# Voltage divider with diode

I'm having trouble analysing this circuit and was hoping someone could help me. The circuit in question is below.

I want to find the voltage drop and current across each of the parts. So far I've found the thevenin equivalent:

I'm a little confused as to how to find the current. Would I simply use $$\frac{11.2V}{13.33k\Omega}?$$

Or does the diode subtract 0.6 or so volts from the total voltage to be $$\frac{10.6V}{13.33k\Omega}?$$

• The second is correct – Simus994 May 2 '17 at 20:42
• And the next step would be to verify this current will actually make the diode into a region where it has the forward voltage drop you have assumed. – Eugene Sh. May 2 '17 at 20:44

Follow the arrows below. The first transition is due to making a Thevenin equivalent out of $V_1$, $R_1$, and $R_2$. The second transition is just summing the remaining two resistors:
There will therefore be about $600\:\textrm{mV}$ subtracted away from the Thevenin voltage, leaving $5\:\textrm{V}$ across the total resistance of $10\:\textrm{k}\Omega$. About $500\:\mu\textrm{A}$.