RonH's equations

Discussion in 'The Projects Forum' started by Mr Smiley, May 1, 2010.

1. Mr Smiley Thread Starter New Member

May 1, 2010
6
0
Hi Everybody,

Does anybody have the equations for the circuit from Ron H

Many thanks

Last edited: May 1, 2010

Apr 20, 2004
15,815
293

Feb 17, 2009
4,513
1,272
4. Ron H AAC Fanatic!

Apr 14, 2005
7,049
674
I'm confused. Post the schematic.

EDIT: If you mean this one, the output voltage is

Vout=Vin*R4/(R3+R4)*(1+(R2/R1))-Vcc*(R2/R1)

In this circuit, Vcc=15V.

• mikewilliams problem sch.PNG
File size:
17.8 KB
Views:
31
Last edited: May 2, 2010
5. Mr Smiley Thread Starter New Member

May 1, 2010
6
0
Many thanks Ron,

That's the one, i added the above question to the original thread and someone kindly started a new one for me.

Last time i had a problem like this was around 15 years ago and i've been racking my memory about how i did it . It was in a book in my office at the time that belonged to someone else, i still know where that book is (can't remember title or author ) but can't get access to it anymore. I did copy the section but thats around here somewhere and is alluding every search.

The example link given by Jony130 ( many thanks Jony130 ) is close. In his example he sets vd to the lowest of the input voltages. In your example you don't seem to do that, the Vout starts at 13v when R2 is tied to 15v. If i remember rightly it means setting the inverting and non-inverting equations equal to one another and going from there. Could anyone clear up my confused understanding.

Many thanks

Mr Smiley

Last edited: May 2, 2010
6. Ron H AAC Fanatic!

Apr 14, 2005
7,049
674
Are you wanting a general procedure for designing a level shifter? There are several variables involved, such as
1) do you want the same peak-to-peak output swing as the input, or different?
2) What power supplies do you have? Are they stable, or will you need a separate reference voltage?

My procedure in this case was to write the transfer function of the circuit:

Vout=Vin-13.
I started with the -13. I'm not sure why I used a 15V supply - perhaps because BeenThere had used it in his proposed solution. Anyhow, to get from +15V to -13V, all you need is an inverting amp with a gain of 13/15, hence the choices of R1 and R2.
OK, we've taken care of the level shifter. Now we need noninverting, unity gain for the Vout=Vin part. If we connect Vin directly to the noninverting input, we will get a gain of ((R1+R2)/R1). To get unity gain, we need an attenuator in front of the op amp.
We need attenuation*gain=1, so

$\frac{R4}{R3+R4}*(\frac{R1+R2}{R1})=1$

Rearranging,

$\frac{R4}{R3+R4}=\frac{R1}{R1+R2}$

By inspection we can see that if we make R1=R4 and R2=R3, the equation will be satisfied.

7. Mr Smiley Thread Starter New Member

May 1, 2010
6
0
Many thanks Ron,

I understand it more clearly now thanks to yourself ,

Just as a side question, how would a different gain between input and output be applied, in the inverting ( level shifter section ) or the non-inverting section.

Once again, many thanks

Mr Smiley

8. Ron H AAC Fanatic!

Apr 14, 2005
7,049
674
In The example I used, you would change the gain equation:

$\frac{R4}{R3+R4}*(\frac{R1+R2}{R1})=gain$

And then solve for R3 and R4. Since you have 2 unknowns, but only one equation, you have to come up with a value for one of them, and there are several things you might have to consider, depending on the application: Op amp input bias and offset currents, source loading, high frequency rolloff due to stray capacitance, resistor noise, ... Have I forgotten anything?
EDIT: i forgot to point out that the maximum attainable gain with the circuit configuration used here is (R1+R2)/R1. This is easily fixed.

Last edited: May 3, 2010