# Diode question help here please

We have the circuit in the figure.I have to find the currents in the diodes and in the resistances R1 and R2. The diodes are real and Vd=0.7 V

Awesome.

Here is the original figure

I have drawn the circuit and I have replaced the diodes with the voltages

I apply the KVL in the first loop and I have -20+ 0.7+0.7 - I2R2=0 so I2 here is negative...what am I doing wrong?

• what are the diodes? are the arrows supposed to be the diodes? Aug 24, 2015 at 10:34
• Are "awesome" diodes also invisible? Aug 24, 2015 at 10:37
• I have replaced the diodes in the figure with their voltages of 0.7 Volts
– Xizi
Aug 24, 2015 at 10:38
• The arrows show the direction of the currents
– Xizi
Aug 24, 2015 at 10:39
• I added the original figure
– Xizi
Aug 24, 2015 at 10:41

Assuming all the diodes take 0.7V.

VR1 is parallel over D2 = 0.7V

VR2 = 20V- 2* 0.7V = 18.6V

With these values known you can calculate the currents

• I don't understand why is VR1 0.7 Volts?
– Xizi
Aug 27, 2015 at 20:26
• over the diode there can be no more than 0.7V of voltage the resistor is connected parallel over the max voltage of the diode 0.7V so the voltage will be the same Aug 28, 2015 at 0:28
• I understand.I want to know,what Vi interval should I study to know when the diodes are on or off?
– Xizi
Aug 28, 2015 at 10:36
• If there a voltage below the forward voltage (0.7V) then that voltage will also be over the diode If the voltage rises above 0.7V there will only be 0.7V over the diode. A diode is on (closed switch) if there is a voltage of 0.7V across the diode. if it's less than 0.7V or a negative voltage the diode blocks current. Aug 28, 2015 at 10:55
• Does 20 Volt voltage influence that?
– Xizi
Aug 29, 2015 at 10:45

In the direction of the current I2 is selected as you have drawn, It just should be negative. Your 20 volt battery (which is much higher than 0.7) drives all the currents clockwise.