Register to reply

Irritatingly failed resistance calculations

Share this thread:
prush242
#1
Jan28-14, 01:40 PM
P: 5
Gentlemen/Ladies,

Long time reader, first time posting. I have a few projects that I would like to get done, and I realized if I don't start getting them done, I never will. Heh.

I'm making mistakes calculating the resistors I would need for the following. I've done it 5 times now because each time I do the math, I'm coming to different conclusions. This will be my first learnings on surface mount soldering, so I want to be sure the bits are correct, then I can only blame myself for any soldering mistakes. So:

The board has R1=7500ohm, R2=7870ohm, and R3=2100ohm, as a voltage divider sampling filtered output from the 53355 buck inductor (0.6V), and runs at 1.0514V low voltage. At high voltage, R3 is not used, and it runs at 1.172V.

What I would like to do is replace R1 and R2, such that the board, using R1, R2, and R3, runs at 1.172V low voltage, and then using only R1 and R2, runs at 1.25V high voltage. The use of higher voltage is not in question as far as the board is concerned.

I think I'm screwing up the calculations between the different options for R1 and R2:

Vo = 1.17V on low voltage.

R7 = (1.17V - 0.6V) * (7870+2100) / 0.6V

R7 = (0.57V)*(9970)/0.6V

R7 = (55/60)*9970 = 9139ohm for low voltage.

But it has been QUITE some time and I'm not quite sure I am using the formulas correctly. Either way, I think too many numbers are leading me astray.

Anyone have any thoughts? And thank you, in advance!
Phys.Org News Partner Engineering news on Phys.org
UConn makes 3-D copies of antique instrument parts
Amazon worker piloted drone around Space Needle
Five next-generation technologies for positioning, navigation and timing
berkeman
#2
Jan28-14, 04:08 PM
Mentor
berkeman's Avatar
P: 40,659
Quote Quote by prush242 View Post
Gentlemen/Ladies,

Long time reader, first time posting. I have a few projects that I would like to get done, and I realized if I don't start getting them done, I never will. Heh.

I'm making mistakes calculating the resistors I would need for the following. I've done it 5 times now because each time I do the math, I'm coming to different conclusions. This will be my first learnings on surface mount soldering, so I want to be sure the bits are correct, then I can only blame myself for any soldering mistakes. So:

The board has R1=7500ohm, R2=7870ohm, and R3=2100ohm, as a voltage divider sampling filtered output from the 53355 buck inductor (0.6V), and runs at 1.0514V low voltage. At high voltage, R3 is not used, and it runs at 1.172V.

What I would like to do is replace R1 and R2, such that the board, using R1, R2, and R3, runs at 1.172V low voltage, and then using only R1 and R2, runs at 1.25V high voltage. The use of higher voltage is not in question as far as the board is concerned.

I think I'm screwing up the calculations between the different options for R1 and R2:

Vo = 1.17V on low voltage.

R7 = (1.17V - 0.6V) * (7870+2100) / 0.6V

R7 = (0.57V)*(9970)/0.6V

R7 = (55/60)*9970 = 9139ohm for low voltage.

But it has been QUITE some time and I'm not quite sure I am using the formulas correctly. Either way, I think too many numbers are leading me astray.

Anyone have any thoughts? And thank you, in advance!
Welcome to the PF.

Can you upload a schematic? It's hard to follow what you are trying to do.
prush242
#3
Jan28-14, 04:28 PM
P: 5
Quote Quote by berkeman View Post
Welcome to the PF.

Can you upload a schematic? It's hard to follow what you are trying to do.
Surprisingly enough, yes. Does it help?
Attached Thumbnails
schematic.GIF  

prush242
#4
Jan28-14, 10:07 PM
P: 5
Irritatingly failed resistance calculations

Now, as it happens, I have some 8800ohm and 9100ohm resistors. So I tried to calculate the values if I were only to replace R1, leaving R2 and R3 alone, and came up with this:

Vo = 0.6V * [8800/(7870+2100)] + 0.6V

Vo = 0.6V * 0.8826 + 0.6V

Vo = 0.5296V + 0.6V = 1.130V, and:


Vo = 0.6V * [9100/(7870+2100)] + 0.6V

Vo = 0.6V * 0.9127 + 0.6V

Vo = 0.5476 + 0.6V = 1.150V, thus:


Vo = 0.6V * (7500/7870) + 0.6V

Vo = 0.6V * .9530 + 0.6V

Vo = 0.5718V + 0.6V = 1.172V, therefore:

Q3 = 1.172V = operating point at low voltage.


Thus:

Vo = 1.172V low voltage.

R1 = (1.17V - 0.6V) * (7870+2100) / 0.6V

R1 = (0.57V)*(9970)/0.6V

R1 = (55/60)*9970 = 9139ohm for low voltage.

and

Vo = 1.25V at high voltage.

R1 = (1.25V - 0.6V) * (7870+0) / 0.6V R3 is zero, because it isn't used at high voltage.

R1 = (0.65V)*(7870)/0.6V

R1 = (65/60)*7870 = 8525ohm


So, averaged: (8525+9139)/2 = 8832, and I have 8800s and 9100s.


Still going... with 9100ohm:

Low voltage, target is 1.17V

Vo = 0.6V * [R1/(R2+R3)] + 0.6V

Vo = 0.6V * [9100/(7870+2100)] + 0.6V

Vo = 0.6V * 0.913 + 0.6V

Vo = 0.5476V + 0.6V = 1.148V, 1.9% difference from target.


High voltage, target is 1.25V

Vo = 0.6V * [R1/(R2+R3)] + 0.6V

Vo = 0.6V * [9100/(7870+0)] + 0.6V

Vo = 0.6V * 1.156 + 0.6V

Vo = 0.694V + 0.6V = 1.294V 3.46% difference from target


Then with 8800ohm

Low voltage, target is 1.17V

Vo = 0.6V * [R1/(R2+R3)] + 0.6V

Vo = 0.6V * [8800/(7870+2100)] + 0.6V

Vo = 0.6V * 0.8826 + 0.6V

Vo = 0.5296V + 0.6V = 1.130V, 3.48% difference from target.


High voltage, target is 1.25V

Vo = 0.6V * [R1/(R2+R3)] + 0.6V

Vo = 0.6V * [8800/(7870+0)] + 0.6V

Vo = 0.6V * 1.12 + 0.6V

Vo = 0.671V + 0.6V = 1.271V 1.67% difference from target.

I have zero faith in those numbers, given my differing results, but what I get is I could possibly just replace R1 with the 8800ohm resistor. I don't think that's a good idea, because it seems it is too high for low voltage and too low for high voltage.

Hence, I would rather replace both of them but it's getting Lost in Translation.

I hope that helps, as I am at an impasse.

Thanks again anyone, for any thoughts. 8^]
prush242
#5
Jan31-14, 02:15 AM
P: 5
Anyone? Anyone? Bueller? 8^]
berkeman
#6
Jan31-14, 12:40 PM
Mentor
berkeman's Avatar
P: 40,659
I'm having a bit of trouble tracking your work... What does this mean:

Q3 = 1.172V = operating point at low voltage.
And what is the purpose of this circuit? Why do you need such low output voltages, and with that precision? What is Rds(on) for the FET Q3? The Vfb = 0.6V when the buck DC-DC output is in regulation, right?
prush242
#7
Feb3-14, 07:17 PM
P: 5
Thanks for your response, Berkeman!

Purpose: The circuit runs one of two oscillators, presently 12.0MHz and 14.318MHz. When all three resistors are in use, the circuit runs the 12.0MHz oscillator (low voltage). When only two are in use (R3 is unused), the circuit powers the 14.318MHz oscillator (high voltage).

Low voltages and precision: Because that is what’s there on the board. I didn’t make it from scratch. 8^]

Q3 is the feedback divider that samples the output voltage. With Q3 off/open, the VRM is set to output 1.05V. The same logic signal that triggers the higher oscillator enables Q3 through R25, and it shorts around R3, removing it from the string. I think Rds(on) is less than 1 ohm, which means the combo of (1 ohm)||(R3) is a few orders of magnitude smaller than the other pieces.

Vfb: Correct, 0.6V when the buck DC-DC output is in regulation.

I want to change the two oscillators to 14.318MHz and 16.0MHz for low and high respectively. Though it just dawned on me. Once I change the board, it will never run at 14.318, it will always run at 16.0MHz, which changes how I'm thinking this can be done.


Register to reply

Related Discussions
Roll resistance calculations error?! Mechanical Engineering 1
How to treat internal resistance in calculations...? Introductory Physics Homework 75
Projectile calculations including air resistance and weight General Physics 2
Air resistance calculations Advanced Physics Homework 1
Rocket Calculations, wind resistance, Rectangular hyperbola General Math 11