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Registered Member #3859
Joined: Sun May 01 2011, 03:47PM
Location:
Posts: 179
I built a circuit that i ripped off the data sheet. Why do I get a 1.9V on the output when there is no input? Are -input and the resistor acting as a voltage divider and giving me this voltage? How do I solve this problem? Yet the overvoltage works.
Registered Member #3859
Joined: Sun May 01 2011, 03:47PM
Location:
Posts: 179
The two capacitors are combined into one between power and ground. I am not sure why, but i see it on many circuits like that. So why do I get voltage when there is no input?
Registered Member #3610
Joined: Thu Jan 13 2011, 03:29AM
Location: Seattle, WA
Posts: 506
Whoa, how about posting once and waiting for a response...
It's difficult to read the sideways schematic, but it looks like you're using a 3V power supply? I just had a look at the datasheet and it lists the absolute minimum as +/- 3V, so 6V total, with ground in the middle. I would start by making sure you have a proper power supply.
The capacitors are for decoupling. They should be placed as close as possible to the power pins they connect to. The idea is to counter the effects of parasitic resistance in the traces when the IC draws spikes of current.
Registered Member #3859
Joined: Sun May 01 2011, 03:47PM
Location:
Posts: 179
James wrote ...
Whoa, how about posting once and waiting for a response...
It's difficult to read the sideways schematic, but it looks like you're using a 3V power supply? I just had a look at the datasheet and it lists the absolute minimum as +/- 3V, so 6V total, with ground in the middle. I would start by making sure you have a proper power supply.
The capacitors are for decoupling. They should be placed as close as possible to the power pins they connect to. The idea is to counter the effects of parasitic resistance in the traces when the IC draws spikes of current.
The reason I repost is because I keep working on it.
Here is the datasheet:
I am testing this device with DC only. When I build this circuit, I will put caps next to the pins. The amp is supposed to work with DC and signals.
I have a problem that the -input pin and the output are connected through a resistor, therefore, I suspect that -input circuit and that resistor form a voltage divider that causes this voltage to appear on the input pin.
They should act as a negative feedback? What do you think it the problem? Maybe there is a better circuit to use.
Registered Member #3610
Joined: Thu Jan 13 2011, 03:29AM
Location: Seattle, WA
Posts: 506
You can edit an existing post.
The resistor from output to negative input provides negative feedback, you need that otherwise you have nearly infinite gain and the op-amp will act as a comparator.
What is your power supply? As I said, if you are running it only from 3V it won't work.
Registered Member #3859
Joined: Sun May 01 2011, 03:47PM
Location:
Posts: 179
As far as I remember from the datasheet, this opamp is designed to run of the 3 volt supply.
The reason I picked it was so a microcontroller can run off the same 3 volts and the input can be limited to +3 volts for the ADC input. The ADC doesn't except negative voltages as I understand. And the microcontroller is: STM32F103RET6 I think that few generic microcontrollers accept -voltages on their ADCs. Is that true? Why wouldn't it work?
Registered Member #3610
Joined: Thu Jan 13 2011, 03:29AM
Location: Seattle, WA
Posts: 506
I looked at the datasheet and it said the minimum supply was +/- 3V. I could be wrong, but that implies that you need a bipolar supply, with 3V on either side of ground, in other words a split 6V supply. There are op-amps designed specifically to use a single ended supply, but in that case you need to create a false ground.
Try a split supply with your circuit and see what it does.
Registered Member #3859
Joined: Sun May 01 2011, 03:47PM
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Posts: 179
So I should use 6 volts with a voltage divider? I have another opamp that can clamp input that i might end up using. But thank you for explaining the +-.
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Problems with a circuit prototype for voltage clamping amp with 1 gain based on AD8036 amp.
