Guess thats it?

 

Hello,

Yes, a switch can be used to separate parts from a circuit.

You now can complete the both answers.
You already have all values for the open switch.

Bertus

 

Got it Thanks for your patience and help.

Soon i'll upload another task to be evaluated Hope you're willing to take a look at it.

 

Hello,

I am happy that I could be of any help.
I will see your other task soon.

Bertus

 

Sorry, I'll make sure to add units

Assuming the gain is 40.
Base current is: emitter devided by 40
118 mA / 40 = 2.95 mA

Close, but the base current is the emitter current divided by the gain. For high gains, the difference is pretty insignificant, but power transistors have low gains to begin with and if you are driving a transistor into saturation (using it as a switch, for instance) then you can easily get into current gains in the range of 5 or even less.

And as far as i know the collecter current is approximately the same as emitter.

True -- under conditions where the current gain is sufficiently high.

You often start a design assuming infinite gain such that the collector and emitter currents are equal and there is no base current. For many, many designs you don't ever need to go beyond this assumption.

Doesn't the base current determine if the transistor is closed or open?

Yes and no. Remember that the transistor has three possible states -- cutoff, linear, and saturation. Having a forward-biased base-emitter junction with non-zero base current is required in order to be in either linear or in saturation.

 

Here we go again

Lets decide the switch is closed
Q1 gain is 50
Q2 gain is 20
Q3 gain is 110

That's it. already stucked.
I know I need at least 1.8 V after the R1 junction to activate Q1, Q2 and Q3.

 

What's the voltage at the top of R3 (let's use the bottom node, STEL, as the 0 V reference node)?

Assuming that the base current into Q3 is negligible, what if Uout?

What is the voltage at the base of Q1?

 

R3 would be 0.6V which is just enough to activate Q3.
Uout would be 2.5 times R3
0.6 V x 2.5 = 1.5V

Base of Q1 is calculated by Uout subtract the two diodes in Q1 and Q2
1.5V - 0.6V - 0.6V = 0.3V

Am i on to something? :b

 

R3 would be 0.6V which is just enough to activate Q3.

Right idea except that the emitter of Q3 isn't at 0 V. What is the voltage at the top of the zener diode D1?

 

Right idea except that the emitter of Q3 isn't at 0 V. What is the voltage at the top of the zener diode D1?

The zener diode require 5.6V but i thought it would take what it needs from the R2 connection?
How does that junction work?

 

Each node can only have one voltage. So if the zener has to have 5.6 V across it to allow current, then the cathode has to be 5.6 V higher than the anode. If the anode is connected to the node we choose to call 0 V, then the cathode must be 5.6 V higher than 0 V. If the cathode is connected to the emitter of Q3, then what must Q3's emitter voltage be? If the base of Q3 must be 0.6 V higher than the emitter in order to get base current flowing, what must Q3's base voltage be?

 

If all voltage required in the Zener diode is to be taken from Q3 then the emitter has to be 5.6 V like the zener diode.
Q3 base is base-emitter diode added to emitter
5.6 V + 0.6 V = 6.2 V

Uout is Q3 emitter times 2.5
6.2 V x 2.5 = 15.5 V

Q1 base is Uout subtract 2x diodes
15.5 V - 0.6 V - 0.6 V = 14,3 V

R1 is Uin+ subtract Q1 base
24 V - 14.3 V = 9.7 V

Current in R1 is V/Ω
9,7 V / 4700Ω = 2.1 mA

 

Current in R4 and R3 is Uout divided by resistance.
15.5 V / (3.3k Ω + 2.2k) Ω = 2.8 mA

Current in RL is Uout devided by resistance
15.5 V / 150 Ω = 103 mA

Uout total current
2.8 mA + 103 mA = 105.8 mA

 

If all voltage required in the Zener diode is to be taken from Q3 then the emitter has to be 5.6 V like the zener diode.
Q3 base is base-emitter diode added to emitter
5.6 V + 0.6 V = 6.2 V

Correct.

Uout is Q3 emitter times 2.5
6.2 V x 2.5 = 15.5 V

6.2 V is the Q3 base (which is what you want)
Other than that, it is correct

Q1 base is Uout subtract 2x diodes
15.5 V - 0.6 V - 0.6 V = 14,3 V

But, for Q2 to be on, doesn't Q2 base need to be 0.6 V higher than Q2 emitter?

Similarly, doesn't Q1 base need to be 0.6 V higher than Q1 emitter?

Your on the right track -- just need to exercise a bit more care.

 

Haha guess it was late last night

I had to ADD on the diodes to find Q1 base, not subtract them.
15.5 V + 0.6 V + 0.6 V = 16.7 V

In my post above i calculated the currrent through R4 and R3 but i have a feeling its wrong.

The current that runs to R3 has to run through R4. In which case current in R4 is Uout devided by R4?

So the total current would be Uout devided by R4 parallel with RL?

 

I've attached some calculated voltage and current values.

 

Try again with R2 voltage and current.

 

Would that be 24 V - 5.6 V = 18.4 V
18.4 V / 1000 ohm = 18.4 mA
?

 

Haha guess it was late last night

I had to ADD on the diodes to find Q1 base, not subtract them.
15.5 V + 0.6 V + 0.6 V = 16.7 V

In my post above i calculated the currrent through R4 and R3 but i have a feeling its wrong.

I generally stop reading as soon as we get to something that is incorrect, since it commonly messes up much of what follows, and then focus on getting that point corrected.

The current that runs to R3 has to run through R4. In which case current in R4 is Uout devided by R4?

R4 also has what ever current is flowing to the base of Q3, which we are assuming is negligible.

R4 is NOT Uout divided by R4, because Uout is not the voltage that appears across R4.

Remember that Ohm's Law is a relationship between the resistance of a resistor and the voltage across THAT resistor and the current through THAT resistor.

So the total current would be Uout devided by R4 parallel with RL?

Depends on what you are including in "total current" -- is it the total current leaving the output of this regulator, or the total current draw from the supply? Need to work on being explicit and clear in your terms.

 

I've attached some calculated voltage and current values.

View attachment 101626

Most of these are good, but look at R2 more carefully, remembering that Ohm's Law requires the use of the voltage ACROSS the resistor -- you can't just pick the voltage on one end of it.

Once you have the emitter currents of the transistors you can determine the base currents and then compare those to the currents in the branches that are feeding them to decide if the assumption that the base currents are negligible was warranted. You also need to determine the Vce (collector-emitter) voltage drops across each transistor to confirm that the assumption that the transistors are on and in the active region was valid. Finally, you need to determine the current in the zener to ensure that it is within the allowable limits.

You are making good progress.