I understand that but not even my college we carry the units during the calculations,its no so hard to do it here because the math its simple and there are few steps,but in other problems.....My teachers only demand the units at the end of the calculations,unless the as you said the quantities are dimensionless

And your teachers are doing you a serious disservice. I understand that many textbook authors and many teachers are extremely sloppy with units. That is often a reflection of the fact that they have not worked in an environment in which failing to track and check units results in situations in which people can literally die. To them, a wrong answer is just a wrong answer and you lose some points for it. In the real world, wrong answers have real consequences and, as an engineer, you have an ethical, moral, and frequently legal reponsibility to exercise due diligence to ensure your answers are correct. Do you really want to be the defendent in a civil or criminal proceeding after a few hundred people have been killed by the failure of something you designed and your defense comes down to, "I couldn't be bothered to track and check my units on such a simple calculation that only had a few steps."?

It doesn't matter whether or not the calculation is simple with few steps -- after all, we've seen you make mistakes that units were able to catch in simple calculations with few steps in some of the other threads you have going on. I make those kinds of mistakes all the time. So does everyone else.

What i need to find is the IC of Saturation that is the current on the Load that saturates the transistor,rigth?

We know the current on the load (should be "current through the load") as the transistor goes into saturation. It's 0.5A. This occurs when the load resistance is 29Ω. If the resistance goes above that, the transistor moves deeper into saturation and we have less than 0.5A in the load. This is why we set the maximum load resistance to 29Ω. Because anything higher will result in the circuit not being able to maintain 0.5A of current in the load.

Let's come at this from a different angle. We know that the maximum value of the load resistance is 29Ω. What is the power that is being dissipated by the transistor when we have that as our load?

Can the transistor handle that? If the answer is "yes", then the circuit is clearly "well dimensioned" for a 29Ω load.

Does the power in the transistor go up or go down as the load is reduced from 29Ω to something a bit smaller? If so, then is there a value of the load resistance that results in the transistor dissipating 1W and, if we reduce the load resistance below that value the power in the transistor will exceed the 1W maximum?

I did not understood this very well:

Hopefully I've addressed this above.

But i also need to calculate the minimum value of the resistor so that the circuit function properly that is VCE=>0.5V,is that it?

The minimum value of the load resistor has NOTHING to do with Vcesat. That was what the maximum value of the load resistor was all about. As long as Rload is less than 29Ω, we KNOW that the transistor IS NOT in saturation and that Vce > Vcesat.

We already know that R<=29 Ohm.So the worst case would be like Rload=1 Ohm

Why would that be worse than Rload = 0.1Ω?

Vcc = R1*Ic-Vce - Vload
20=5 V-0.5 V-1 Ohm*ImaxLoad

Well, you're trying to use units. Look at the left hand side. It's 20V, not 20.

Vce is NOT 0.5V EXCEPT when Rload is EQUAL OR GREATER than 29Ω!

ImaxLoad=14.5A !!!!I must be thinking wrong because this current would do some pretty serious damage on the transistor and the rest of the circuit..

Forget what damage it might or might not do, ask where the hell it is coming from! In your equation the 5V came from Ic=0.5A through Rc=10Ω. So if Ic is half an amp, where is the other 14A coming from?

 

And your teachers are doing you a serious disservice. I understand that many textbook authors and many teachers are extremely sloppy with units. That is often a reflection of the fact that they have not worked in an environment in which failing to track and check units results in situations in which people can literally die. To them, a wrong answer is just a wrong answer and you lose some points for it. In the real world, wrong answers have real consequences and, as an engineer, you have an ethical, moral, and frequently legal reponsibility to exercise due diligence to ensure your answers are correct. Do you really want to be the defendent in a civil or criminal proceeding after a few hundred people have been killed by the failure of something you designed and your defense comes down to, "I couldn't be bothered to track and check my units on such a simple calculation that only had a few steps."?

It doesn't matter whether or not the calculation is simple with few steps -- after all, we've seen you make mistakes that units were able to catch in simple calculations with few steps in some of the other threads you have going on. I make those kinds of mistakes all the time. So does everyone else.



We know the current on the load (should be "current through the load") as the transistor goes into saturation. It's 0.5A. This occurs when the load resistance is 29Ω. If the resistance goes above that, the transistor moves deeper into saturation and we have less than 0.5A in the load. This is why we set the maximum load resistance to 29Ω. Because anything higher will result in the circuit not being able to maintain 0.5A of current in the load.

Let's come at this from a different angle. We know that the maximum value of the load resistance is 29Ω. What is the power that is being dissipated by the transistor when we have that as our load?

Can the transistor handle that? If the answer is "yes", then the circuit is clearly "well dimensioned" for a 29Ω load.

Does the power in the transistor go up or go down as the load is reduced from 29Ω to something a bit smaller? If so, then is there a value of the load resistance that results in the transistor dissipating 1W and, if we reduce the load resistance below that value the power in the transistor will exceed the 1W maximum?



Hopefully I've addressed this above.



The minimum value of the load resistor has NOTHING to do with Vcesat. That was what the maximum value of the load resistor was all about. As long as Rload is less than 29Ω, we KNOW that the transistor IS NOT in saturation and that Vce > Vcesat.



Why would that be worse than Rload = 0.1Ω?


Well, you're trying to use units. Look at the left hand side. It's 20V, not 20.

Vce is NOT 0.5V EXCEPT when Rload is EQUAL OR GREATER than 29Ω!



Forget what damage it might or might not do, ask where the hell it is coming from! In your equation the 5V came from Ic=0.5A through Rc=10Ω. So if Ic is half an amp, where is the other 14A coming from?

So to calculate the dissipated power, i have a to consider a situation where Ic=0.5A.
Now for VCE i know that it has to be greater or equal than 0.5 V.This only ocurs when the load is equal or greater than 29 Ohm..Therefore the dissipated power would be P=VCE*IC=0.5 V*0.5 A=0.25 W,unless someone puts a load on the circuit that has resistance minor than the 29 Ohm,as you said " If the resistance goes above that, the transistor moves deeper into saturation and we have less than 0.5A in the load", so the VCE will remain 0.5 V and the IC will decrease so as the dissipated power so i think that the worst situation really is when we have VCE=0.5V and IC=0.5 A,rigth?

Thanks

 

So to calculate the dissipated power, i have a to consider a situation where Ic=0.5A.

Yes, at least for the first pass, because if Ic is not 0.5A then the circuit already is not functioning as intended and something else is wrong. In practice, we would also consider other cases from the standpoint of minimizing damage even in situations in which the circuit is already not functioning as desired.

Now for VCE i know that it has to be greater or equal than 0.5 V.This only ocurs when the load is equal or greater than 29 Ohm..

NO! It only occurs when the load is LESS THAN or equal to 29Ω. If the load is greater than that, then the transistor is saturated, VCE is 0.5V, and the current is less than the desired 0.5A.

Therefore the dissipated power would be P=VCE*IC=0.5 V*0.5 A=0.25 W,unless someone puts a load on the circuit that has resistance minor than the 29 Ohm,

Yes.

as you said " If the resistance goes above that, the transistor moves deeper into saturation and we have less than 0.5A in the load", so the VCE will remain 0.5 V and the IC will decrease so as the dissipated power so i think that the worst situation really is when we have VCE=0.5V and IC=0.5 A,rigth?

No. What if the load resistor is reduced to, say, 20Ω? We will staill have Ic=0.5A and Vc will still be Vcc-VRC = 20V-5V = 15V. But what will the voltage at the emitter, Ve, be? What will Vce be? What will the resulting power in the transistor be?

If Ic=0.5A and the maximum power dissipated in the transistor is 1W, what is the largest value of Vce that can be tolerated?

 

Yes, at least for the first pass, because if Ic is not 0.5A then the circuit already is not functioning as intended and something else is wrong. In practice, we would also consider other cases from the standpoint of minimizing damage even in situations in which the circuit is already not functioning as desired.



NO! It only occurs when the load is LESS THAN or equal to 29Ω. If the load is greater than that, then the transistor is saturated, VCE is 0.5V, and the current is less than the desired 0.5A.



Yes.



No. What if the load resistor is reduced to, say, 20Ω? We will staill have Ic=0.5A and Vc will still be Vcc-VRC = 20V-5V = 15V. But what will the voltage at the emitter, Ve, be? What will Vce be? What will the resulting power in the transistor be?

If the load had 20 Ohm,Ic=0.5 A Vc=15 V.We Know that VCE=VCB+VBE

VC=VCE-20 Ohm*0.5 A

15V=VCE-10V

VCE=5V

VCE=VC-VE
5 V=15 V-VE

VE=10 V-->Why its relevant to know VE in this case?

So the maximum power in this case would be P=5 V*0.5 A=2.5 W,is that it?

So we have to consider a case where Rload=0.00000000000000(.....)1 Ohm to predict the worst case?

If Ic=0.5A and the maximum power dissipated in the transistor is 1W, what is the largest value of Vce that can be tolerated?

If we have IC=0.5A ,the maximum value that VCE could assume not to exced 1W of dissipated power would be VCE=2 V.

 

Yes, you are starting to get it.

It's not that we *need* to know VE, it is just a convenient intermediate point in finding VCE. If we know VE and we already have Vc, then VCE=VC-VE. If we know the load current and we know the load resistance, we can find VE very easily.

Yes, "worst case" would be if the load were shorted. But that may or may not be a case we are particularly interested in. If the user shorts the load, the transistor will get hot and may get damaged. Well, there are lots of circuits that we use all the time that are not going to behave well if they are shorted. I think the question of how to deal with the shorted-load problem is not within the scope of the problem (but I'm guessing on that, so I could be wrong). It is worth notiing in your write up what the short circuit power dissipation is and mentioning that if the design is supposed to be short-circuit safe that the transistor needs better heat sinking (or some other method used to shut down the circuit if the transistor gets too hot).

But I think the better point of the problem is to determine the range of load resistances for which the circuit will function as intended. We know that the load can't be any bigger than 29Ω. As you have correctly determined, the power dissipated when the load is reduced to 20Ω is 2.5W, and so that is already too small of a load value. The question is, what load value will result in the power dissipation in the transistor being 1W?

You have all the pieces to answer that question. You've already determined, again correctly, that the maximum VCE that can be tolerated is 2V. If VCE is 2V, what is VE? If you know VE and Ic, what is the corresponding value of Rload?

 

Yes, you are starting to get it.

It's not that we *need* to know VE, it is just a convenient intermediate point in finding VCE. If we know VE and we already have Vc, then VCE=VC-VE. If we know the load current and we know the load resistance, we can find VE very easily.

Yes, "worst case" would be if the load were shorted. But that may or may not be a case we are particularly interested in. If the user shorts the load, the transistor will get hot and may get damaged. Well, there are lots of circuits that we use all the time that are not going to behave well if they are shorted. I think the question of how to deal with the shorted-load problem is not within the scope of the problem (but I'm guessing on that, so I could be wrong). It is worth notiing in your write up what the short circuit power dissipation is and mentioning that if the design is supposed to be short-circuit safe that the transistor needs better heat sinking (or some other method used to shut down the circuit if the transistor gets too hot).

But I think the better point of the problem is to determine the range of load resistances for which the circuit will function as intended. We know that the load can't be any bigger than 29Ω. As you have correctly determined, the power dissipated when the load is reduced to 20Ω is 2.5W, and so that is already too small of a load value. The question is, what load value will result in the power dissipation in the transistor being 1W?

You have all the pieces to answer that question. You've already determined, again correctly, that the maximum VCE that can be tolerated is 2V. If VCE is 2V, what is VE? If you know VE and Ic, what is the corresponding value of Rload?

If VCE=2V and VC=15V and IC=0.5

VCE=VC-VE

2 V=15 - VE

VE=13 V

VE=Iload*Rload

13 V=0.5 A*Rload

Rload =26 Ohm

The question is, what load value will result in the power dissipation in the transistor being 1W?

Therefore Pdissipated=VCE * IC
= 2 V *0.5 A=1 W


Thanks

 

Correct.

So now you know that the range of load values for which the circuit will operate properly is 26Ω to 29Ω. Less than that and the transistor will dissipate more than 1W, while above that the transistor will saturate and less that 0.5A will be delivered to the load.

 

Correct.

So now you know that the range of load values for which the circuit will operate properly is 26Ω to 29Ω. Less than that and the transistor will dissipate more than 1W, while above that the transistor will saturate and less that 0.5A will be delivered to the load.

Thanks for all the help