# Conditioning Electronics

Discussion in 'General Electronics Discussion' started by Gon Says, Jun 11, 2015.

1. ### Gon Says

1
0
Jun 11, 2015
Hello guys,

I'm having a problem understanding the equations for the circuit analysis of the figure below taking into account the question made in order to obtain the dimensions of the transistors and voltages:

Thank you very much in advance! Any help is appreciated

5,165
1,087
Dec 18, 2013
Hello
Is this the circuit and values they say will work? Or is this something you got of the internet?

3. ### hevans1944Hop - AC8NS

4,593
2,149
Jun 21, 2012
The input voltage, Vo, varies from 50 x 100 mV/psi = 5 V at 50 psi to 150 x 100 mV/psi = 15 V at 150 PSI. Since you want the output, Vout, to vary from 0 to 2.5 V, the range of the input signal, which is 10 V (V150 psi - V50psi), must be attenuated by a factor of 2.5/10 = 0.25 and then the input signal must be offset to zero at 50 psi.

The first op-amp is wired as a unity-gain inverter, meaning Vx varies from -5 V to -15 V over the 50 psi to 150 psi range as Vo varies from +5 V to +15 V over that range. The second op-amp stage is wired as an inverting summing amplifier with a gain of Ry/R = 625/2500 = 0.25 for each of the two inputs, Vref and Vx. Thus when the input is at +5 V, Vx = -5 V and when this is summed with Vref = +5 V the result is Vout = 0. When the input is at +15 V, Vx = -15 V and when this is summed with Vref = +5 V and the difference multiplied by the gain of 0.25 and inverted the result is Vout = +2.5 V.

Note that the first op-amp must be capable of a negative output of -15 V and the second op-amp must be capable of a positive output of only +2.5 V. Normally you would use a pair of dual op-amps with ±15 VDC power rails, each op-amp capable of ±15 V full-scale output. This is not common (rail-to-rail output). Most op-amps with ±15 V supplies will only swing their outputs ±10 V. This can be accommodated by decreasing the gain of the first stage, making the feedback resistor, Rx, smaller to limit the output Vx to -10 V at 150 psi. You will then have to adjust the resistor values in the second stage to restore the 0 V to +2.5 V output range. I will leave those calculations to you now that I have explained the principle..

Is this a homework problem for use in a class you are taking?

Last edited: Jun 12, 2015
Arouse1973 likes this.

5,165
1,087
Dec 18, 2013
Hi Hop
Isn't the output approx. +2.5 Volts.

5,165
1,087
Dec 18, 2013
Having another quick look. I see when the input is 15 Volts the output of the first buffer is -15 Volts. This puts -5 Volts on the input of the last op-amp. For 2.5 Volts out this would need a gain of 2 (inverting). This is (2500/2)/625 = 2. I think that's right.

6. ### hevans1944Hop - AC8NS

4,593
2,149
Jun 21, 2012
Yes, it is. Thanks for noticing that. I edited my post #3 to correct it.

No, it puts -15 volts on the input resistor, R, of the last op-amp. After the offset necessary to make the output zero at 50 psi (-5 V) is removed by Vref, there is effectively -10 V on the input of the second op-amp and a gain of -0.25 to produce +2.5 V output. Gets a little trickier to calculate if both R's on the input (signal and Vref offset) are not equal. For either case, equal or not equal input resistors, you can just treat the inputs separately and sum their contributions to the output. For Vref, that's Vref (-R/Ry) and for Vx it's Vx (-R/Ry), or (Vref + Vx) (-R/Ry). Since Vx is a negative voltage and Vref is a positive voltage, you can take the positive difference in their absolute values and multiply by the gain factor (-R/Ry). Or algebraically add Vref and Vx and multiply by the gain factor (-R/Ry) because both have the same value input resistor.

5,165
1,087
Dec 18, 2013
Ah yes your right, I was looking at the actual non-inverting pin voltage which is -5 Volts.

8. ### hevans1944Hop - AC8NS

4,593
2,149
Jun 21, 2012
The non-inverting (+) pins on both op-amps are grounded (at least according to the schematic) which means the inverting pins (-) are maintained at a "virtual ground" by feedback from the op-amp outputs.

5,165
1,087
Dec 18, 2013
Blonde moment, I meant inverting sorry. I worked out the voltage from the divider right on the input pin duhh. It was late at night, that's my excuse

10. ### hevans1944Hop - AC8NS

4,593
2,149
Jun 21, 2012
No excuses needed here, Adam. We all make mistakes. Like me forgetting (or ignoring) for at least a month that the PIC10F206 uses pins for multiple purposes, and then not programming accordingly. Once I spent an entire day (with pay!) "troubleshooting" a complex op-amp circuit, wondering why none of the signals were at their proper levels... changing the op-amps numerous times (thank God I put them in sockets!) and even pulling out and measuring resistors to make sure I was using the correct values. Nada. Nothing I tried solved the problem. Until... I finally noticed the ±15 V DC bench supply I was using to power everything was not actually plugged into a power outlet. Arrgh. (other comments deleted because they may offend some people reading this post) After providing power to the op-amps everything worked as it should, restoring my faith in my analog design skills.

Hop

5,165
1,087
Dec 18, 2013
LOL been there done that

Martaine2005 likes this.
12. ### CDRIVEHauling 10' pipe on a Trek Shift3

4,960
651
May 8, 2012
Ha! There's nothing more refreshing than self deprecation to keep one humble.

On another note... I think a blind, deaf and dumb squirrel would conclude that this is a homework assignment.

Chris

hevans1944 and Arouse1973 like this.
13. ### davennModerator

13,798
1,939
Sep 5, 2009
I had considered (briefly) moving it