# Operation point at rest of transistors Q3 and Q4

Discussion in 'Homework Help' started by Schmitt_trigger, Feb 8, 2013.

1. ### Schmitt_trigger Thread Starter New Member

Feb 5, 2013
20
0
I am trying to find the operation point at rest(Ic,Vce) of Q3 and Q4 when V1=V2=0V

First i think i have to determine the Ic of Q2 and Q1.$Ib=I(14 K resistor)/2$

So I will have Ib aprox 0.43 mA.Then i think that Ie Q3 and Ie Q4 will also be equal to 0,43 mA

I also know that the gain of the transitors is 300 so considering Ie aprox equal to Ic i know that Ic of Q3 and Q4 will also be equal to 0,43 mA.
I can determine Ib of Q3 and Ib of Q4 which will be equal to $Ib=Ic/beta$ so $Ib=0,43/300=0,00143mA$

Then Vce=Vcb+Vbe i know vbe i have determined amd i think i can determine Vcb through the following equation

$12=6,8k*Ic+Vcb+10k*Ib=0$ .Are my thoughts correct?(This for Q4) for Q3 i would have $12=Vcb+10*Ib$

But i think i might have done a mistake considering that IcQ3=IcQ4 because Q3 doesn't have a colector resistor.

Thank you

File size:
30.6 KB
Views:
31
2. ### #12 Expert

Nov 30, 2010
16,033
6,545
Your mistake is at Q1 and Q2. There is 24 volts to account for.
In the real world, Ib Q2 is negligible...or you can acount for it with a gain of 300.
I don't know what teachers expect nowadays.

Last edited: Feb 8, 2013
3. ### WBahn Moderator

Mar 31, 2012
17,461
4,701
Huh? You seem to be saying that the base current (of some transistor) is half of the current in the 14kΩ resistor. Does that make sense? How many paths can the current leaving the bottom of that resistor take?

How do you come up with 0.43mA? Show your work so that we can see how you are arriving at these numbers.

So you seem to be claiming that if the base current is 0.43mA (base current of which transistor?), that the emitter currents of Q3 and Q4 will each be 0.43mA, meaning that the collector current in Q2, which is the sum of the emitter currents in Q3 and Q4, must be 0.86mA, right?

Unless you are given the Early voltage for the transistors (or equivalent information regarding their output impedance) you have to assume that the collector current is independent of colector-emitter voltage as long as the device is not in saturation.

4. ### Schmitt_trigger Thread Starter New Member

Feb 5, 2013
20
0
The current leaving the 14k resistor it can either to the base of Q1 or base of Q2

The 0.43mA i got by doing $I (14k resistor)=(12)/(14*10^3)=0.857 mA$

which dividing by two gives 0.43 mA which is the base current of Q2 and colector current of Q1

yes i did meant that the collector current in Q2, is the sum of the emitter currents in Q3 and Q4, must be 0.86mA

No they didn't gave the Early voltage the only thing i know is the gain and Vbe=0.7V

Can i go this way to find Vce of Q3 and Q4?

5. ### WBahn Moderator

Mar 31, 2012
17,461
4,701
Why can't it enter the collector of Q1?

How have you determined that the voltage across the resistor is 12V?

Upon what do you base the conclusion that the current splits evenly?

If the base current in Q2 really is 0.43A, then how are you arriving at the collector current in Q2 being 0.86A?

Okay, so this makes the analysis quite a bit easier but reduces the accuracy of the results. Still, it is good enough to get results that are pretty close.

Partly. You are making a lot of bad assumptions, but the general path you are taking is pretty close.

6. ### Schmitt_trigger Thread Starter New Member

Feb 5, 2013
20
0
I thought that the current split evenly because the resistance offered by the base of the transistors is the same since the transistors are of the same type.In the colector os Q1 i think that there isn't any current because the in the equivalent model i know the colector acts as a current source and current sources offer infinite resistance.

$I(14 k resistor)=(24)/(14*10^3)=1,71mA$ therefore the Ib's of Q1 and Q2 is equal to 0.85 mA then i get $IcQ2=beta*IbQ2=300*0,85 mA$ which gives $IcQ2=0,25A=250mA$

Once i know this i know $IEQ1=IEQ2=ICQ2/2$.Then using those equations that i mention i can find Vcb for both transistors and then find Vce for Q1 and Q2,right?

Thank you

7. ### WBahn Moderator

Mar 31, 2012
17,461
4,701
By this reasoning, then there can't be any current flowing into the collector of Q2, either. Also, by this reasoning, where does the emitter current of Q3 and Q4 come from? It can't come from the collector because that has infinite resistance. So does it come from the base? If so, doesn't it have to pass through the 10kΩ resistors? What would the voltage drop across those resistors be?

Doesn't it stand to reason that, assuming all of the transistors are in the active regions, that if the base currents of Q1 and Q2 are the same that thair collector currents would be the same?

In the active region, what do you know about the relationship between the base current and the collector current of a transistor?

Why are you saying that there is 24V across the 14kΩ resistor?

If IcQ2 is 300 times its base current, then why isn't IcQ1 equal to 300 times its base current?

This much is correct.

You are still quite a ways from getting to that point.

The first hurdle that you have got to overcome is your inconsistent views on the collector current of a transistor.

Let's first work with a simplified version of the circuit in which we severe the connect to the base of Q2 and replace Q1 with a simple diode with the cathode connected where the emitter presently is and the anode connected where the base/collector presently are.

Write KVL for this circuit and determine the diode current.

Get that far and then well proceed from there.

8. ### Schmitt_trigger Thread Starter New Member

Feb 5, 2013
20
0

In the ative region i know that $Ic=Beta*Ib$ ,the transistor acts like a current source controled by voltage

The voltage across the 14k resistor is 24 V because we have -12 V and 12V feeding the circuit so the potencial is 24 V

Yes the IC1 is also equal to $300*Ib1$

You say to disconect Q2 and Substitute Q1 by a diode?
KVL: $24V=14k*I(14kresistor)+Vdiode+500 Ohm*I(500 Ohm resistor)

24V=14k*I14k+0.7V+500*0.0024 A

I14k=1,47mA$
i think...

9. ### WBahn Moderator

Mar 31, 2012
17,461
4,701
If it's controlled by voltage, then why doesn't the equation you give involve a voltage?

So, by that reasoning, the voltage across the collector and emitter of Q3 and Q4 must also be 24V. After all, we have -12V and 12V feeding the circuit, so the potential must be 24V. If this isn't the case, then what's the difference.

If the 14kΩ has 24V across it, then what is the voltage at the bottom of that resistor? It must be -12V, right? What is the voltage drop across the base-emitter junction of Q1? You original post (or a subsequent follow-up) said that it was 0.7V, so doesn't that make the voltage at the emitter of Q1 equal to -24.7V, which is also the voltage at the top of the left hand 500Ω resistor. So that means that the voltage at the bottom of the resistor is 0.7V higher than the voltage at the top, meaning that there is a current of 0.7V/500Ω, which is 1.4mA, flowing upward in the 500Ω resistor. Do you buy that? If not, what is wrong?

So what's that say about the claim that the current in the 14kΩ resistor splits evenly between the two base currents?

Where'd you get the 2.4mΩ for the current in the 500Ω resistor?

Does that make sense? If you have 1.47mA flowing in the 14kΩ resistor and 2.4mA flowing in the 500Ω resistor, where is the difference coming from?

10. ### Schmitt_trigger Thread Starter New Member

Feb 5, 2013
20
0
The equation does not involve voltage because i am using a simplified model.To make sense of what i said we would have to consider the Ebbers-Moll model which states $Ic=Is*[exp(Vbe/Vt)-1]$but thats not what i intend here.

and yes about this "So, by that reasoning, the voltage across the collector and emitter of Q3 and Q4 must also be 24V. After all, we have -12V and 12V feeding the circuit, so the potential must be 24V."

yeah they have also 24V of pontential

2.4 m ohm current?Thats a bit hard to get

The 2.4 mA current i got by doing tht 12V/500 Ohm,which is must be wrong because of what you mention..So i do not know what the current in the 500 Ohm resistor..

11. ### WBahn Moderator

Mar 31, 2012
17,461
4,701
A big part of your problem -- probably the biggest part -- is that you appear to have some pretty weak fundamentals that you need to overcome.

Let's use as an example the circuit I asked you to analyze in which you have a +12V supply going to a 14kΩ resistor and, from there, through a diode and then through a 500Ω resistor to a -12V supply.

The analysis of that circuit should have proceeded something along the following lines:

The total voltage dropped across the three components is

Vtot = (+12V) - (-12V) = 24V
The current in all three components has to be the same because they are in series, so let's call that Itot.
The voltage drop across the 14kΩ resistor is Itot*14kΩ, the voltage drop across the forward-biased diode is 0.7V, and the voltage across the 500Ω resistor is Itot*500Ω.

The sum of the voltage drops has to equal the total voltage, so:

24V = Itot*14kΩ + 0.7V + Itot*500Ω
24V - 0.7V = 23.3V = Itot*(14kΩ + 500Ω)
23.3V = Itot*14.5kΩ

Itot = 23.3V/14.5kΩ

By the time you started working with transistors, you should have been able to write that last equation down by inspection and gotten a total current of 1.607mA. The entire process should have taken no more than 20 seconds. You need to ask yourself why it didn't and consider the likely ramifications going forward if you continue not being able to do so.

First, let's consider Ohm's Law. The resistance, R, of a resistor is the ratio of the voltage across THAT resistor to the current through THAT resistor. The fact that the voltage at some other point in the circuit happens to be -12V doesn't matter. You need the voltage across THAT resistor. In this circuit, the 14kΩ resistor has 22.5V across it. That it is even in the ballpark of the 24V total across all three is a consequence of it being so much larger than the other resistor, which only has
0.8V across it.

Second, you don't seem to understand Kirchhoff's Voltage Law. I asked you to use it earlier and you didn't even try.

Third, you seem to have a reasonable awareness of Kirchhoff's Current Law, but don't seem willing to consistently apply it.

Until you get a firm grasp on these fundamentals, you really have no business trying to work with transistor circuits because the poor fundamentals will defeat you every time. So take a step back and work with circuits having just DC sources and resistors until you get KVL, KCL, Mesh, Nodal, and Superposition analysis methods down pat.

Check out the E-book on this site (see Vol I - DC) and work through it being sure that you understand the material at each stage. Ask questions here and we will be more than happy to work with you.

12. ### Schmitt_trigger Thread Starter New Member

Feb 5, 2013
20
0
LOL..I have litlle experience on electronics altough i know the concepts you mention when working on a circuit like this i am most worried in understanding how the cirucuit works then thinking that the elements are in series and that the current is the same in all of them.Besides we are moving from my questions which have no nothing to do it this.

I don't think my reasoning was rubbish so i what i needed was some help to see what i was doing wrong, i didn't need that someone came and asked me now change the circuit like this so that i can see if you still remember the KCL,KVL and other stuff and if you don't i am gona tell you that you do not have the basis...lol

13. ### WBahn Moderator

Mar 31, 2012
17,461
4,701
You say that you are most worried in understanding how the circuit works, yet you lack the most basic fundamental skills and knowledge needed to determine how the circuit works, as you amply demonstrated repeatedly. You say you wanted help to see what you were doing wrong. I spent a considerable amount of time pointing out where you were going wrong and asking basic questions so that you could see where you were going wrong, but you lacked the fundamentals to answer them. I pointed that out and made it clear that I was willing to help you come up to speed. But you really weren't looking for help to see where you were going wrong because you weren't really interested in finding out where you were going wrong. You only wanted someone to work the problem for you and give you the answer.

Please go into some field other than engineering. You lack the basic skills and clearly have intention of gaining them and quite apparently don't see that as a problem. With that attitude, sooner or later you will get a passle of people killed, so please go into some other field.

thatoneguy and Ron H like this.
14. ### #12 Expert

Nov 30, 2010
16,033
6,545
When I get that irritable, I take a break for a few weeks.