Simple Diode Qs: +3v/+1v & 3mA/4mA - Max

[maxsthekat]

Homework Statement

Here is a link to two figures I drew from a book I am using http://www.flickr.com/photos/96575810@N00/4339202724/sizes/o/

These figures are assuming ideal diodes (no voltage drop, no resistance).

In this example, on the left hand circuit, the voltage is listed as +3v and the current is 3mA. For the right hand circuit, it's +1v and 4mA.

What I don't understand is why this is... For the left circuit, this seems to make sense, as +3 to GND is the largest difference. So, does this mean in multiple voltage scenarios like this we can simply ignore the other (lesser) voltages?

For the right hand circuit, why would the voltage at that point be +1?

I know these are really basic, but I don't quite understand it. Any help is appreciated :)

-Max

 

[maxsthekat]

Sorry, correct link: http://farm3.static.flickr.com/2751/4339202724_87ac1be74c_o.png

 

[cepheid]

On the the left-hand circuit, all three diodes are initially forward biased, because all three cathodes are tied to ground, and all three anodes are at voltages higher than ground. HOWEVER, once a current develops, the properties of ideal diodes mean that the diodes behave like a short. There is no voltage drop across them. Considering the top diode, what this means is that the current will flow across the diode and then through the resistor. All of the voltage will develop across the resistor. Hence, the top end of the resistor must be at +3 V. Therefore, since the top end of the resistor is connected to all three cathodes, the bottom two diodes become reverse biased and don't conduct. That's why they don't "matter" (in the sense of affecting the output).

In the right hand circuit, it's a similar situation, except that this time it is the lowermost diode that becomes the only one to conduct, and for things to be consistent, the voltage drop across the resistor must be such that you're down to +1 V at the end of it.

 

[vk6kro]

In the right hand drawing, the bottom diode will conduct because there is enough voltage on the anode of this diode to make it forward biased. The voltage at the anode will be 1 volt because the diode is perfect.

The other two diodes have +1 volt on their anodes and +2 or +3 volts on their cathodes, so they are reverse biased and cannot conduct. So, V is 1 volt and the current will be 4 mA if the resistor is 1000 ohms.


Yes, you are right... any non-conducting diodes, and the voltage sources they are connected to, can be removed if you like because they have no effect on the circuit operation.

 

[DeeNos]

I can explain that the voltage listed in the left circuit (+3v) is the potential difference between the diode's anode and cathode. This means that the anode is at a higher potential (+3v) compared to the cathode, which is at ground (0v). This potential difference allows the diode to conduct current, resulting in a current of 3mA flowing through the circuit.

In the right circuit, the voltage listed (+1v) is also the potential difference between the anode and cathode. However, in this case, the anode is at a lower potential (+1v) compared to the cathode, which is still at ground (0v). This means that the diode is reverse biased and will not conduct current.

To answer your question about ignoring other voltages, in an ideal diode scenario, we can assume that the diode will only conduct current when the anode has a higher potential than the cathode. In multiple voltage scenarios, this means that we can ignore any lower voltages and only consider the highest potential difference. However, in real world scenarios, diodes do have a voltage drop and some resistance, so it is important to take those into account when analyzing circuits.

I hope this helps to clarify the concept of diode behavior in these simple circuits. If you have any further questions, please let me know.