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dimpledur said:Remember KVL? ƩV=0 around a loop. Apply this fact to solve for unknown voltages.
gneill said:This is one of those questions where the answers will depend upon the level of detail and the assumptions that you make; a simply posed question that could take pages to answer!
For example, is Vcc > Vc or is Vc > Vcc? (are the diodes forward or reverse biased by the supply voltages?). What about the region where the diodes are just beginning to turn on or off and their I vs V curves are nonlinear? What sort of model are you going to use for the diodes? Real diodes have small reverse bias leakage currents, will that be taken into account?
wtfman said:Well let's just say its not that complicated. Let's put Vc as 12 V and Vcc as 24. The diodes are forward basis where D1 is Silicon and D is Germanium. Thats all the details given. Just straightforward.
wtfman said:lol not for me. But still any ideas on the problem?
ok I am going to use the total voltage for the voltage drops in each diode. So for D1, the voltage drop would be Vd1 = VR - .7. Which is Vd1 = 36-.7 = 35.3gneill said:Sure. Since they're forward biased, pencil in the forward voltage drops for the diodes. What does that leave for the voltage drop across the resistor? (KVL)
wtfman said:ok I am going to use the total voltage for the voltage drops in each diode. So for D1, the voltage drop would be Vd1 = VR - .7. Which is Vd1 = 36-.7 = 35.3
For D, the voltage drop would be Vd = VR - .3. Which is Vd = 36-.3 = 35.7
Does that help?
gneill said:No, in fact it's rather bizarre The voltage drop across the silicon diode is 0.7V. That's it, that's all. The voltage drop across the germanium diode is 0.3V. That's it, that's all. So what's left for the resistor?
gneill said:Are you familiar with KVL? Can you write the KVL equation for the loop?
wtfman said:VR = Vcc -.7 -.3 + Vc ?
gneill said:Almost. Watch your polarity on Vc; when you pass through it on your way around the loop, you're going from its + terminal to its - terminal.
wtfman said:So VR= Vcc - .7- .3- Vc?
gneill said:Is that a guess? Why don't you explain your reasoning? Take me on a "round the loop tour" of the circuit.
wtfman said:ok well the voltage for the resistor is the Vcc added to the silicon diode. It is in forward basis, so its 12-.7 and then past the resistor into the second diode, again with forward basis, and then the last terminal where it goes from the + terminal to its - terminal into the ground.
Hence reasoning: VR= Vcc- .7-.3- Vc
gneill said:Well, that's more or less it. For KVL work it's better to deal with individual potential differences across the components rather than the potential at various nodes around the circuit with respect to ground.
So I would say: beginning the loop at point C (at the right end of the resistor) and proceeding clockwise, There is a voltage drop of 0.3V for the germanium diode, followed by a drop of 12V for Vc, followed by a rise of 24V for Vcc, and then a drop of 0.7V for the silicon diode. The sum is then: -0.3V - 12V + 24V - 0.7V = 11V. This remaining 11V will be the potential drop across the final component in our loop, the resistor, bringing us back to point C where we started. Thus the potential drop across R is 11V.
wtfman said:ok fair enough but what about the voltages ( A,B,C and D) across the different points or is it the same for the voltage drop across R?
Thanks again for all the help
gneill said:After you've established all the individual potential differences, then you can go back and sum them from the ground node to the individual nodes (you can work in either direction around the loop).
gneill said:Potential at point B with respect to ground: Starting with the ground node and heading to point B: Go from ground through Vcc through diode 1: +24V - 0.7V = 23.3V.
wtfman said:So at C would be 23.3V + 11 V = 34.3 V with respect to ground?
gneill said:Almost. The resistor has a voltage drop as you pass through it in the direction of current flow. That is, going from left to right through the resistor you have an 11V drop.
gneill said:Just so.
A diode is a specialized electrical component that allows current to flow in only one direction. When a diode is forward biased, it has a low resistance and allows current to flow through the circuit, resulting in a drop in voltage. When a diode is reverse biased, it has a high resistance and does not allow current to flow, resulting in no voltage drop.
The voltage drop across a diode can be calculated using Ohm's Law, V=IR, where V is voltage, I is current, and R is resistance. The resistance of a forward biased diode is typically around 0.6-0.7 volts. For example, if the current through a diode is 10 mA, the voltage drop would be 0.6-0.7 volts.
A diode can serve several purposes in a circuit, including rectifying AC signals to DC, protecting other components from reverse current, and regulating voltage levels. In some circuits, diodes can also be used as voltage references or temperature sensors.
Yes, multiple diodes can be used in a circuit. In series, they can be used to increase the voltage drop, while in parallel, they can be used to increase the current carrying capacity.
When incorporating diodes into circuit calculations, it is important to consider their voltage drop and current carrying capacity. This can affect the overall voltage and current in the circuit, and may require adjustments to the circuit design. It is also important to consider the direction of current flow and how the diodes are connected in the circuit.