re , you would need > 1 mF (huge) at 100 Hz. and a 50-ohm source is not low enough for Av > 300.

#4 Do not expect Av > 100 for a linear large signal unless negative feedback is used then Av is reduced. H biased designs are primitive and very nonlinear - due to the square law of Vbe vs Ic with a bypassed Re. Gain of 50 is more reasonable per stage with Rc/Re ratios

#5 When estimating THD harmonic distortion on a sine wave, use AC coupling and compare the peak difference to Vpp or (Vp+ - Vp-)/Vpp *100%. Another method is just to scope a triangle wave, which is more visual.

Next design improvement shorts out Re and bypass cap, reduces Vcc to 12V, uses DC bias to Vcc/2 on collector or 6V/10k = 600 uAdc. Negative feedback (NFB) ratio of 50% of Rcb/Rsb (series Rb) will be your approximate Av.

web SIM on Falstad

Gain and Distortion are tradeoffs. Here back to H bias but raised input impedance by moving Re to series R and thus reducing Vbe variations but this is with constant hFE = 150 which is unrealistic.

 

[MrAl]

Hello there
I am faced with such a question


1) Calculate and determine the value of the resistor RE that will ensure that VC=6 V.

2) Redraw the transistor model of the equivalent circuit for small amplitude AC signals. Determine the voltage gain Av of the circuit.

View attachment 146355

I solved the question as follows ;

VB = [24K / 24K+82K] x 16V = 3,62 Volt

RB = 24Kx82K / 24k+82k = 18,56K

6 = 16 - ICx5,6k
Ic = 1,78mA

IC = ßxIb = 150xIb
Ib = 11,86 microamper

IE = 11,86x10^-6 x 151 = 1,79mA
re = 26mV / 1,79mA = 14,5
VB - VBE - IBxRB - IExRE = 0
3,62 - 0,7 - [ (11,86 x 10^6) x 18,56K ] - 1,79mA x RE = 0
RE = 402K

Av = -ß x Rc / ßre + (ß+1)RE
Av = -150 x 5600 / 2,175 + 0 (The reason why it is 0 is because the Re resistor and capacitor are in parallel, I know that it is an ineffective element in AC analysis.)

This is my solution, are my solution steps correct? or if there is a point where I made a mistake when converting units, can you warn me, I need to correct my shortcomings and learn the right one. thanks a lot in advance

Hi,

I get around 1500 for RE.
That's considering Vbe=0.65v which is typical, and Beta can spread a bit starting with 100, maybe 50 to 200.
For increasing Vbe the ideal value for RE would go down to as low as 1400.
The base current comes out somewhere around 18ua with Beta=100.


This is after doing a full network analysis which includes Vbe and Beta.
Usually these questions include more information like the values to use for Vbe and Beta.

Note: a more exact value for R4 is 1464 Ohms.
Also, sensitivity due to any internal rb is minimal.
Solution comes from Nodal Analysis.