Most say Rbb = Vcc * R2 / (R1 + R2) but, does emitter side resistor(R4) also affect Vbb like posted picture?

In the below picture, I anticipated the voltage divider provide 0.83V (5V * 2 / 12) but, Vbb is 0.61V.

Above is just a simulation.

but the expected result also did not appear when I made a circuit and experimented.

 

Vbb=Vcc*R2/(R1+R2)-R1||R2*ibase=R1||R2*(Vcc/R1-ibase)

 

Vbb=Vcc*R2/(R1+R2)-R1||R2*ibase=R1||R2*(Vcc/R1-ibase)

Is I-base ignored when calculating Vbb because I-base is very tiny in most circuits?

 

The correction is -R1||R2*ibase.
In correctly designed circuits, the current through the divider is 4 to 7 times the base current, and this can change the base voltage by 15 to 25%. Increasing the current reduces the input impedance of the amplifier. It is necessary to calculate the minimum gain of the transistor.

 

The correction is -R1||R2*ibase.
In correctly designed circuits, the current through the divider is 4 to 7 times the base current, and this can change the base voltage by 15 to 25%. Increasing the current reduces the input impedance of the amplifier. It is necessary to calculate the minimum gain of the transistor.

Thank you@@

 

Most say Rbb = Vcc * R2 / (R1 + R2) but, does emitter side resistor(R4) also affect Vbb like posted picture?

In the below picture, I anticipated the voltage divider provide 0.83V (5V * 2 / 12) but, Vbb is 0.61V.

Above is just a simulation.

but the expected result also did not appear when I made a circuit and experimented.

Thank you@@

Most say Rbb = Vcc * R2 / (R1 + R2) but, does emitter side resistor(R4) also affect Vbb like posted picture?

In the below picture, I anticipated the voltage divider provide 0.83V (5V * 2 / 12) but, Vbb is 0.61V.

Above is just a simulation.

but the expected result also did not appear when I made a circuit and experimented.

Most say Rbb = Vcc * R2 / (R1 + R2) but, does emitter side resistor(R4) also affect Vbb like posted picture?

In the below picture, I anticipated the voltage divider provide 0.83V (5V * 2 / 12) but, Vbb is 0.61V.

Above is just a simulation.

but the expected result also did not appear when I made a circuit and experimented.

 

Hi: I am not certain of what you are trying to accomplish with the circuit you show. This is a typical class A bias scheme as you might find in say an audio amplifier. I'm simple terms, the first thing to do is to choose an operating point. Lets assume we have a transistor of Beta = 100, and we are going to use a 12 volt supply. We want the collector voltage to be about 7 volts with a collector current of 5 ma. This would indicate a 1K resistor in the collector and using a factor of .1, the emitter resistor would be 100 ohms. With a Beta of 100 and 5 ma collector current you would need to provide 50ua of base current. Using a factor of 10x for stability, the current in the divider should be about 250ua. WE now need to calculate the base voltage. The total current in the emitter resistor will be 5ma plus the 50ua base current. The drop across the emitter resistor will be .505 volts coupled with approximately .7 volts across the base emitter diode for a total of 1.205 volts. Next find the total resistance value to allow 250ua in the divider network. With a 12 volt supply that would be 12/.000250 = 48K. The voltage at the junction of the two resistors should be set to 1.205 volts. So 1.205/.000250 = 4.8K for the lower resistor and 48000 - 4800 = 43K for the upper. The reason for keeping the emitter resistor low Is that it adds negative feedback. Likewise the current in the divider is set to 10x or more to provide stability of the operating point.

 

Hi: I am not certain of what you are trying to accomplish with the circuit you show. This is a typical class A bias scheme as you might find in say an audio amplifier. I'm simple terms, the first thing to do is to choose an operating point. Lets assume we have a transistor of Beta = 100, and we are going to use a 12 volt supply. We want the collector voltage to be about 7 volts with a collector current of 5 ma. This would indicate a 1K resistor in the collector and using a factor of .1, the emitter resistor would be 100 ohms. With a Beta of 100 and 5 ma collector current you would need to provide 50ua of base current. Using a factor of 10x for stability, the current in the divider should be about 250ua. WE now need to calculate the base voltage. The total current in the emitter resistor will be 5ma plus the 50ua base current. The drop across the emitter resistor will be .505 volts coupled with approximately .7 volts across the base emitter diode for a total of 1.205 volts. Next find the total resistance value to allow 250ua in the divider network. With a 12 volt supply that would be 12/.000250 = 48K. The voltage at the junction of the two resistors should be set to 1.205 volts. So 1.205/.000250 = 4.8K for the lower resistor and 48000 - 4800 = 43K for the upper. The reason for keeping the emitter resistor low Is that it adds negative feedback. Likewise the current in the divider is set to 10x or more to provide stability of the operating point.

See

 

http://www.learningaboutelectronics.com/Articles/How-to-calculate-vbb-of-a-transistor

You calculate it as a simple voltage divider.
And to calculate the current, as if R1 and R2 would be grouped in parallel between base and Vbb.

 

See
View attachment 133594

I have some questions about your answer.

1. Did you consider the current through the base-emitter when calculating Vbb?
2. Is not the low side current in the divider 200uA If the upper side current in the divider is 250uA?
3. the simulation result was different from what I calculated myself like posted picture. I guess it is because the hfe(beta) is different. So do I have to adjust components manually when I make a circuit?
4. what is the negative feedback of emitter resistor? I've known the larger emitter resistor the stronger the stability about hfe variability.
http://www.learningaboutelectronics.com/Articles/Voltage-divider-bias-of-a-BJT-transistor

 

See
View attachment 133594

Hi: You just need to increase R3 slightly to provide a little more base current.

 

The purpose of my calculation is to show that it is necessary to take into account the base current. I have calculated for my life more than a thousand schemes. And I know how to calculate the working point of circuits, especially such a simple one. I showed that phranzdan is not right.
I calculated the circuit with he's resistor values.
The value of the upper voltage divider resistor (12V-1.205V)/(250uA+50uA)=36k.
With such a resistor (if Beta = 100), the current of the transistor will be 5 mA.

 

The purpose of my calculation is to show that it is necessary to take into account the base current. I have calculated for my life more than a thousand schemes. And I know how to calculate the working point of circuits, especially such a simple one. I showed that phranzdan is not right.
I calculated the circuit with he's resistor values.
The value of the upper voltage divider resistor (12V-1.205V)/(250uA+50uA)=36k.
With such a resistor (if Beta = 100), the current of the transistor will be 5 mA.

How about my calculation in the posted picture above?
And I adjusted the R2 value to get the desired value(10 times voltage gain).
Is this the way transistors are designed?

 

Hi guys: Sorry I messed up. I have been away from circuits since 2000 so I am a little fuzzy. Remember I said the divider current should be at least 10X the base current. As is evident, I used a factor of 5X. To correct, set the divider current at 500ua. Keep the base bias voltage the same around 1.3 volts. For a drop of 1.3 volts the bottom resistor needs to be 1.3/.0005 or 2.6K. The upper resistor would be (12-1.3)/.0005 or about 21K. I realize I have not compensated for the 50ua of base current the lower resistor will not see. For the correction, use 450ua which would yield a resistor of 2.8K. This illustrates the point of making the current in the divider at least 10X the base current. For a multiple of 20X, the divider current would be 1ma then the lower resistor would be 1.3K and the compensation would result in an error of about 68 ohms. Unfortunately, in my day we did not have simulators to get it exact on the first cut and calculations were difficult when at least at the start of my career we had nothing more than a slide rule. Go to your desk and come up with a first cut then go to the lab and tweak it. There are trade since as you increase the divider current you reduce the lower resistor value and hence the input impedance. The comment on the emitter resistor causing negative feedback has nothing to do with the D.C. bias but rather with an A.C. signal. This problem can be corrected with a bypass capacitor across the emitter resistor. See, I don't even call them condensors anymore.

 

Hi guys: Sorry I messed up. I have been away from circuits since 2000 so I am a little fuzzy. Remember I said the divider current should be at least 10X the base current. As is evident, I used a factor of 5X. To correct, set the divider current at 500ua. Keep the base bias voltage the same around 1.3 volts. For a drop of 1.3 volts the bottom resistor needs to be 1.3/.0005 or 2.6K. The upper resistor would be (12-1.3)/.0005 or about 21K. I realize I have not compensated for the 50ua of base current the lower resistor will not see. For the correction, use 450ua which would yield a resistor of 2.8K. This illustrates the point of making the current in the divider at least 10X the base current. For a multiple of 20X, the divider current would be 1ma then the lower resistor would be 1.3K and the compensation would result in an error of about 68 ohms. Unfortunately, in my day we did not have simulators to get it exact on the first cut and calculations were difficult when at least at the start of my career we had nothing more than a slide rule. Go to your desk and come up with a first cut then go to the lab and tweak it. There are trade since as you increase the divider current you reduce the lower resistor value and hence the input impedance. The comment on the emitter resistor causing negative feedback has nothing to do with the D.C. bias but rather with an A.C. signal. This problem can be corrected with a bypass capacitor across the emitter resistor. See, I don't even call them condensors anymore.

Thank you very much!!