Resistive Potential Divider Problem - Page 1
Forum section: electronics / projects

author: jonmarx
author_rank: Contributor
timestamp: August 31, 2013, 02:41:56 pm
content: I'm almost embarrassed to ask this, but I have what appears to be a quite simple resistive potential divider circuit problem that I can't manage to solve.The circuit comprises an input voltage VIN, output voltage VOUT, four resistors, and a fixed +1.23V bias; hopefully the ASCII schematic below is legible. I want to derive an equation for VOUT as a function of VIN and the four resistor values.I've worked out plenty of equations for the currents, voltage drops etc. in the circuit but can't seem to find a way to eliminate all the currents and reduce them to the basic equation I want. I'm sure there is a simple solution or I've missed something, or perhaps my brain is simply getting too old for the algebra needed.Any anyone help?(PS This is not a school question - I'm 49 years old! FYI it's a real-world design problem relating to the control of the output voltage from a Vicor PSU module. VOUT is fed into a control pin on the PSU and sets its output voltage; R4 and the +1.23V bias are internal to the PSU.)Code:VIN ---- +1.23V| |R1 R4| ||----R3--------> VOUT|R2|------------------------ GND

author: Kjelt
author_rank: Super Contributor
timestamp: August 31, 2013, 02:57:39 pm
content: I think you can use Kirchhoffs lawshttp://en.wikipedia.org/wiki/Kirchhoff's_circuit_laws

author: Unknown
author_rank: Unknown
timestamp: August 31, 2013, 03:05:50 pm
content: This is a linear circuit, so easiest way might be superposition: first short the +1.23V node to ground and calculate Vout if only Vin was turned on. Then short the Vin node to ground and calculate the Vout as result of the 1.23 V source. Then add these two together and you have the output voltage with both voltage sources turned on. Shorting the other voltage sources to ground (as opposed to making them open circuit) is critical here.The other option would be to write Kirchoff's current law (the sum of all currents flowing in or flowing out of a node is zero) for the two unknown nodes. Let's label the node connecting R1, R2 and R3 V1. We can write the voltages across these resistors as functions of the node voltages and apply Ohm's law to find the current. KCL for the V1 node is:(V1-Vin)/R1 + (V1-Vout)/R3 + V1/R2 = 0KCL for the Vout node is:(Vout-1.23 V)/R4 + (Vout-V1)/R3 = 0You now have two linear equations with two unknowns. Solve for Vout. The outcome should be identical to superposition.

author: jonmarx
author_rank: Contributor
timestamp: August 31, 2013, 03:10:13 pm
content: Hi,Thanks for the quick responses, and for the reminder about superposition and Kirchoff's law (something I don't think I've had cause to use since university!).I'll grab a pen and paper and start working it out....

author: IanB
author_rank: Super Contributor
timestamp: August 31, 2013, 03:22:31 pm
content: As alm did, but slightly different. A current balance around the unknown node and then a potential divider to get the output voltage:Unknown voltage, VX:(VIN - VX) / R1 + (0 - VX) / R2 + (1.23 - VX) / (R3 + R4) = 0Rearrange:VX = [ VIN * R2 * (R3 + R4) + 1.23 * R1 * R2 ] / [ (R1 + R2) * (R3 + R4) + R1 * R2 ]Voltage divider:VOUT = VX + R3 / (R3 + R4) * (1.23 - VX)You should check my algebra as I did it in my head and I might have made a mistake...

author: ivaylo
author_rank: Frequent Contributor
timestamp: September 11, 2013, 03:16:47 pm
content: QuoteCan you really add voltages like this? The Superposition Theorem talks about currents, no? Not trying to correct you, just trying to wrap my mind around this as well.

author: Unknown
author_rank: Unknown
timestamp: September 11, 2013, 09:57:08 pm
content: For linear circuits (no semiconductors or non-ideal passive components etc) you can. In general the superposition theorem talks about responses, which can be either voltages or currents. Both KCL and KVL apply for each individual independent source. I'm sure you can derive the additivity of voltages from the additivity of currents, given the linear voltage-current relationship.