Sample exam problems (diodes)

In summary, the conversation revolved around solving sample problems given by an instructor for an upcoming exam. The first problem involved analyzing a circuit with two diodes and determining the output voltage for different input voltages. The second problem caused confusion due to the use of a 500 ohm resistor, but it was eventually solved by considering the voltage drop across the diode and choosing the appropriate voltage value. The third problem involved a zener diode and analyzing the current flow for different input voltages. The conversation also touched on the importance of understanding the problem statement and choosing the correct values for voltage drops.
  • #1
asdf12312
199
1

Homework Statement


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Homework Equations





The Attempt at a Solution


My instructor gave us some sample problems he said might be on the exam, I tried working them out but I was having difficulty with 3 of them:

For the 1st problem, when there is positive value for voltage, I got the 1st diode as off (open circuit) and the 2nd diode as on (short circuit). This would mean Vout is always 0 for positive voltage. When the voltage becomes negative, the 1st diode becomes a short and the 2nd becomes open circuit. So 20||20=10 and Vo=(0.5)Vin using voltage division. Graph for Vout would be 0 when the 5V peak sine wave is positive curve, and a curve of half the amplitude when it is a negative curve (min=-2.5).

For the 2nd problem, I was used to working with 1K ohm resistors so the 500 ohm kinda confused me. But if the value of resistor doesn't matter, I find that diode starts conducting when Vi=2.5 and conducts fully when Vi=2.7 (+2V). This would mean graph is a linear line (Vo=Vi) until Vi=2.7 when it becomes a horizontal line. For the -3V, diode starts conducting when Vi=-2.5 and conducts fully when Vi=-2.3. Graph would look similar to the +2V one.

For the 3rd one, I had a zener diode for another problem so I am assuming there is positive value at the negative end. Also VZ=VZ0 because rz is very small. For +20V, I got current flowing through diodes B, C, D. the voltage is 20-2Vd-Vz=10.4V. For +5V, I got a negative voltage after the voltage drop across zener diode so I found no current would flow through the loop. For -5V, I got a positive voltage after the voltage gain in zener diode, so zero current again. For the last one (-20V), I found current flowing through doides A, C, E if I assume zener diodes can conduct voltage in both ways. Then the voltage is -20+Vz+2Vd=-10.4V.
 
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  • #2
(a) is A-OK.

For (b), the resistance certainly does matter.
When Vin = 0 what is the current flowing thru the lower diode? You have to try both Vd = 0.5V and Vd = 0.7V and pick the right one.

Then, run Vi from -5V to +5V to get the output. What is Vout when Vin = -1.9V? When Vin = -2.1V?

Does the top diode ever turn on?

You gt (c) right also.

Why would the zener not conduct current the same way for a negative 20V input as for a positive 20V input? The zener current is in the same direction for both +20 and - 20V inputs.
 
  • #3
(b) I don't really understand this type of circuit, usually with the previous problems i had only 1 diode and voltage battery. When vi=0, the current through lower diode would be (0-Vd+3) diovided by resistance. I'm not really sure what you mean by choosing Vd=0.5 or 0.7 though. I get I=5 or 4.6mA depending on this.

When Vi=-1.9 and -2.1, current has to flow through bottom diode only. I think vout would be -2.3V for both because that's when the diode finishes conducting.

am I doing this right so far? Maybe the vout graph has to include both values then, for both the +2 and -3V sources. So I should graph both for vi=-5 to +5V. But I think my only problem is what vout would be at Vin=0V, -2.3V or just 0V.

also, for (c), would I be right in thinking that if I graphed VR, it would look like the sin wave input except with amplitude of 10.4V? And V0 would be a constant value instead of a curve, +/- 9.6V depending on the sine value.
 
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  • #4
If the input is -1.9V, and the voltage drop across the bottom diode is 0.5V, what would be the current?

And what about the top diode? What input voltage would turn it on?
 
  • #5
asdf12312 said:
(b) I don't really understand this type of circuit, usually with the previous problems i had only 1 diode and voltage battery. When vi=0, the current through lower diode would be (0-Vd+3) diovided by resistance. I'm not really sure what you mean by choosing Vd=0.5 or 0.7 though. I get I=5 or 4.6mA depending on this.

Remember what the problem stated: if i < 1 mA, Vd = 0.5 and if i > 1 mA then Vd = 0.7v. So which do you think you should choose?
 
  • #6
I see now. If input is -1.9V, the current through the bottom diode is (-1.9-Vd+3)/500=1.2mA. But this is >1mA..so I should choose the bottom diode Vd=0.7V right? For the top diode using a similar problem I was assigned I found it starts conducting at 2.5V.
 
  • #7
For the top diode: what's the highest voltage the anode can assume?
 

1. What are diodes and how do they work?

Diodes are electronic components that allow current to flow in only one direction. They have two terminals, an anode and a cathode, and are made of a semiconductor material. When a voltage is applied in the correct direction, the diode conducts current. However, when the voltage is applied in the opposite direction, the diode blocks current flow.

2. How are diodes used in circuits?

Diodes have a variety of uses in circuits, including rectifying AC signals to DC, protecting circuits from reverse voltage, and regulating voltage levels. They can also be used in combination with other components, such as capacitors and resistors, to create different types of circuits like filters and oscillators.

3. What is the difference between a forward biased and reverse biased diode?

A forward biased diode is when the voltage is applied in the direction that allows current to flow. In this state, the diode has a low resistance and acts like a closed switch. On the other hand, a reverse biased diode is when the voltage is applied in the opposite direction, and the diode has a high resistance, acting like an open switch.

4. How do you calculate the current and voltage in a diode circuit?

The current and voltage in a diode circuit can be calculated using Ohm's Law. The current can be calculated by dividing the voltage by the resistance, or I = V/R. The voltage can be calculated by multiplying the current by the resistance, or V = IR. It is also important to take into account the diode's forward voltage drop and reverse breakdown voltage in these calculations.

5. What are some common failure modes of diodes?

Some common failure modes of diodes include open circuit, short circuit, and high reverse leakage current. Open circuit failure occurs when the diode breaks and current cannot flow through it. Short circuit failure happens when the diode's anode and cathode are connected, causing a high current flow. High reverse leakage current occurs when the diode fails to block current flow in the reverse direction, which can cause damage to other components in the circuit.

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