Solving Opamp Circuit Problem w/ LM358

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The discussion revolves around issues faced with an LM358 op-amp circuit intended for simple voltage gain with inputs from 0-5V. The user reports unexpected behavior, where smaller inputs yield higher outputs and larger inputs result in lower outputs, contrary to expected gain behavior. Key problems identified include a capacitor in series with the power supply, a resistor incorrectly placed, and a lack of negative feedback in the circuit design. After reevaluating the circuit, the user describes a revised setup involving a differential amplifier and a Wheatstone Bridge configuration, aimed at controlling heat output through a transistor. Ultimately, the user decided to revert to a previous working design after realizing the new circuit had significant flaws.
Sirius24
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I am also having an op-amp problem. I want a simple voltage gain through an op-amp with inputs ranging from 0-5V, which will then be fed through an integrator. I am using the LM358 op-amp. The datasheet and the diagram are attached. For some reason, though, I am getting a gain far exceeding expectations. Also, the smaller the input, the greater output I receive out, and larger inputs are resulting in lower outputs. To my understanding, there should be an overall gain, not a lower output for a high input. I don't understand what is wrong, and was hoping someone could help me.
 

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Sirius24 said:
I am also having an op-amp problem. I want a simple voltage gain through an op-amp with inputs ranging from 0-5V, which will then be fed through an integrator. I am using the LM358 op-amp. The datasheet and the diagram are attached. For some reason, though, I am getting a gain far exceeding expectations. Also, the smaller the input, the greater output I receive out, and larger inputs are resulting in lower outputs. To my understanding, there should be an overall gain, not a lower output for a high input. I don't understand what is wrong, and was hoping someone could help me.

Your circuit makes almost no sense at all. Perhaps you should describe what you are trying to achieve, because whatever that is it's almost certain that your circuit won't do it.

Some major issues with this circuit are :

1. Capacitor in series with the positive power supply.
2. Resistor from balance input to non-inverting input.
3. Lack of any negative feedback.

What are you trying to achieve with the above?
 


uart said:
Some major issues with this circuit are :

1. Capacitor in series with the positive power supply.
2. Resistor from balance input to non-inverting input.
3. Lack of any negative feedback.

What are you trying to achieve with the above?

My partner recently re-drew the circuit diagram, and I guess I should have taken a closer look at it before posting.

As it is currently set-up on the circuit board is as follows:
1. A differential amplifier.
2. Rails at 0V and 12 V.
3. Virtual ground at 6V. The positive op-amp input goes through a resistor to this virtual ground.
4. The inputs come from a voltage difference across a Wheatstone Bridge, moderated by a potentiometer adjustment and a thermistor.
5. There is currently no capacitor.

The output of this op-amp will go through an integrator and a transistor with the end result (hopefully) heating up a wire to keep a container warmed. The heat would be adjustable by the posteniometer and the voltage difference across the Wheatstone bridge would decrease as the resistance of the thermistor works to match that of the poteniometer.

I hope this makes more sense.

~Sirius
 
Please post the redrawn circuit diagram.
 
skeptic2 said:
Please post the redrawn circuit diagram.
The new one is incorrect. I'll fix the mistakes and post it as soon as I can.
 
We ended up scraping our design and revertin gack to our old one, which works well enough for its purpose. Thank you for the help, though.
 
I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

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