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You're correct that the zener 2 is not in zener mode.
But if the voltage is 0.7V, then wouldn't zener 1 also have a voltage of 0.7V?
In which direction do you get a voltage drop of 0.7V?
Is that the same direction as for the zener mode?
Circuit with zener Diodes
- Thread starter AD633
- Start date
But if the voltage is 0.7V, then wouldn't zener 1 also have a voltage of 0.7V?
In which direction do you get a voltage drop of 0.7V?
Is that the same direction as for the zener mode?
I get 230.8Ω. Generally it is reasonable to use three significant figures, so you could call this 231Ω. You answer is close enough that I doubt you would lose points.
You need to spend some time reviewing what you write. You meant to say
\( P{_Z_1}=P{_Z_2}=0.8 W\)
and even this is not correct and has caused you to get a wrong answer.
What you should have said was
\( P_{Z_1max}=P_{Z_2max}=0.8 W\)
This is important because just because the two zeners each have a maximum current they can handle does NOT mean that they are each handling that maximum current. In general, as you increase the supply voltage you will reach the current limit of ONE of the zeners first.
\(
So what is the voltage across R2?
Does this make sense?\)
The drop voltage of 0.7 V occurs from the anode to the cathode,so its in the oposite direction of the drop voltage in zener mode.
But if the potential in the mesh of the diode < than 0.7 V ,we would still have a drop voltage of 0,7 across the diode?
I can't write the equation of the mesh
\(
0.7 V=500 Ohm*I500 Ohm
I 500 Ohm=0.0014 A
V 500 Ohm=(0.0014 A)*(500 Ohm)=0.7 V \)
Seems wrong to me....
No, you are forgetting something. For a regular diode, the forward voltage drop is 0.7V and the reverse (breakdown) voltage is -50V (or less). There is (essentially) no current flowing in either direction between these two voltages.
For a zener diode, the forward voltage drop is 0.7V and the reverse voltage is specified (in this case 5V, actually it should be -5V). So ...
For a zener diode, the forward voltage drop is 0.7V and the reverse voltage is specified (in this case 5V, actually it should be -5V). So ...
No, it is not correct.
The nice thing about circuit analysis is that you can almost always check your work to verify that the answer is, indeed, correct.
If you supply 98.826V to the circuit, then (assumiing the zeners don't vaporize), the current flowing through R1 is
I_R1 = (98.826V - 10V)/230.1Ω (using your value for R1)
I_R1 = 386mA
The current through R2 is going to be
I_R2 = (10V-5V)/200Ω
I_R2 = 25mA
The current through the 1kΩ is 10mA.
Therefore, the current through Z1 is
I_z1 = 386mA - 25mA - 10mA = 351mA
and the power is Z1 is
P_z1 = I_z1 * Vz1 = 351mA * 10V = 3.5W
Iz1 could not have such a high value.Its maximum value should be 80mA
So the problem its in the values of the currents I500 Ohm,IR2,Iw and IR1,rigth?
I think that the values for R1 and R2 are correct.
So the problem can only be in the values of the currents.
\(
Iz1max=80mA
Iz2max=160mA
I500 Ohm =10 mA
IR2=Iz2max+I500Ohm=170mA
Iw=I1kOhm+IR2=10mA+170mA=180mA
IR1=Iw+Iz1max=180mA+80mA=260mA
\)
I am not seing what current values are wrong..
Thanks
If you don't want it to have that current, then don't apply 98V to the input!
We've already established that your values for R1 and R2 are correct.
\(
Did you not even bother to read my earlier post on this point?
http://forum.allaboutcircuits.com/showpost.php?p=625799&postcount=22
What is the voltage across R2 if it has 170mA flowing in it? Does this make sense?
On what basis do you possibly claim that the actual current in Z2 is Iz2max?
Think of it this way.
Take a plank and set it on top of two cans. The first can will support 50lb before crushing and the second can will support 100lb before crushing. You now load the plank right in the middle (which means that an equal amount of weight is applied to each can) and ask someone to tell you what the maximum weight is that you can put on the plank. They come back and say that it is 150lb because the first can will support 50lb and the right can will support 100lb. What would you tell them?\)
Sorry i hadn't seen your previous post.
VleftR2-VRightR2=R2*IR2
10 V-5 V=R2*IR2
5V=200 Ohm*170mA
5V =! 34 V \) ,
it doesn't make sense to have a 170 mA current on R2.
So if we increase the voltage we will reach the limit of current of one of the zeners first.I suppose that will be Z1,since it can only support a maximum current of 80 mA.So i have to start from there and them figure what are the values of the other currents in the circuit.
\(
IR1=Iz1max+Iw
\)
The problem is that Iw depends on the value of Iz2,because IR2=Iz2+I500Ohm and i don't know the value for Iz2 in this case
I supose i have to something like this,since i know VleftR2 and VrightR2
\(
10 V-5 V=R2*IR2
5 V=200 Ohm*IR2
IR2=25mA
IR2=IZ2max+I500Ohm
25 mA=Iz2max+10mA
Iz2max=15 mA
Iw=25mA+80mA=105mA
IR1=80 mA+Iw=80mA+105mA=185mA
\)
\(
Vsupplymax=R1*IR1+R2*IR2+500Ohm*I500Ohm
Vsupplymax=230,1 Ohm*180mA+200Ohm*25mA+500Ohm*10mA=51,418V \)
Checking the current that goes through R1
\(
IR1=(51,418 V-10 V)/(230,1 Ohm)=180mA.<185mA\)
It may be right now....
Thanks
\(
VleftR2-VRightR2=R2*IR2
10 V-5 V=R2*IR2
5V=200 Ohm*170mA
5V =! 34 V \) ,
it doesn't make sense to have a 170 mA current on R2.
That is wrong because one the cans can only support 50lb of pressure so it will crack when the pressure exceed that value.
So if we increase the voltage we will reach the limit of current of one of the zeners first.I suppose that will be Z1,since it can only support a maximum current of 80 mA.So i have to start from there and them figure what are the values of the other currents in the circuit.
\(
IR1=Iz1max+Iw
\)
The problem is that Iw depends on the value of Iz2,because IR2=Iz2+I500Ohm and i don't know the value for Iz2 in this case
I supose i have to something like this,since i know VleftR2 and VrightR2
\(
10 V-5 V=R2*IR2
5 V=200 Ohm*IR2
IR2=25mA
IR2=IZ2max+I500Ohm
25 mA=Iz2max+10mA
Iz2max=15 mA
Iw=25mA+80mA=105mA
IR1=80 mA+Iw=80mA+105mA=185mA
\)
\(
Vsupplymax=R1*IR1+R2*IR2+500Ohm*I500Ohm
Vsupplymax=230,1 Ohm*180mA+200Ohm*25mA+500Ohm*10mA=51,418V \)
Checking the current that goes through R1
\(
IR1=(51,418 V-10 V)/(230,1 Ohm)=180mA.<185mA\)
It may be right now....
Thanks
And this kind of check falls under the category of "always, always, always ask if the answer makes sense".
If you are saying that the load would only be 50lb, then that is wrong because the load is shared equally between the cans. So the max load is 100lb because that is the load at which at least ONE of the cans reaches its max load.
Right answer, but completely wrong reasoning. If Z1 could support 8000mA it would still reach its limit first.
Think about it.
Once the supply voltage is high enough to turn on the 10V zener, the rest of the circuit becomes static and does not change any more as the supply voltage continues to increase. Why is that?
NO!
Iz2max is 160mA. It is a parameter. The maximum current that can be allowed to flow in Z2. What you have found here is the ACTUAL current in Z2, which is Iz2. You then need to verify that Iz2<Iz2max.
You've defined Iw to be the sum of the current in R2 and the current in the 1kΩ resistor. Why are you all of a sudden claiming that the current in the 1kΩ resistor is 80mA. What would the voltage across this resistor be if that were true? Does this answer make sense?
Before you said that IR1=185mA. Now you are using 180mA. Why?
Keep checking.
Q1) Of the 180mA, how much goes through R2?
Q2) Of the 180mA, how much goes through the 1kΩ resistor?
Q3) How much is left that has to go through Z1?
Q4) Is this equal to Iz1max? Not just less than, but equal to it (within roundoff error) since the point was to find the MAXIMUM supply voltage that could be used without exceeding the power ratings on the zeners.
Because if the zener Z1 is on,the voltage on the rest of the circuit is imposed by the Vz of the zener.The voltage applied on the rest of the circuit does not chage anymore.(nor the current that goes through Iz2)
Yes i shouldn't have wrote Iz2max.What i meant was that it is the maximum value of current allowed in Z2,so that Iz1<= Iz1max,which is also < Iz2max
It would be 37 V,which does not make sense.
I made mistake when calculating Iw
\(
Iw=I1kOhm+IR2=10mA+25mA=35mA
IR1=35mA+80mA=115 mA
\)
\(
Vsupplymax=(230,1 Ohm)(35mA)+(200 Ohm)*(25mA)+500Ohn*(10mA)=36,46V \)
\(
IR1=(36,46 V-10 V)/(230,1 Ohm)=115 mA
IR2=(10 V - 5 V)(200 Ohm)=25mA
I10kOhm=(10V)/(1kOhm)=10mA\)
So the current that goes through
\( IZ1=115mA-25mA-10mA=80mA=IZ1max \)
which i suppose this means that i have found Vsupply max.
For zener 2
\( IR2=IZ2+I500 Ohm
IZ2=25mA-10mA=15mA<160mA \)
It feels right now.
For question 3) when Vsupply is equal to 5 V
Do i have to find all values for the currents again?
Can i apply KVL to determine IR1?
\(
IR1=(5 V-0,7 V)/(230,1 Ohm)=18,7 mA \)
\(
Iw=IR1+I1k
I1kOkm=(0,7V)/(1kOhm)=700 uA
Iw=18,7mA+700uA=19,4mA
IR2=Iz2+I500Ohm \)
,since i don't know neither of them i supose i can't go this way
\( IR2=Vleft R2-VRight R2=0,7 V-0,7 V=0V \) so there isn't current flowing through R2,wich means i am not seing this well
Thanks
Yes, you have question 2 correct now. Now see if it is clear that you could right down a simple equation, by inspection, to solve for the maximum supply voltage. You know the following:
1) R1 has (Vsupply-10V) across it.
2) R2 has 5V across it.
3) The 1kΩ resistor has 10V across it.
4) Z1 has 80mA through itt.
KCL allows you to then write down, by inspection:
\(
\frac{V_{supply}-10V}{231\Omega}=\frac{5V}{200\Omega}+\frac{10V}{1k\Omega}+80mA
\)
It takes practice and experience to get to this point, but you will get there a lot quicker if you develop the habit of using 20/20 hindsight to look at problems after you have solved them and see if parts of the solution are actually reasonably obvious. And by "habit" I mean that you get to the point where you just automatically do it in practically every problem you work as part of your ritualistic "does the answer make sense" check.
1) R1 has (Vsupply-10V) across it.
2) R2 has 5V across it.
3) The 1kΩ resistor has 10V across it.
4) Z1 has 80mA through itt.
KCL allows you to then write down, by inspection:
\(
\frac{V_{supply}-10V}{231\Omega}=\frac{5V}{200\Omega}+\frac{10V}{1k\Omega}+80mA
\)
It takes practice and experience to get to this point, but you will get there a lot quicker if you develop the habit of using 20/20 hindsight to look at problems after you have solved them and see if parts of the solution are actually reasonably obvious. And by "habit" I mean that you get to the point where you just automatically do it in practically every problem you work as part of your ritualistic "does the answer make sense" check.
In general, no. You are looking for the voltage across the 500Ω resistor (which is labeled Node C in the drawing I provided previously), so you only have to do as much work as is needed to find that voltage.
Possibly. It depends on how you apply it.
Why are you saying that the zener voltage is 0.7V? It is still reverse biased and unless the reverse bias voltage exceeds the zener voltage, it looks like a reverse-biased diode, meaning only that it has no current flowing in it.
As yourself this: If a diode is reverse biased such that it has no current flowing in it, does it have any impact of the circuit analysis?
Zener 1 is no longer reversly biased,since it has Vz=10 V and Vsupply is equal to 5 V.
About zener 2 i don't know if it is directly or reversly biased,since i don't the current that goes through IR2,so i cant find the tension in node C.
Thanks
The bias of a diode is the anode voltage minus the cathode voltage. Notice that there is nothing in that about the zener voltage.
If it is forward biased, that means that current must be flowing from bottom to top through it. Does that make sense?
Go back and review what the zener voltage means.
In order to be forward biased, the voltage at the cathode would be 0.7V LESS than the voltage on the anode. The anode of both zeners is tight to ground (0V), so that would require that Vsupply by -0.7V or less.
If you agree that the voltage will be less than the zener voltage, then it will behave just like a normal diode. How would you analyze the circuit if you replaced the zeners with normal diodes (and be sure to keep the orientation the same, namely with the anodes tied to ground)?
Zener Voltage is the voltage imposed by the diode when it is inversly polarized.
But for the zener to be reversly polarized the voltage from the cathode to the anode,Vsupply as to be >= Vz or not?
I would replace the diode by a 0,7 V voltage source,with minus on the anode of the diode and plus on the cathode,that is the drop voltage would be 0,7 V from the anode to the cathode.
\( 10 V=0,7V+R1.IR1
IR1=46,5 mA
\)
From this i can compute Iw,rigth?
Iw=47,2 mA.And how can i calculate IR2,now?
Thanks
Not quite. The zener voltage is a limit on the amount of reverse bias voltage that can exist. If you reverse bias a 10V zener by, say, 6V then the voltage will be 6V and no current will flow. Just like a normal diode. But as soon as you try to exceed a reverse voltage of more then 10V, the diode will start conducting and will allow as much current to flow as necessary to keep the voltage from rising above 10V.
Sounds just like a description of a forward-biased diode except with a 10V knee instead of a 0.7V knee, right? That's a good way to think of it. The mechanism is different, but the effect is largely similar.
No. To be reverse polarized, the voltage from the anode to cathode simply has to be negative, just like any other diode. The magnitude of the reverse diode has to be (or try to be) greater than the zener voltage in order to place the diode into zener breakdown.
Two things: (1) If the minus is on the anode, then the drop would be 0.7V from cathode to anode. (2) You ONLY replace the diode with a 0.7V source IF the diode is forward biased enough to place it into forward conduction. This requires that there be a current flowing from anode to cathode.
I don`t know if - in the discussed context - it is important for AD633, however for the sake of exactness I think we can agree that the voltage across the device will not exactly be constant (say 10volts).
As far as I remember, the nominal voltage Uz,n is defined as a voltage corresponding to 50% of the maximum allowed current Iz.
That means: For other currents the voltage across the Z-diode is slightly above or slightly below this nominal voltage - eqivalent to a dynamic (differential) resistance r,z that is rather low (some ohms).
To be exact: r,z contains also a temperature dependent part r,z,th .
It is clear that this dynamic resistance must be taken into account when the quality of any voltage stabilization is to be calculated.
Definitely agree, though I made a point of not mentioning it because (1) the problem does not give enough information to account for that behavior, and (2) I think it is better not to muddy the waters to much right now and to keep things simple.
As with the forward conduction characteristic, the reverse conduction characteristic has a voltage-current relationship and is NOT just a constant voltage at any current. But the dynamic resistance is generally considerably smaller than the dynamic resistance when forward biased.
Interestingly, there are two modes of breakdown - zener and avalance. Below 5.6V the zener dominates and above it the avalance dominates. But one of them (zener, I think) has a negative temperature coefficient while the other has a positive tempco. But at 5.6V the two modes are pretty much balanced and the device is nearly temperature independent.
As with the forward conduction characteristic, the reverse conduction characteristic has a voltage-current relationship and is NOT just a constant voltage at any current. But the dynamic resistance is generally considerably smaller than the dynamic resistance when forward biased.
Interestingly, there are two modes of breakdown - zener and avalance. Below 5.6V the zener dominates and above it the avalance dominates. But one of them (zener, I think) has a negative temperature coefficient while the other has a positive tempco. But at 5.6V the two modes are pretty much balanced and the device is nearly temperature independent.
Yes - I also definitely agree.
And - to complete the picture - the temperature-dependent part of the differential resistance is, of course, practically zero for 5.6 V.
For all other cases it is
* either positive or negative, and
* frequency dependent because of the thermal capacity of the device. That means: The influence of this effect is larger for very slow current (power) changes and rather small for quick changes.
