HOW the D1 is OFF and How the circuit is flowing for the values -10<= Vs<=10?
i would be glad if explained with image of the circuit and the direction of the current flowing...

diode is ideal with Vf= 0.7 V. The solution given here is tough to understand please give me a easier explanation. and plot for the Vin vs Vout.

Moderators note: reduced font size, as large fonts are like shouting

 

Try picking some voltages for Vs and seeing what Vo turns out to be.

Also, where did this explanation come from? It says that D1 is OFF if Vs > 0 V. Do you agree with that?

Start with Vs << 0 V and determine which diodes are ON and OFF and what the output is.

Then determine what the lowest value of Vs will be that causes one (or possibly more) of the diodes to change stage. This is a critical point. Determine what Vo is at that value of Vs.

Keep repeating this until you are at a high enough value of Vs that no higher value can cause a diode to change state.

Since the circuit is piecewise linear, the transfer characteristic will be a series of straight lines connecting the Vs,Vo pairs at the critical points.

 

Hello,

With Vs open circuited Vo should be 0v. That's because the two 10k resistors and pairs of two series diodes are the same top and bottom, so the result is +10v across the top section and -10v across the bottom section so the output is zero. That means Vs is also at ground, zero volts. When Vs goes slightly positive, diode D1 becomes reverse biased.
Do the math over again to prove this and see what you can find out.

 

Amarnath, try to visualize it this way. Imagine that there is no Vs for a moment. Current will flow from the +10 V through the resistor, through all four diodes, through the bottom resistor, and into the -10 V source. Can you see that?
The amount of current flowing through the top resistor is just enough to keep 0.7 volts at the connection of D1 and D2. Do you see that?

Now add in Vs, and make it a positive value, like +1 volt. Can you see how D1 becomes reversed biased? No current flows through a diode that is revered biased.
Now think about where the current that was flowing through D1 now goes.
Does that help?

 

I would also like to participate in this discussion!

Why everybody says that D1 is reverse biased for Vs > 0 V???

Let's picture this:
Let's say there is a voltage drop at top resistor of 3 V. Then, D1 cathode voltage would have to be 0.7 V lower so that D1 remains ON, hence 2.3 V. So if Vs is like 2 V, then D1 will still be forward biased, no?

 

When Vs is zero, the amount of current through the 20K resistor at the top ensures that the voltage at the anode of D1 is 0.7 V, therefore if Vs goes up, D1 becomes reverse biased. So you can't just say that there's a voltage drop of 3V. Only Vs is a variable in this setup.

 

When Vs is zero, the amount of current through the 20K resistor at the top ensures that the voltage at the anode of D1 is 0.7 V, therefore if Vs goes up, D1 becomes reverse biased. So you can't just say that there's a voltage drop of 3V. Only Vs is a variable in this setup.

Think about this. You seem to be assuming that the voltage at the top of D1 remains at 0.7 V even if Vs goes up. It doesn't.

What if Vs = 3 V? Then the current through the top 20 kΩ resistor will be such that the voltage at the anode will be 3.7 V.

On the face of it, it might seem that D1 will be forward biased until Vs = 9.3 V. However, what also needs to be taken into account is the ability of the circuit to source current into D1.

 

Think about this. You seem to be assuming that the voltage at the top of D1 remains at 0.7 V even if Vs goes up. It doesn't.

What if Vs = 3 V? Then the current through the top 20 kΩ resistor will be such that the voltage at the anode will be 3.7 V.

On the face of it, it might seem that D1 will be forward biased until Vs = 9.3 V. However, what also needs to be taken into account is the ability of the circuit to source current into D1.

Yes, agreed. If Vs is 3.0 V, then it works as you say. But that's different than just making the resistor node 3 V out of thin air.

I don't believe the explanation of the circuit is entirely correct. The diodes are all floating in a balanced arrangement, so a change in Vs gets reflected at all four nodes of the diode array and the diodes remain in a state of "on the edge" of conduction.....until we get to something like Vs = 4.3V.

 

Yes, agreed. If Vs is 3.0 V, then it works as you say. But that's different than just making the resistor node 3 V out of thin air.

I don't believe the explanation of the circuit is entirely correct. The diodes are all floating in a balanced arrangement, so a change in Vs gets reflected at all four nodes of the diode array and the diodes remain in a state of "on the edge" of conduction.....until we get to something like Vs = 4.3V.

What do you mean by the diodes are all "floating in a balanced arrangement"?

How are the diodes "on the edge of conduction"?

If Vs = 0 V, then Vo is also 0 V and no current flows through the load. But each diode has 233 uA of current flowing in it. That's not "one the edge of conduction" and I don't see anything floating.

If Vs = 2 V, then there is 100 uA of current in the load which means that there is 100 uA more current in D2 than in D4. Similarly, there is 100 uA of current being supplied by Vs resulting in 100 uA more current in D3 than in D1. That's not what I would call "floating in a balanced arrangement".

What we don't know is how much current if flowing in D1 or D2 (as long as we use an ideal constant-Vd diode model). But if we use the exponential model, then we can determine the actual currents (which will also show that vo is not quite the same as Vs, too).

 

WBahn, I agree with you. I mistook the 0.7 volts as "on the edge" without thinking about the current. I thought about it later and decided that the diodes are doing just fine. Your description is now my description. I think I was trying to make the textbook explanation make sense. It doesn't.

 

I would like to see what conclusion do we get from this circuit.

I don't fully understand all your thoughts but I was trying to make my own.

I was wondering if Vo is not only influenced by the 10 V power supply, the top 20 kΩ resistor, the D2 and the bottom right 20 kΩ resistor (assuming a constant voltage drop for diodes of 0.7 V)?

Can't I write the following equation:

10 V = 40 kΩ*I + 0.7 V, where 20 kΩ*I is Vo?

 

I would like to see what conclusion do we get from this circuit.

I don't fully understand all your thoughts but I was trying to make my own.

I was wondering if Vo is not only influenced by the 10 V power supply, the top 20 kΩ resistor, the D2 and the bottom right 20 kΩ resistor (assuming a constant voltage drop for diodes of 0.7 V)?

Can't I write the following equation:

10 V = 40 kΩ*I + 0.7 V, where 20 kΩ*I is Vo?

No. You are making the classic mistake of throwing any I and any R at Ohm's Law in order to find some unrelated V.

Ohm's Law relates a resistance to the voltage across THAT resistance and the current flowing through THAT resistance.

Your equation uses the same I flowing through the to resistor and the load resistor. Yet there is clearly another path available to the current after it flows through the top resistor and D2. So why would you assert that it must ALL flow through the load resistor and that NONE of it can flow through D4 and then through the bottom resistor and to the -10 V supply?

 

I see your point... I have just simulated the circuit in falstad site and I see now that there is a situation when all diodes are ON and contributing to Vo!

But in the simulation I was not expecting that for Vs >= 0 V, the diodes were immediately ON. I was expecting that they were ON only when Vs was lower than the voltage "after" the top resistor, I mean the voltage measured at the node connecting D1 and D2 wrt GND.

Spoiler: Link

http://www.falstad.com/circuit/circ...1953125+0+-1 o+9+64+0+550+5+0.000390625+1+-1

 

Think about this. You seem to be assuming that the voltage at the top of D1 remains at 0.7 V even if Vs goes up. It doesn't.

What if Vs = 3 V? Then the current through the top 20 kΩ resistor will be such that the voltage at the anode will be 3.7 V.

On the face of it, it might seem that D1 will be forward biased until Vs = 9.3 V. However, what also needs to be taken into account is the ability of the circuit to source current into D1.

Hi there,

I think using your method brings in the question of what effect the 20k load resistor has on the biasing of diode D1. As the input voltage (on the left) rises, the load resistor has more and more of an effect because the current through that resistor has the secondary effect of pulling the node at the top of the bridge down. That combined with the fact that the input voltage is rising eventually reverse biases diode D1 at a voltage much lower than 10v. It depends on the load too. Picture a 1 ohm resistor where the 20k load resistor is now. The only difference is the 20k takes longer to show the effect of the current draw on the top node.

But we can probably sweep all this aside anyway, because for this problem the answer is already given and so they are banking on the load resistor drawing the top node down at least a little immediately. That would cause a reverse bias as soon as the input went above zero by even 1 microvolt. Do i agree with that? Not really, but they did show the view they want to take on this already and although it is oversimplified, that's the way they wanted it. Giving a more reasonable answer like "3.1v" would probably cause a failure of that question unless the instructor was given advanced notice of the alternate interpretation.

We could probably prove that the top node voltage does not follow the input for just any input voltage level but stops short of 10v, much short of that most likely, because the output impedance (load) is not infinite.

To visualize the effect, simply remove the load resistor and think about the circuit. Then once you've reached a certain input voltage, say +5v, replace the load resistor and think about what that extra current draw does to the top node voltage. It MUST pull it down at least a little. All other things equal, D1 stops conducting.

 

I see your point... I have just simulated the circuit in falstad site and I see now that there is a situation when all diodes are ON and contributing to Vo!

But in the simulation I was not expecting that for Vs >= 0 V, the diodes were immediately ON. I was expecting that they were ON only when Vs was lower than the voltage "after" the top resistor, I mean the voltage measured at the node connecting D1 and D2 wrt GND.

Each diode IS only on when the voltage drop across them is 0.7 V (using the simple model that we are).

D1 is only on when Vs is 0.7 V lower than the voltage "after" the top resistor.
D2 is only on when Vo is 0.7 V lower than the voltage "after" the top resistor.
D3 is only on when Vs is 0.7 V greater than the voltage "before" the bottom resistor.
D4 is only on when Vo is 0.7 V greater than the voltage "before" the bottom resistor.

This is why I have been telling people ever since Post #2 that they need to start with Vs at a large negative voltage and then identify the critical points at which one or more diodes change state. There aren't that many such points. At most there are four (one for each diode).

But no one seems willing to do that.

 

Each diode IS only on when the voltage drop across them is 0.7 V (using the simple model that we are).

D1 is only on when Vs is 0.7 V lower than the voltage "after" the top resistor.
D2 is only on when Vo is 0.7 V lower than the voltage "after" the top resistor.
D3 is only on when Vs is 0.7 V greater than the voltage "before" the bottom resistor.
D4 is only on when Vo is 0.7 V greater than the voltage "before" the bottom resistor.

This is why I have been telling people ever since Post #2 that they need to start with Vs at a large negative voltage and then identify the critical points at which one or more diodes change state. There aren't that many such points. At most there are four (one for each diode).

But no one seems willing to do that.

Hello again,

We can do that if you like, but what are your thoughts on post #14 ?

I also am including a diagram to illustrate.
In figure 1 we have the circuit with no load and zero volts in.
In figure 2 we have the circuit with zero volts in and the 20k load.
In figure 3 we remove the load, then apply +0.1 volts input.
In figure 4 we add a load resistor of unspecified value, but we should think about what happens to the top bridge node that was +0.8 volts before we put the load back. We can think of the load as going from say 1 megohm down to 1 ohm for example, passing through 20k. We can then set it to 20k to find the final voltage.

Note however this still ignores the fact that the answer was already given, but we still want to see what happens in a more non-ideal case. We can probably come up with a more realistic answer, although for any exam the OP would have to follow the text more verbatim.

I think i get reversed bias at 3.1 volts, but i'll have to go over this to double check. You might also want to check my results for the currents and voltages shown. Note i left some of the #4 circuit node voltages up for discussion.
Also keep in mind that the left hand node is an INPUT, while the right hand node is always an OUTPUT.

LATER:
Using the corrected equation i get close to 3.1v for Vin now, and that agrees with a simulation of the circuit using the assumptions mentioned.

Here's an analysis of the circuit for the top node voltage. This analysis also inserts resistances R1 in series with each diode to sense current through the diode. Without at least some resistance the diode acts as a pure voltage source so i dont think that's the best idea, although it is an alternate.

Equation for Vtop:
[Temporarily removed for correction...will replace later with the corrected version]
[The corrected version is almost ready and yields 3.1v]

Notes:
Vp is the positive voltage source value, and the absolute value of the negative source voltage value (so this is 10).
Vtop is the voltage at the top node of the diodes.
Rtop is the top resistor value, which is 20k, and that's the bottom one too.
RL is the load resistor, which is 20k in the schematic.
Vd is the diode voltage, which would be 0.7 volts.
Rd is the small resistance in series with each diode. I used 1 microhm for numerical calculations while using 128 digits of numerical precision. This resistor is used to sense current as well as ensure that the diode acts as a passive device and not a pure voltage source.

What we can do with this equation...
1. Calculate Vtop, obviously.
2. Calculate Vtop-Vin, which gives us diode D1 voltage. When this voltage drops below exactly 0.7v the diode no longer conducts.
3. Calculate diode D1 current.
4. Determine when the diode stops conducting, given the assumption that there must be some finite, non zero positive resistance in each diode. Since this violates the premise that the original question assumes we will get a more realistic result but it will not match the results of the text exactly.

It should be noted that using real diode models will give us different results too.

Also of interest is if we set R1=0 that means the diodes have no series resistance at all and so act as pure voltages sources and this leads to Wbaln's solution where:
Vtop=Vin-Vd

up to the point where the top resistor current goes to zero.

 

Hello again,

This came out very interesting. Using the correct version of the network equation, i get Vin=3.1v with a limited precision of 16 digits, but when i go higher, i get very nearly zero volts! And this is using the same equation with the same values for components.

I'll post the full equation a bit later hopefully after i have had time to go over it again. It gives the same results as the simulator though using 16 digits of precision, but going with more digits gives what must be the more accurate result which very nearly equals the original text (0v).

After the component values given for the circuit are inserted into this equation, the result is:
13333333334*Vin/13333333335+46666666688/66666666675-0.7=0

and solving for Vin we get -2.325e-10 which is very small, nearly zero. A plot of the above would show that as Vin goes from some negative value to some positive value, we get zero when Vin is nearly zero, and the line would be a straight, diagonal line that passes through zero.

Note that this result can not be confirmed with a circuit simulator because most of them use 16 or 17 digits of precision for the calculations. The only way is to write the equations and solve them symbolically as much as possible, then finally insert the component values.

I'll also try inserting a very tiny resistance in series with the input source Vin a little later and see what happens.

In general this means this is a very stiff circuit where the results depend highly on nearly exact values and assumptions.

 

Hello again,

Here's the final equation for the top node at the top of the diode bridge:

Vtop=(2*Rd^2*Rtop^2*Vp+4*Rd^3*Rtop*Vp+Rd^4*Vp+4*
Rd*Rtop^3*Vin+4*Rd^2*Rtop^2*Vin+Rd^3*Rtop*Vin+4*Rd*Rtop^3*
Vd+8*Rd^2*Rtop^2*Vd+2*Rd^3*Rtop*Vd)/(4*Rd*Rtop^3+10*Rd^2*
Rtop^2+6*Rd^3*Rtop+Rd^4)

You can experiment with this if you like by changing some values and see what happens.
This is the actual network equation that is exact in itself, but it is a little mysterious too because accuracy may depend highly on the precision used to calculate values and on what assumptions about the diode are acceptable.

The components in the circuit are as follows:
Vp is the top source voltage (10v in the circuit), and the absolute value of the negative source voltage at the bottom of the circuit,
Vd is the fixed diode voltage (0.7v in the circuit),
Vin is the input voltage assumed to be a constant source with zero impedance,
Vtop is the voltage at the top of the diode bridge junction of D1 and D2,
Rtop is the resistor at the top of the circuit (20k in the circuit) and also the load resistance (20k in the circuit) which makes the equation a little simple, and also the resistance at the bottom of the circuit (20k also). Calling all three the same variable makes the equation simpler, since they are the same anyway, and
Rd is the small resistance placed in series with each diode to sense current and force the overall diode to be a passive element not an active element.

Again we can calculate the voltage at the top node with this equation, and we can calculate the current through the diode D1 with:
i1=(Vtop-Vin-Vd)/Rd

This allows us to sense the direction of the current through D1, and if it reverses then we know it must be reversed biased.

Now when i insert any 'real' values for Rd like 1, 0.1, etc., i see a balance that occurs at a particular input voltage Vin, where the current through the load starts to load down the top node voltage Vtop. This voltage again came out to 3.1 volts, but what is with a real world value for R1. This has been verified with a circuit simulator for several values of components.
Using a non real world value, we get either 0v or that the top node voltage follows the input voltage much higher as Wbahn suggested, depending on how you want to interpret the resulting equation when setting Rd equal to zero:
Vtop=Vin+Vd

There are two ways to interpret this. Either (A) the Vtop voltage follows the input voltage, or (B) the Vtop voltage minus the input minus Vd solves for the voltage across Rd, and the result is zero:
i1=(Vtop-Vin-Vd)/Rd
i1=(Vin+Vd-Vin-Vd)/Rd
i1=0

So we reach a physical dilemma when we try to define the diode as a pure voltage source. Thus, the paper must be implying that there is at least some resistance internal to the diode in order to be able to call it a passive element.

This was quite interesting