I ve got this circuit
,Assump that both D1, D2 conduct, V cut in all diodes 0.6V, I calculate V1=5-0.6=4.4V, then VR1 = 10-0.6-V1=10-0.6-4.4=5V, =>Ir1 =5/10=0.5mA positive means it follows the forward bias D1,is this correct?

 

Assuming D1 and D2 conduct. .. but how much do they conduct? What sort of diode are they? How warm are they?

 

Assuming D1 and D2 conduct. .. but how much do they conduct? What sort of diode are they? How warm are they?

V cut in 0.6V, all of them.. , assump that they totally conduct.

 

What you are supposed to be calculating?

 

What you are supposed to be calculating?

calculate ID1

 

calculate ID1

And you were given those resistor values? - I thought perhaps you have calculated them from somewhere.
Yes - there is 5V across R1, so the current is 5V/10k

 

And you were given those resistor values? - I thought perhaps you have calculated them from somewhere.
Yes - there is 5V across R1, so the current is 5V/10k

so both D1, D2 fully conduct?

 

hi lin,
Disconnect the D2 diode what is the voltage at V1?
E

 

hi lin,
Disconnect the D2 diode what is the voltage at V1?
E

d2 off, V1=10-0.6-Vr1.

 

hi lin,
Please post the actual numeric voltage value, not the equation

E

 

hi lin,
Please post the actual numeric voltage value, not the equation

E

cannot, the requirement is finding Id1 or current through R1, my work already at top.

 

Hi lin,
In order to check if the D2 is conducting, you must calculate the value of V1 when D2 is disconnected.

From the Voltages of +10V and -5V at the ends of the resistance network, what do calculate the total voltage across the 16.2k network.?

E

 

Hi lin,
In order to check if the D2 is conducting, you must calculate the value of V1 when D2 is disconnected.

From the Voltages of +10V and -5V at the ends of the resistance network, what do calculate the total voltage across the 16.2k network.?

E

D2 disconnected thus I remove the branch with D2, I also remove branch with D3 because no current through D3, the current through the R1, R2, R3 is = (10-(-5))/ (10+4+2.2) = 15/16.2=0.93mA

 

hi lin,
OK,
So what is the voltage drop across 10k with that current?

From that voltage drop value, you should be able to calculate V1.

E

 

hi lin,
OK,
So what is the voltage drop across 10k with that current?

From that voltage drop value, you should be able to calculate V1.

E

Voltage across 10k = 0.93*10=9.3V, thus V1= 10-0.6-9.3=0.1

 

hi
So if V1= 10V -(10,000 * 0.00093A) = ~0.1V
Will D2 be conducting??

E

 

It seems that it would be easier to calculate if all the voltages were positive. Therefore let us modify the circuit a little:

Lets say, -5V source is the Earth.
Previous Earth is +5V.
In the same way, +5V source is +10V , and +10V is +15V.

Let’s calculate the voltages relative to the ground. The ground is 0V.

Let's disconnect the diodes D2 and D3 at the beginning of the calculation.

Then, the current through the three resistors will be : (15-0.6): ( 10+4+2.2) = 0.889mA

Voltage R3-R2 is : 0.889x2.2= 1.95V
Voltage R2-R1 is : 0.889x(2.2+4)= 5.51V

 

hi
So if V1= 10V -(10,000 * 0.00093A) = ~0.1V
Will D2 be conducting??

E

yes, D2 conducting because +5V more than +0.1V at V1. V1=+0.1V less than +10V thus D1 is also on.

 

hi lin,
Yes,
A point to remember that the D2 diode will create a 0.6V voltage drop from the +5V.
Are there any more questions you need to ask for this topic.
E

 

hi lin,
Yes,
A point to remember that the D2 diode will create a 0.6V voltage drop from the +5V.
Are there any more questions you need to ask for this topic.
E

thus V2 = -5-I3R3=-5-0.93*2.2=-7.046V less than 0V thus D3 on, right??