I was trying to make a howland current pump with my opamp with single supply.
I know its not ideal to use the lm833 like this, but the datasheet says it works reasonably, so I tried it but everytime the output was wrong. Trying to figure out what was wrong, I ended up removing every component and testing is as simple as possible, and thats when I got to this point. (Sorry, I edited the image wrong, the top side -Vee has 0v)
So, the B side of the opamp has 1v on (+), and I connected the (-) and (out) directly, but the output is still completely wrong.
I tested the voltages directly on each pin, and tried with two different lm833, and on a different part of the breadboard, but still got the same results.
I'm really intrigued of why this is working so wrong. With the lm741 I've never got this problem with the same setup. What am I doing wrong?
Please post a hand-drawn schematic diagram of the circuit you tried to build, not a photo of the breadboard with a circuit that obviously will not do anything useful.
Well, it won't do anything useful, but as I mentioned I tested the voltage on each pin and that's all there is to it.
The circuit I tried to build is this one:
But the output was always at 22ma, when it should be about 4ma.
As the circuit was getting too big and I couldn't find the mistake, I took every component off to test the opamp alone, that's when I got to this test circuit.
I mean, the pin 5 has a 1v input, so 6 and 7 on this circuit should also have 1v, but it's outputting 14v. This behaviour is weird. With the lm741 I had no problems with this exact test circuit, that's why I wanted to understand what's the difference, or what I'm doing wrong.
In the circuit in the link I posted, the bottom opamp outputs about 0.8v, but when I did it with the lm833 it was outputting 14v. that's why I tested without every other component and got to this test.
The breadboard photo in the first post does not show any resistors. What does that photo have to do with the schematic posted in reply #4?
If the photo is irrelevant, edit your first post to delete the useless photo. Otherwise, add in the resistors shown in the schematic and remeasure the output voltage.
On this forum, the only information we have on your project is provided by you. Make sure that the information is correct before continuing with this thread.
The point made by herbschwarz in reply #5 is spot on. Never leave op amp inputs floating. Connect unused IN- to Out and connect unused IN+ to some valid voltage between Vee and Vcc. Also, add the required decoupling capacitor (minimum 100 nF) between Vee and Vcc, close to the chip.
The resistors are in other breadboard, the white wire is releasing 1v and I did measure it.
If the + side has 1v and I measured it, does it really matter how I got to that voltage? As I said, I measured it on the IC pin.
I just did draw it on the schematics because I thought it could help, but if you want exactly like the breadboard I can just remove the resistors and add a 1v input. It was obvious that there was something else outputting 1v to the (+) side.
Anyway, why are your answers so hostile? You could give some insight instead.
herbschwarz:
With one of the op amps not connected
to any input, having both inputs floating,
maybe it is oscillating.
I'll try to close the A side to see if the output change, thanks!
You realize that the lm833 can't output a voltage that goes all the way up to the positive supply voltage or all the way down to the negative or ground supply voltage, right? - see the DC Electrical Characteristics table in the datasheet, common mode output range - with +/- 15v, the output swing is +/- 12v minimum, 13.x v typical. If you need to output voltage all the way to the power rails, you need to use an opamp with "rail-to-rail output".
Good catch Dr. Azzy !
But from what I understood , the OP wanted to know why the voltage follower had 14V out instead of the 1V input ( although I don't understand
the load)
I think he can confirm that by running the
circuit off a 9V PP3 smoke alarm battery or
a 12V battery, or pu a 10k/5W dropping
resistor between the 24V and the op amp Vdd
pin.
DrAzzy:
You realize that the lm833 can't output a voltage that goes all the way up to the positive supply voltage or all the way down to the negative or ground supply voltage, right?
Yes, I realise that, but in this case shouldn´t it at least go closer to 1v? Say... 3v or something? As I mentioned, doing it with the old lm741 did output exactly 1v.
raschemmel:
I think he can confirm that by running the
circuit off a 9V PP3 smoke alarm battery or
a 12V battery, or pu a 10k/5W dropping
resistor between the 24V and the op amp Vdd
pin.
I'll try that as soon as I get home. I have a 12v regulator, if I input 12v on (+) side then it should get the output right. (I hope so)
TomGeorge:
Hi,
If you are supplying 24Vpos to pin 8.
Why are you using pins 2 and 3 as the opamp -IN2 and OUT2, when they are -IN1 and +IN1?
Tom....
I made the mistake when drawing it. The pin 1 is to the right and at pin 4 it was supposed to be written 0v.
I´ll try the same circuit again with the lm358. This lm833 is acting too weird for me.
The voltage divider output is 1V and that
is connected to the V_ IN"+" pin input
of the op amp.
As you have not stated it, I assume you
do not know that particular op amp configuration
is called a Voltage Follower because the output
"follows" whatever is input on the VIN_+ pin,
which in this case is 1 V.
It is NOT an amplifier !
It is a buffer , which , by definition is a
device with High Input Impedance and low
output impedance (50 ohms or less) ,
which makes it ideal as voltage to current converter.
I_out= V_out/R_out
The Output Load device (a resistor)
functions as a current to voltage converter,
converting I_out to V_out= I_out * R_out
(Ohm's Law).
I don't understand the circuit changes you
mentioned so post an updated correct
schematic along with your Project Objective
(what you are trying to do and what you
expect the result to be)
You have not explained this yet so please
do so.
Also, you have nothing connected to the
output (you have NO LOAD!)
That nonsense about a Howland Current Pump
is just that.
You circuit is not even remotely similar to a
Howland Current Pump:
What's going on ?
Are you leading us on a "wild goose chase" ?
What's going on ?
Are you leading us on a "wild goose chase" ?
The circuit I posted from falstad is the one in the figure 6 of the file from TI you mentioned. but with the resistors changed so 4v leads to 20ma.
I aso put the trim on the negative side instead of the positive side. The result is exactly the same.
I don't think you understand op amps.
You need a digital multimeter to measure
the output with No load, (NOT an analog voltmeter)
A voltage follower outputs the input voltage,
with or without a load so no load is needed to test
it.
Ah, yes, I used a digital multimeter. But what´s the difference?
Yes, a voltage follower outputs the input voltage. This is exactly what I did to test the lm833. The output should be 1v when the non inverting input is 1v, but the measured output was 14v.
Well, I understand the theory behind opamps, I did draw the circuit and made the calculations for non symetric improved howland current pump. With the lm741 all my calculations were spot on.
The moment I changed to lm833 everything went wrong. that´s why I tested it with the voltage follower and nothing worked as it should, that´s why I posted here. I wondered if there was something about the component that didn´t work as I expected.
raschemmel:
Why did you choose 4V @20mA instead of 5V ?
Because the resistors of the howland current pump has 1% tolerance. If I output 5v from the microcontroller I may reach a number lower than the 20ma needed. With 4v outputting 20ma I can reach 20ma even in the worst case scenario by adjusting the output accordingly.
