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* you have one battery with 5V,
* you have one battery with 8V (mind the direction of the voltage vector)
* you have a 10 ohms resistor
* you have the diode.
General solution:
* (First sight) assume the voltage across the resistor to be 0V, what voltage is across the diode?
* how much current is flowing through the resistor?
* in what direction?
* calculate the voltage drop at the resistor using ohm´s law
* the sume of voltage in a circuit (circle) is 0V. Knowing that you can calculate the voltage drop across the diode.
right picture:
* (first sight) assume the diode current is 0mA, what voltage is across the diode?
* with this voltage: do you expect current flowing through the diode?
--> if no: solved.
--> if yes: look into the datasheet of the diode and get a clue what voltage across the diode you can expect.
* With that value calculate all currents within the circuit.
* again look into the datasheet with the calculated diode current an see if your voltage drop meets your expectations.
* If there is a deviation, then do the calculations with the new diode voltage.
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Other solutions:
For the right picture you can paint and calculate an equivalent circuit with only one resistor and one battery.
With that you can
- either: precisely calculate the currents and voltages
- or: build a graphical solution wher you paint: battery voltage, resistor U-I, diode U-I, and find the crossing between the U-I lines.
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Did you expect values?
;-)
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If you need more help, then show us your thoughts and calculations...
Usually a practical diode model is a series of a battery generating Vd, a resistor Rd and an ideal diode. The ideal diode has no voltage drop when forward biased while no current is flowing through it if reverse biased. You can see the model in the attachment.
If the voltage between anode and cathode is less than Vd (anode more positive than the cathode), the ideal diode will be reversed biased then no current will be flowing into Rd (so in this case you can consider the whole circuit as an open switch)
So first of all remove the diode from you circuits and calculate the voltage between the nodes where anode and cathode have to be connected. If it is less than Vd or negative (this last means the cathode node is higher in voltage than the anode) then this value will be the voltage across the diode since it will be reversed biased, since it will behave as an open switch. This is the case of your circuit on the left
If instead the voltage at the node where the anode has to be connected is higher than that on the node that has to be connected to the cathode and the difference is higher than Vd, replace the diode of your circuits with Vd and Rd of the practical model (it is not necessary to place also the ideal diode, since in this case it will behave as a closed switch), then solve the circuit following the normal rules. This is the case of your circuit on the right. Notice that you have to know the values of Vd (usually 0.6 V for a Si diode) and Rd to obtain a numerical solution
Left circuit - clearly the diode is reverse biased by upto 3v, so only leakage currents will flow and voltage across diode will be ~ -3v
Right circuit - 10v batt will reverse bias it, while 20v batt (thru the 10K) will forward bias it. So voltage at the resistor/ diode node will be minus (-) 0.7v w.r.t Gnd which is the voltage across the diode.
@klaus
Thanks for the advise
Direction of the Green symbol is the +ve direction
It's National Instrument (MultiSim + Ultiboard) for design , simulation and PCB
zeyad
- - - Updated - - -
@ klausst
the direction of current if +ve it means it's the direction of the green arrow ,-ve means opposite direction here another view of simulation with a multimeter, note the sign of (+/-) on it.
best regards
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