I was playing around with my voltmeter and I came across something I don't know how to work the math of.
The circuit built was the one below, the right power supply was a 9V battery and the left was a six volt. Both resistors R1 and R2 are 1K ohms.
What is the math to determine how much current will go through (certain points on the circuit) R2 and D1?
Diodes are non-linear devices so we need to make some assumptions (a.k.a. guess) about what operating state they are in.
With \$V_D = 9V\$ and \$V_A = 6V\$, I'm going to assume that D2 is forward biased and D1 is reversed biased. The arrows show the assumed direction of current flow. I'm calculating with \$R1 = R2 = 1k\Omega\$ and I'm assuming there's a 0.7V forward voltage drop for each diode.
That means there is a near-zero current flow through D1 so effectively \$V_A\$ is not part of the circuit. That means we're left with D2, R1, and R2 in series.
We can assume that D2 has some fixed forward voltage drop. The actual amount depends on what diode you have, though a basic silicon diode will typically drop \$V_{D2} = ~0.7V\$. Different diodes will have different voltage drops, for example a small red LED can have a 2V forward voltage drop and Schottky diodes can have a ~0.2V forward voltage drop. The best bet is to look at the datasheet for whatever diode you are using.
So \$V_C = V_D - V_{D2} = 8.3V\$
To solve for \$V_B\$ we just have a simple voltage divider circuit.
\$V_B = V_C \cdot \frac{R2}{R2 + R1} = 4.15V\$
Now that we've solved for all the voltages, we must check our assumptions.
Namely, \$V_D > V_C\$ and \$V_A < V_B\$
The assumption about D2 was correct (\$V_D > V_C\$), but the assumption about D1 was not.
That means we must change our assumptions and re-solve the problem.
This time I'm going to assume that both diodes are forward-biased.
Like before, \$V_C = 8.3V\$ (assuming 0.7V diode voltage drop). However, this time \$V_B\$ is also dictated by a diode voltage drop:
\$V_B = V_A - V_{D1} = 5.3V\$
Now that we know the nodal voltages, we can solve for currents.
\$I_{D2} = I_{R1} = \frac{V_C - V_B}{R1} = 3mA\$
\$I_{R2} = \frac{V_B}{R2} = 5.3mA\$
\$I_{D1} = I_{R2} - I_{R1} = 2.3mA\$
Time to check our assumptions. The current flowing through both diodes are in the forward bias direction so both of our assumptions were correct and we're done.
I'm going to call V1 the voltage on the left in series with D1 and V2 the voltage on the right in series with D2.
The first thing to note is that the forward voltage drop for each diode is VF(D1) and Vf(D2). These values are usually 0.6V to 0.7V for a silicon diode.
Here, I drew it out for clarity:
The voltage between A and GND (VAGND) is V2 - VFD2
Similarly, VBGND = V1 - VFD1
Since we know the voltage either side of R1, we can calculate IR1 (and IV2)
IR1 = (VAGND-VBGND)/R1
IR1 = ((V2-VFD2)-(V1-VFD1))/R1
IR1 = (V2 - VFD2 - V1 + VFD1) / R1
Also, We know the voltage across R2 also:
IR2 = VBGND/R2
IR2 = (V1-VFD1)/R2
IV1 = IR2 - IR1.
IV1 and IV2 are the currents sourced by V1 and V2 respctively
Its become easier to analyze the circuit when you forget the forward bias voltage for the diodes. ( Once you understand the trip points, you can then consider them later ).
Lets call the left Cell as V1 and the right one as V2.
So lets replace diodes by switch. Which is closed when forward biased and open when reverse.
Now if the voltage at the junction of R1 and R2 would be half of V2 ( V2 * (R1 + R2) / 2 ) when D2 is forward biased. Just keep that in mind.
Case 1: V2 < V1
No current flows through R1 since D2 is open. ( reverse biased ). Current through R2 is entirely supplied by V1 so it could be
Ir2 = V1/R2
adjusting for assumptions:
Ir2 = (V1 - Vdf ) / R2; where Vdf is the forward drop across Diode.
Ir1 = 0; ( since D2 is reverse biased ).
Case 2: V2 > 2 times V1
Now V2 is powerful enough to create a voltage at the node of R2 and R1 so as to shut down D1.
Ir1 = Ir2 = V2 / (R1 + R2); ( since both are in series ).
Thus adjusting for assumptions,
Ir1 = Ir2 = (V2 - Vdf) / (R1 + R2);
Case 3: V2 > V1 but V2 < 2 times V1
This is the case where both the diodes are forward biased. For example as in your case, where V1 = 6V and V2 = 9 V.
Its quite easy to solve when you look at the current flowing through resistors R1 and R2 and the voltages across them.
Lets assume Id1 and Id2 are the currents flowing through the diodes. so we have,
Id2 is flowing thorugh R2, and Id1 + Id2 flowing through R1.
So we have two equations.
V1 = R2 * ( Id1 + Id2 );
V2 = V1 + ( Id2 * R2);
adjusting for forward drops;
V1 - Vdf = R2 ( Id1 + Id2);
V2 - Vdf = (V1 - Vdf ) + ( Id2 * R2 );
Since we have the values for R1 and R2 as R = 1K, putting them in the equation we have;
V1 - Vdf = R ( Id1 + Id2); -----( 1 )
V2 - Vdf = (V1 - Vdf ) + ( Id2 * R ); -----( 2 )
Solving for Id2 in equation ( 2 ) we have;
Id2 = ( V2 - V1 ) / R;
Substituting Id2 in the equation ( 1 ) we get,
V1 - Vdf = R ( Id1 + ((V2 - V1 ) / R ) );
that boils down to;
Id1 = ( 2V1 - V2 - Vdf ) / R;
Examples.
Solving for V2 = 9V and V1 = 6V , R = 1K and Vdf = 0.7Volts,
Id2 = (9 - 6 ) / 1000 A = 3mA;
Id1 = ( 2 * 6 - 9 - 0.7 ) / 1000 = (12 - 9 - 0.7 ) / 1000 = 2.3mA;
Solving for V2 = 9V and V1 = 7.5V , R = 1K and Vdf = 0.7Volts,
Id2 = (9 - 7.5 ) / 1000 A = 1.5mA;
Id1 = ( 2 * 7.5 - 9 - 0.7 ) / 1000 = (12 - 9 - 0.7 ) / 1000 = 5.3mA;
