# What is the reason for getting 7.8 V as output in between two circuits with two voltage sources?

Since I am beginner to circuit design, I am in progress of understanding different basic circuits behaviour.

I designed the below switching circuit with clear transistor calculation. I am using this circuit to run a fan, the circuit in right is an internal circuit from fan, but in order to understand its complete behavior when I try to simulate it in LTspice. I am getting 7.8 V as collector output. I would like to understand what is the reason.

Since there is two voltage sources 12 V as transistor circuit, and 7.5 V as fan internal circuit. I expected 12 V as output but why 7.8 V.

Circuit:

Output :

Thank you

• When Q1 is off there's current flowing into the R6, R7, R8 junction, and there is a voltage drop across R3. There are several different ways of calculating the drop. Aug 15, 2023 at 13:13
• If you're truly switching the fan on and off with M1, then you shouldn't need the bias network formed by R6 and R7, or the series resistor R8. Just run the collector of Q1 to the M1 gate. You'd only need that bias network if you're trying to run M1 in a linear mode. Aug 15, 2023 at 15:10

When you have multiple sources of voltage and you want to calculate the voltage on a node, you need to add the effect of each voltage source. This means, first, you need to calculate the voltage of this point, assuming you only have the 12V source and the other one is zero. Then calculate the voltage of this point, assuming you only have the 7.5V, and the other one is zero. Then add these two values.

Let's assume we only have the 12V source. By putting the V1 to zero, it means that the terminal is connected to the ground. Now, the voltage of the point of interest is simple divider be R3 and (R6 || R7). This will lead to 12*(R6||R7)/(R3+(R6||R7)) that will result to 12*6.66666/16.66666 that is equal to 4.8V.

Do the same for the 7.5V source, and you will end up with 3.0V. This means the voltage of this point is 4.8+3.0=7.8V.

Superposition theorem statement

• 7.8V is correct. Something is wrong with your calculation. The question is using LTspice - it properly shows 7.8V. The same circuit in CircuitLab also shows 7.8V.
– JRE
Aug 15, 2023 at 13:47
• You are right. The error came from me rounding up the values. By using more precision, it turned out the same correct value. Aug 15, 2023 at 15:12
• @Saadat. Your answer is correct. However, you are using superposition, not Thevenin's Theorem. Superposition is another fine tool for solving circuits. Aug 15, 2023 at 19:01
• Thanks for pointing out the mistake. I updated the answer. Aug 16, 2023 at 8:48

Certain parts of the circuit can be ignored at certain points of analysis. What they are, comes with experience. But when the transistor Q1 is off, the circuit you have is essentially this:

simulate this circuit – Schematic created using CircuitLab

Then there is a very useful tool for analyzing circuits called Thevenin Equivalents. R6, R7, and V2 can be replaced by a single voltage (Thevenin equivalent voltage) and a single resistor (Thevenin equivalent resistance).

simulate this circuit

Solving this circuit gives 7.8 V for Vtest!

To find the Thevenin equivalent of a voltage divider, such as R6 and R7, the Thevenin equivalent resistance is just the equivalent resistance of the resistors in parallel. The Thevenin equivalent voltage is just the voltage one would see at the output of the voltage divider if it were open circuited. In this case, you would see 7.5V*(20k/30k) = 5V.

When the input transistor is switched off, all you have in your circuit to dictate the voltage at its collector are these components: -

Do you know how to manipulate the circuit using theorems to prove that the voltage is 7.8 volts? Can you do that yourself?

My simulation: -

Can you see that the circuit can be transformed to this equivalent circuit: -

And that is equivalent to 1.95 mA flowing into a 4 kΩ resistor. Have you been schooled in using these types of transformation?

Analysis when Q1 is off. in this case Q1 behaves like an open switch, whereby the circuit consists of the generator V1, the resistance R3 and the voltage divider R1 and R2 powered by V2. Millman's theorem is applied to determine the voltage at node A.