# Problem with transistor

#### zugfxxroa

Joined Apr 5, 2021
8
Hello,

in the following image you can see a circuit with a npn transistor. I marked the given values with red and the unknown with blue. Below the circuit you can see my calculations and solutions. I'm stuck at finding I1. I would be very thankful for some guidance.

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#### ericgibbs

Joined Jan 29, 2010
18,221
hi z,
Welcome to AAC.
Are you sure your circuit drawing is correct.?
Looks as though R1 and R2 maybe drawn wrong.??

E

#### Papabravo

Joined Feb 24, 2006
20,600
The current I1 will flow from the power supply through R3 and across the CE junction of the transistor to GROUND. (Hint: Add a GROUND symbol to the schematic). In saturation (transistor is on) the Vc(sat) is 0.043 Volts. If you subtract that from 12 V, that will be the voltage drop across R3. You have the voltage drop, you have the resistance and the current through R3 and the transistor must be.....?

#### zugfxxroa

Joined Apr 5, 2021
8
My idea was to work with a mesh, but I think that doesn’t work.. but I don’t know why.

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#### Papabravo

Joined Feb 24, 2006
20,600
hi z,
Welcome to AAC.
Are you sure your circuit drawing is correct.?
Looks as though R1 and R2 maybe drawn wrong.??

E
Only if voltage divider bias was the intention. I agree it looks strange, but it is not necessarily wrong. In either case it has no effect if the transistor is in saturation the current in R3 will be determined.

#### Papabravo

Joined Feb 24, 2006
20,600
My idea was to work with a mesh, but I think that doesn’t work.. but I don’t know why.
The mesh goes from the positive terminal of the 12V source through R3, across the CE junction of the transistor, back to the negative terminal. What's the problem?

#### zugfxxroa

Joined Apr 5, 2021
8
I built this circuit beforehand and determined these values in reality:
U1=10.95V
I1=11.98mA
U2=0.043V
I2=1.02mA

#### zugfxxroa

Joined Apr 5, 2021
8
The current I1 will flow from the power supply through R3 and across the CE junction of the transistor to GROUND. (Hint: Add a GROUND symbol to the schematic). In saturation (transistor is on) the Vc(sat) is 0.043 Volts. If you subtract that from 12 V, that will be the voltage drop across R3. You have the voltage drop, you have the resistance and the current through R3 and the transistor must be.....?
Are they the same?

#### Papabravo

Joined Feb 24, 2006
20,600
Are they the same?
Yes the current in the resistor and the collector current in the transistor must be the same. They are both part of a series branch. One end of the branch is at +12Volts and the other end of the branch is almost at Ground. The currents flowing in and out of the battery will be different because there is a parallel branch going through R1, R2, and the base emitter junction. This current will split at the junction of R1 and R3 and recombine at the emitter where the very small base current is added to the much larger collector current to form the emitter current which is then returned to the battery.

For basic calculations we consider the collector current and the emitter current to be the same. In practice the ratio of the two currents very close to 1. If you know the beta or hfe of the transistor this ratio is computed as β/(β+1). For a transistor with a β of 100, this ratio is 100/101 = 0.9901. For transistors with a higher value it will be even closer to 1.

#### zugfxxroa

Joined Apr 5, 2021
8
Yes the current in the resistor and the collector current in the transistor must be the same. They are both part of a series branch. One end of the branch is at +12Volts and the other end of the branch is almost at Ground. The currents flowing in and out of the battery will be different because there is a parallel branch going through R1, R2, and the base emitter junction. This current will split at the junction of R1 and R3 and recombine at the emitter where the very small base current is added to the much larger collector current to form the emitter current which is then returned to the battery.

For basic calculations we consider the collector current and the emitter current to be the same. In practice the ratio of the two currents very close to 1. If you know the beta or hfe of the transistor this ratio is computed as β/(β+1). For a transistor with a β of 100, this ratio is 100/101 = 0.9901. For transistors with a higher value it will be even closer to 1.
Thank you for your reply. This is my solution, which differs from my measured value... Furthermore, are my other solutions in your opinion correct?

#### zugfxxroa

Joined Apr 5, 2021
8
Thank you for your reply. This is my solution, which differs from my measured value... Furthermore, are my other solutions in your opinion correct?
Nevermind, I made a dumb mistake. I now have the correct solution. Thank you very much!

#### zugfxxroa

Joined Apr 5, 2021
8
Thank you for your reply. This is my solution, which differs from my measured value... Furthermore, are my other solutions in your opinion correct?
But I'm still curious why I don’t get the correct solution if I use the mesh between the resistors..

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#### Papabravo

Joined Feb 24, 2006
20,600
But I'm still curious why I don’t get the correct solution if I use the mesh between the resistors..
There is no "mesh" between the resistors. a mesh is a complete loop in a circuit that begins and ends at the same place.
If you add the voltage drops and voltage rises around the loop, the sum must be zero
Similarly there are "node" equations which com from the principle that the sum of the currents into and out of a node must be zero.

#### zugfxxroa

Joined Apr 5, 2021
8
There is no "mesh" between the resistors. a mesh is a complete loop in a circuit that begins and ends at the same place.
If you add the voltage drops and voltage rises around the loop, the sum must be zero
Similarly there are "node" equations which com from the principle that the sum of the currents into and out of a node must be zero.
Pardon my English, I expressed myself not good enough. My question is why do I get a false solution if I choose the loop in the image below.

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#### Papabravo

Joined Feb 24, 2006
20,600
No problem, I'm just trying to help you get the terminology correct
The diagram where you have indicated a loop is not really a loop, it is a pair of parallel branches, and neither branch contains an independent voltage or current source.
There is really only one loop which includes the independent source of +12V, with two parallel branches One branch goes through the Base Emitter (BE) junction, and the other goes through the Collector Emitter (CE) junction. In order to balance the voltage drops in a loop, there must be an equal voltage rise.

When there is a branch, as there is at the junction of R1 & R3, you can analyze the branches separately because they have the same voltage across both branches. The currents can be different, but the voltage across both branches remains the same.

I hope that is clear and that I have not created further confusion. I'm pretty sure your English is better than my (pick any language, except maybe Latin).

#### BobaMosfet

Joined Jul 1, 2009
2,102
Hello,

in the following image you can see a circuit with a npn transistor. I marked the given values with red and the unknown with blue. Below the circuit you can see my calculations and solutions. I'm stuck at finding I1. I would be very thankful for some guidance.
Title: Understanding Basic Electronics, 1st Ed.
Publisher: The American Radio Relay League
ISBN: 0-87259-398-3

#### MrAl

Joined Jun 17, 2014
10,898
If the voltages are all given you can use Ohm's Law to calculate the currents and thus any node voltage between two resistors.