Hi.I have the following circuit and i am trying to answer the following questions.

1)Prove that there is negative feeback in the circuit.

2)Calculate Vz and R2 in order that the current in the load is equal 0.5A

3)If the load is reisistive what are the possible values for the load?

I am not seing how there is negative feedback when the loop is done by the non inverting input?

Thanks

 

The transistor is an inverter, and the feedback is taken from its output (collector) that means that + input acts as a - input and the - input acts as a + input with respect to the output of Q1.

 

The transistor is an inverter, and the feedback is taken from its output (collector) that means that + input acts as a - input and the - input acts as a + input with respect to the output of Q1.

The transistor is an inverter because the signal enters by the base and gets out by the emiter,rigth?I mean what you said is not true for all transistors with has to due with this specific configuration.

Now for calculating Vz and R2

I have the folowing data:

R1=10 ohm Pmáx(Zener)=0.2W Vsat=0,5V Vsupply=20V

For Vz i made this equation:


Vsupply=Vz+R2*Iz

Pzener=Vzener*Izener

I think that the current in the zener has to be same as in the load so

0,2=Vzener*0.5

Vzener=0.4V


Is this right?Or should i have done something like

Vsupply=R1*Ic+Vce+Vload

The transistor is suposed to be satured?Otherwise i do not know how to calculate VCE and Vload because i do not know the resistance of the load..

Thanks

 

The transistor is an inverter because the signal enters by the base and gets out by the emiter,rigth?

No, the transistor is an inverter because the signal enters by the base and gets out by the collector and then its feedback to op amp non inverting input.

Let as assume that the op amp inverting input voltage rise.
If so the op amp output voltage must decrease (fall). This reduce the BJT base and collector current. So the voltage drop across R1 is smaller. And this increase voltage at non inverting input and causes op amp output voltage to rise. And this rise in the op amp output voltage increase the base and collector current and also R1 voltage drop is larger. So the non inverting input drop.
As you can see the negative feedback tends to the previous state (to equilibrium).

 

The Zener voltage need to be equal to
Vz = R1 * I_Load = 10Ω * 0.5A = 5V
Because this circuit work as a constant current source.

 

Vsupply=Vz+R2*Iz

So 20=5+R2*0,5

R2=30 ohm

Now for question 3)If the load is reisistive what are the possible values for the load?

Vsupply=R1*Ic+Vce+Vload

Ic aprox Iload

20=10*0.5+0+Rload*0.5

So the maximum value of Rload=30 ohm,is this rigth?

Thanks

 

If Pzmax = 0.2W and Pz = Iz * Vz then
Iz = ??

 

Pz=Vz*Iz

0.2=5*Iz

Iz=0,04 A


Vsupply=Vz+R2*Iz

R2=375 ohm,right?

 

Yes, R2 shoudl by larger or equal to 375 ohm

 

Yes, R2 shoudl by larger or equal to 375 ohm

For question 3)

Vsupply=R1*Ic+Vce+Vload

Ic aprox Iload

20=10*0.5+0+Rload*0.5

So the maximum value of Rload=30 ohm


I have a question 4 that states:

Whereas the transistor can dissipate up to a maximum power of 1W, and tell
justified if the mount is well dimensioned.

PTransistor=Vce*Ic

PTransistor=15*0.5=7.5W,so the circuit is not well dimensioned since when we have Vce=15V, Pdissip=7,5W,i think...

 

But 15V is not the voltage across the transistor. Ignoring the base current (usually a reasonable thing to do, but not always), for power dissipation you want, as for anything, the current through a device multiplied by the voltage drop across that device.

You seem to be grabbing any old voltage or current and assuming that it is the one to use. For instance, the problem said to size things so that the current in the load was 0.5A and so you immediately assumed that this was also the current in the zener and in R2. Why? Then if someone provides the answer for that part of it (which they really shouldn't do, IMO), then you just grab it and go on. But are you understanding WHY they got the answer they did and WHY your initial assumption was wrong? In particular, are you understanding it well enough such that if you were given a different circuit that relied on the same principles you would be able to work it out?

One of the things you need to look at is the power dissipation in the transistor as a function of load resistance. In doing so, you will find a limit on the load resistance in order to keep the power dissipation in the transistor below 1W (assuming an unchanging configuation of 0.5A of collector current).

A good exercise would be for you to write up an explanation (a couple paragraphs) of what this curcuit does, how it does it, and how you size the components. Write this as though someone that is just a little behind you is going to read it to understand how this circuit works.

 

But 15V is not the voltage across the transistor. Ignoring the base current (usually a reasonable thing to do, but not always), for power dissipation you want, as for anything, the current through a device multiplied by the voltage drop across that device.

You seem to be grabbing any old voltage or current and assuming that it is the one to use. For instance, the problem said to size things so that the current in the load was 0.5A and so you immediately assumed that this was also the current in the zener and in R2. Why? Then if someone provides the answer for that part of it (which they really shouldn't do, IMO), then you just grab it and go on. But are you understanding WHY they got the answer they did and WHY your initial assumption was wrong? In particular, are you understanding it well enough such that if you were given a different circuit that relied on the same principles you would be able to work it out?

One of the things you need to look at is the power dissipation in the transistor as a function of load resistance. In doing so, you will find a limit on the load resistance in order to keep the power dissipation in the transistor below 1W (assuming an unchanging configuation of 0.5A of collector current).

A good exercise would be for you to write up an explanation (a couple paragraphs) of what this curcuit does, how it does it, and how you size the components. Write this as though someone that is just a little behind you is going to read it to understand how this circuit works.

I consider VCE=15V because Voltage drop in the resistor R1=5V,so the transistor is in the saturation state so VCE=0V...

 

I consider VCE=15V because Voltage drop in the resistor R1=5V,so the transistor is in the saturation state so VCE=0V...

Tale a look at this. Do you see the contradiction? You say you consider VCE=15 because you have concluded that VCE =0V. Well, which is it? 15V or 0V?

Also, why does a voltage drop of 5V across R1 mean that the transistor is in saturation?

Finally, the information you provided in your second post stated that Vsat was 0.5V. So how can it be 0V.

 

Tale a look at this. Do you see the contradiction? You say you consider VCE=15 because you have concluded that VCE =0V. Well, which is it? 15V or 0V?

Also, why does a voltage drop of 5V across R1 mean that the transistor is in saturation?

Finally, the information you provided in your second post stated that Vsat was 0.5V. So how can it be 0V.

LOL.what a mess

VCE=0.5V,and the transistor is in saturation because we are trying to get a bigget current from the collector is able to give,since the voltage is increasing in the non inverting input and therefore Ib and IC increase also.

 

LOL.what a mess

VCE=0.5V,and the transistor is in saturation because we are trying to get a bigget current from the collector is able to give,since the voltage is increasing in the non inverting input and therefore Ib and IC increase also.

No.

Let's say that we have the circuit operating in a state of equilibrium so that the current flowing through the collector resistor is just the right about to make the voltage at the non-inverting input of the opamp equal to the voltage at the invering input. Now let's say that there is some disturbance in the system -- perhaps the resistor heats up a bit and its resistance increases. The increase in the resistance results in a larger voltage drop across it and this, in turn, results in the collector voltage (and hence voltage at the non-inverting input of the opamp) going down. But if the voltage at the non-inverting input goes down, the voltage output of the opamp will go down. This lower output voltage means a lower base voltage on the transistor, which results in less collector current. But a lower collector current means less current in the collector resistor, whch means a smaller voltage drop across it, which means that the collector voltage will rise and the voltage at the non-inverting input of the opamp will rise back toward equilibrium.

This is the essence of negative feedback -- disturbances in the output result in a chain of events that act to reestablish the equilibrum.

None of this has anything to do with the collector-emitter voltage of the transistor, other than it has to remain above Vcesat in order for the transistor to remain active, which is required in order for the feedback mechanism to function.

If you have a 20V supply and you have 5V dropped across the collector resistor, then you know that Vc (the collector voltage) is 15V. The collector votlage and the collector-emitter voltage are two very different things. The 15V is the voltage across both the transistor AND the load. The collector-emitter voltage is just the voltage across the transistor.

 

No.

Let's say that we have the circuit operating in a state of equilibrium so that the current flowing through the collector resistor is just the right about to make the voltage at the non-inverting input of the opamp equal to the voltage at the invering input. Now let's say that there is some disturbance in the system -- perhaps the resistor heats up a bit and its resistance increases. The increase in the resistance results in a larger voltage drop across it and this, in turn, results in the collector voltage (and hence voltage at the non-inverting input of the opamp) going down. But if the voltage at the non-inverting input goes down, the voltage output of the opamp will go down. This lower output voltage means a lower base voltage on the transistor, which results in less collector current. But a lower collector current means less current in the collector resistor, whch means a smaller voltage drop across it, which means that the collector voltage will rise and the voltage at the non-inverting input of the opamp will rise back toward equilibrium.

This is the essence of negative feedback -- disturbances in the output result in a chain of events that act to reestablish the equilibrum.

None of this has anything to do with the collector-emitter voltage of the transistor, other than it has to remain above Vcesat in order for the transistor to remain active, which is required in order for the feedback mechanism to function.

If you have a 20V supply and you have 5V dropped across the collector resistor, then you know that Vc (the collector voltage) is 15V. The collector votlage and the collector-emitter voltage are two very different things. The 15V is the voltage across both the transistor AND the load. The collector-emitter voltage is just the voltage across the transistor.

Therefore

For question 3)

Vsupply=R1*Ic+VCEsat+Vload

Ic aprox Iload

20=10*0.5+0.5(in the worst case,since we only have to guarantee that VCE>=0.5V) +Rload*0.5

So the maximum value of Rload=29 ohm


I have a question 4 that states:

Whereas the transistor can dissipate up to a maximum power of 1W, and tell
justified if the mount is well dimensioned.

When the transistor is at cut(do not know the term in English but in this case we have IC=0,IB=0 and VCE=VCC

PTransistor=Vce*Ic

PTransistor=15*0.5=7.5W,so the circuit is not well dimensioned since when we have Vce=15V, Pdissip=7,5W,i think..

Thanks

 

Therefore

For question 3)

Vsupply=R1*Ic+VCEsat+Vload

Ic aprox Iload

20=10*0.5+0.5(in the worst case,since we only have to guarantee that VCE>=0.5V) +Rload*0.5

There you go ignoring units again. This is the last time I am going to try to help you if you aren't going to pay attention to units.

Vcc = R1*Ic + Vce + Vload

Vce = Vcc - R1*Ic - Vload

Since Vce >= Vcesat = 0.5V

Vce = Vcc - R1*Ic - Vload >= Vcesat

Therefore

Vload <= Vcc - R1*Ic - Vcesat

Vload = Ic*Rload

Ic*Rload <= Vcc - R1*Ic - Vcesat

Rload <= (Vcc - Vcesat)/Ic - R1

Rload <= (20V - 0.5V)/0.5A - 10Ω = 39Ω - 10Ω = 29Ω

Is it really SO hard to show your work and track your units? If I can do it when it is not even my problem, why can't you?

So the maximum value of Rload=29 ohm


I have a question 4 that states:

Whereas the transistor can dissipate up to a maximum power of 1W, and tell
justified if the mount is well dimensioned.

I'm not even sure what this means, since we are given no information about the intended range of load values. What you need to do is find out what range of resistances you can have as the load without overheating the transistor. Compare that to the limit that you have on the maximum value fo the load. If there is an overlap, then I would say that the "mount" is "well dimensioned" for that range of load values. If there is no overlap, then it isn't.

When the transistor is at cut(do not know the term in English but in this case we have IC=0,IB=0 and VCE=VCC

PTransistor=Vce*Ic

PTransistor=15*0.5=7.5W,so the circuit is not well dimensioned since when we have Vce=15V, Pdissip=7,5W,i think..

Once again, you are being inconsistent. You talk about IC=0 and then immediately use IC=0.5A. Which is it? Is it IC=0 or IC=0.5A?

What is the maximum Vce that can be tolerated without overheating the transistor? What limit does this place on the value of Rload?

 

There you go ignoring units again. This is the last time I am going to try to help you if you aren't going to pay attention to units.

Vcc = R1*Ic + Vce + Vload

Vce = Vcc - R1*Ic - Vload

Since Vce >= Vcesat = 0.5V

Vce = Vcc - R1*Ic - Vload >= Vcesat

Therefore

Vload <= Vcc - R1*Ic - Vcesat

Vload = Ic*Rload

Ic*Rload <= Vcc - R1*Ic - Vcesat

Rload <= (Vcc - Vcesat)/Ic - R1

Rload <= (20V - 0.5V)/0.5A - 10Ω = 39Ω - 10Ω = 29Ω

Is it really SO hard to show your work and track your units? If I can do it when it is not even my problem, why can't you?

That result is what i said in the last post and it has units on it...

I'm not even sure what this means, since we are given no information about the intended range of load values. What you need to do is find out what range of resistances you can have as the load without overheating the transistor. Compare that to the limit that you have on the maximum value fo the load. If there is an overlap, then I would say that the "mount" is "well dimensioned" for that range of load values. If there is no overlap, then it isn't.

Once again, you are being inconsistent. You talk about IC=0 A and then immediately use IC=0.5A. Which is it? Is it IC=0 or IC=0.5A?

What is the maximum Vce that can be tolerated without overheating the transistor? What limit does this place on the value of Rload?

The transistor can be in one of two states in this montage as far i as i have understood.So for determing the maximum amount of power i consider a situation where the load is consuming a current of 0.5 A and we have VCE=15V..Although i think that when we have VCE=15 V that implies Ic=0 A and Ib=0A,right?So the situation where the maximum amount of power is dissipated would be something like:


P(DissipatedTransistor)=VCEsat*Ic
P(DissipatedTransistor)=0.5V *0.5A=0.25W since <=1W the montage is well dimensioned.

Thanks

 

It's not enough to just tack the units you know you want to end up with to the final answer. You need to include units at every place in the computation in which you have a value that has units associated with it. You don't need to put units on R1 or Ic, because the symbols carry the units of resistance and current. But as soon as you write 0.5 there needs to be the proper units (unless it is dimensionless). 0.5A and 0.5V are NOT the same thing.

I'm serious about this. People DIE because of engineers that are too damn lazy to be bothered to carry their units. If you can't be bothered to track your units, then go into something where the carnage is much less, such as medicine. At least incompetent doctors are generally limited to killing people one at a time; incompetent engineers kill people in job lots.

When you say the transistor can be in one of two states, what two states?

You've already found a maximum value of the load resistor for the circuit to function properly based on the fact that any larger value will result in the transistor going into saturation.

Now you need to find a minimum value of the load resistor for the circuit to function properly based on the fact that smaller values will result in the transistor dissipating too much power.

So, again, what is the voltage across the transistor, Vce, when it is dissipating its rated power?

 

It's not enough to just tack the units you know you want to end up with to the final answer. You need to include units at every place in the computation in which you have a value that has units associated with it. You don't need to put units on R1 or Ic, because the symbols carry the units of resistance and current. But as soon as you write 0.5 there needs to be the proper units (unless it is dimensionless). 0.5A and 0.5V are NOT the same thing.

I'm serious about this. People DIE because of engineers that are too damn lazy to be bothered to carry their units. If you can't be bothered to track your units, then go into something where the carnage is much less, such as medicine. At least incompetent doctors are generally limited to killing people one at a time; incompetent engineers kill people in job lots.

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

When you say the transistor can be in one of two states, what two states?

You've already found a maximum value of the load resistor for the circuit to function properly based on the fact that any larger value will result in the transistor going into saturation.

Now you need to find a minimum value of the load resistor for the circuit to function properly based on the fact that smaller values will result in the transistor dissipating too much power.

So, again, what is the voltage across the transistor, Vce, when it is dissipating its rated power?

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

I did not understood this very well:

Now you need to find a minimum value of the load resistor for the circuit to function properly based on the fact that smaller values will result in the transistor dissipating too much power.

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

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

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..