# Need a little help getting my head around this

#### jmclemore7

Joined Oct 3, 2017
3
The output from an analog transducer varies from 2V to 4V over the range of the input signal. This voltage is to be connected to the input of an A-D converter which has an input voltage range of 0-5V. The signal conditioning circuit is to be designed to optimise the resolution of the A-D converter.

An LED indicator is also required which will indicate when the level of the transducer output voltage is at, or above, 30% of the maximum output.

(a) Design an opamp circuit to condition the signal so that it varies in the range 0V to 1V.

(b) Design an opamp circuit to amplify the conditioned signal so that it now varies from 0V to 5V.

(c) Design an opamp circuit to produce an enabled output when the voltage is at, or above, 30% of it’s max.

output.

I appreciate all input, or even a point in the right direction....

#### ericgibbs

Joined Jan 29, 2010
11,127
hi j7,
Welcome to AAC.
Is this a Homework assignment, if Yes, I can move it to the Homework Forum.?
E

#### jmclemore7

Joined Oct 3, 2017
3
hi j7,
Welcome to AAC.
Is this a Homework assignment, if Yes, I can move it to the Homework Forum.?
E
It is homework. Sorry for posting in the wrong place.

#### shteii01

Joined Feb 19, 2010
4,644
Regarding (a). It seems that transducer output is 4-2=2 volts peak to peak. You need output to be 1-0=1 volt peak to peak. So. You need to "shrink" transducer output from 2Vpp to 1Vpp. A simple op amp circuit with gain of 1/2 will do that.

Then you need to move the signal "down". The original transducer output is from 2 to 4 volts, so the middle is 3 volts. Your new transducer output is 2/2=1 volt and 4/2=2 volts, so the transducer output will be from 1 volt to 2 volts, with the signal being 2-1=1 volt peak to peak. The middle is 1.5 volts. You need the middle to be at 0.5 volts so that the 1 volt peak to peak signal will fit into 0 to 1 volt range required by the problem. The easy way to do it is to use a two resistors voltage divider across 1 volt power supply. This way 1 volt is divided into 0.5 volts across each resistor, then you apply output of the op amp circuit to the junction of the two resistors.

It is possible to use op amp bias to provide the 0.5V "middle" for the 1 volt peak to peak signal. This way you combine the two circuits into one and use less parts and less space.

#### jmclemore7

Joined Oct 3, 2017
3
Regarding (a). It seems that transducer output is 4-2=2 volts peak to peak. You need output to be 1-0=1 volt peak to peak. So. You need to "shrink" transducer output from 2Vpp to 1Vpp. A simple op amp circuit with gain of 1/2 will do that.

Then you need to move the signal "down". The original transducer output is from 2 to 4 volts, so the middle is 3 volts. Your new transducer output is 2/2=1 volt and 4/2=2 volts, so the transducer output will be from 1 volt to 2 volts, with the signal being 2-1=1 volt peak to peak. The middle is 1.5 volts. You need the middle to be at 0.5 volts so that the 1 volt peak to peak signal will fit into 0 to 1 volt range required by the problem. The easy way to do it is to use a two resistors voltage divider across 1 volt power supply. This way 1 volt is divided into 0.5 volts across each resistor, then you apply output of the op amp circuit to the junction of the two resistors.

It is possible to use op amp bias to provide the 0.5V "middle" for the 1 volt peak to peak signal. This way you combine the two circuits into one and use less parts and less space.
Thank you so much for the help. I was unsure which opamp to use. I figured I would need a summing opamp initially.

#### shteii01

Joined Feb 19, 2010
4,644
Thank you so much for the help. I was unsure which opamp to use. I figured I would need a summing opamp initially.
Op amp wise you likely will want something single supply and rail to rail so that you can get close to you bottom range of 0 volts.

#### n1ist

Joined Mar 8, 2009
188
Part C is a comparator; You have your 0-5V signal on one input and a voltage divider to set the reference level on the other input.