This is a very quick question, but I'm just in doubt here.
In the circuit below, is the voltage at node A 0 volts? I can see that it is connected to ground, but there is a zener-diode on the way as well. Does that have an impact?
So my question again, is the node A voltage 0V?
\$R_1=1k\Omega\$, \$R_2=5k\Omega\$.
The reverse breakdown voltage for both zener-diodes: \$V_z=1,5V\$
Both zener-diodes are ideal.
We don't know.
Without values for R1, R2 and DZ1 it is impossible to answer.
Quick yes or no? No can do.
From the comments:
With DZ2 removed, the current through R1, DZ1 and R2 must be 10V - 1,5 V / (6kohm) = 1,4166 mA.
That looks good.
The voltage across R1 is 1,416mA * 1000 ohm = 1,416 V.
The voltage across R2 is 1,416 mA * 6000 ohm = 7,08 V.
The voltage at node A must then be: vA = 7,08 V - 5 V = 2,08 V. Is this correct?
Yes.
Now we should be able to see the effect when DZ2 is reinserted. What will happen?
When DZ2 is reinserted I'm not so sure about. My initial thought would be that all the current now runs through DZ2, and none through R2, since current "prefers" a less resistive path.
Yes. DZ2 will be forward biased so it looks like a short circuit. (A real diode would have a forward voltage drop of about 0.7 V.) So the effect will be that A will be pulled down to ground.
With different values of R1 and R2 this might not be the case.
If all the current runs through DZ2, then that must mean that iDZ2 = iR1. Calculating iR1: iR1 = (5V − 1,5V − 2,08 V) / 1 kohm = 1,42 mA. That means that iDZ2 = 1,42 mA, but the correct answer is 2,5 mA. Where am I going wrong?
I've highlighted one error. Since A is now forced to 0 V the voltage across R1 is 5 - 1.5 = 3.5 V. Now you can calculate the current through R1.
But there's more. R2 is still in-circuit and some of R1's current will go through R2. Calculate how much will flow through R2, subtract that from the current through R1 and the remainder goes ...?
This circuit does not make much sense but it is appropriate to demonstrate some techniques for intuitively understanding and explaining circuits. Here is how I have done it in a few steps.
First I have accomodated the circuit diagram according to our needs:
Rearranging. As you can see, I have shown how the two power supplies (shown as real batteries) are connected. This allows to see and draw where currents flow in this circuit by closed paths (loops). The positive supply is drawn above and the negative supply below the zero voltage (ground) as our intuition tells us. So the elements with positive voltage drop across them are above and these with negative voltage - below the zero line.
Visualizing. The voltages (drops) are presented as vertical bars (in red) with proportional height that are on top of each other; thus it is easily seen how their heights (voltages) are summed. All this corresponds to our "gravitation notion" about the voltage source as a water column. The currents are represented by full closed paths in green with marked direction (not only by small arrows with arbitrary directions).
Then, we have to explore the circuit in an appropriate intuitive way:
Splitting. When some element (as Z2 here) is crusial for the circuit operation (and the answer), it is preferable to consider both cases - without and with this element... as though we split the circuit operation in two parts - before and now. That is why I have drawn two pictures - without (1) and with Z2 (2).
Thinking by analogies. Here a popular mechanical analogy can be to think of resistors as of springs (the lower the resistance, the stiffer the springs) and of Zener diodes as of stiff non-extendable rods (zero differential resistance). So the spring elasticity is proportional to its resistance and the length of the stretched spring to the voltage drop across it. Then this arrangement can be thought as of "tug of war" where the positive voltage source "pulls up" the common point A between resistors while the negative voltage source "pulls" it down. They do it through the resistors ("springs"); hence the names of such resistors - pull up resistor R1 and pull down resistor R2.
Without DZ2: So, when turning on the power supply in the first picture (without DZ2), the positive power V+ supply begins "pulling up" the point A through the pull-up resistor R1 and the Zener diode DZ1. At the same time, the negative power supply V- "pulls down" the point A through the pull-down resistor R2. Since R1 is five times stronger (stiffer) than R2, it will stretch five times less... and since DZ1 has zero differential resistance, it will not stretch at all. So the point A will rise with 2 cm (2 V) above ground...
With DZ2: When connecting DZ2, it will be forward-biased; so it "pulls down" the point A to 0.7 cm (0.7 V) above ground.
No, not necessarily. There will be some voltage drop across the zener diode. The data sheet for the zener diode will tell you how large it is.
Edit: This answer is mostly irrelevant now that the question has been updated.
The question was not quite right, but the simple answer is no, it cannot be at 0V. If ideal zener diode means that Vzbreakdown = 1.5V with Iz(min) ~ 0mA, then DZ2 will be forward biased, the voltage of A would be around 0.6V-0.7V.
The Zener diode needs to be biased at right value to have desired result, for example, make sure Dz1 is reverse biased with IDz1 > Iz(min).
