Yes that will work too.
[book=But, In the real world (provided the two units are completely isolated from each other) you can connect the scope directly across the the load]%[/book]
Ever blow a fuse on your scope doing this? I have.
The units are not usually isolated because the ground pins on the power cords of the power supply and the scope are tied together at the service for safety. So the usual way to measure this is with 2 probes (the probes grounds are usually tied to power ground) and the scope is set to measure probe 1 minus probe 2.
Yes ,not a fuse ,I nearly burned up a probe a few times trying to connect the ground until I figured out what was happening.
Back in 1985 when I first got it,I got a loud BANG and pop a breaker and I thought I was out an $1000 scope, but I lucked out and I am still using that same scope today.
I have an Issue with ground loops with my Signal generator and my computer becuase of this.
The cool part is I don't need the extra wire for ground when measuring them with the scope.
Although the extra noise is significant when trying to read tiny level signals.
Since then I have redone the power cable configuration and haven't had any issues so far. jer
Back in 1985 when I first got it,I got a loud BANG and pop a breaker and I thought I was out an $1000 scope, but I lucked out and I am still using that same scope today.
I have an Issue with ground loops with my Signal generator and my computer becuase of this.
The cool part is I don't need the extra wire for ground when measuring them with the scope.
Although the extra noise is significant when trying to read tiny level signals.
Since then I have redone the power cable configuration and haven't had any issues so far. jer
As many of you already stated, both inputs neeed to be biased. Thank you all for pointing me in the right direction.
How ever there is still one problem. The dc voltge at the output. How can I get rid of it without a cap at the output? Looking at the cmoy schematic there is no output cap. What kind of voodo is that or am I missing something.
I reproduced my circuit in the lab and I still get a dc output. I used a lm318 thoughfor the lab.
How ever there is still one problem. The dc voltge at the output. How can I get rid of it without a cap at the output? Looking at the cmoy schematic there is no output cap. What kind of voodo is that or am I missing something.
I reproduced my circuit in the lab and I still get a dc output. I used a lm318 thoughfor the lab.
The output signal ground must be referenced to the same voltage as what the output voltage is biased at.
Then the d.c. voltage will disappear and only the a.c. component (signal) will remain to be measured between the output and the refferenced output ground.
Man try saying that real fast! jer
Then the d.c. voltage will disappear and only the a.c. component (signal) will remain to be measured between the output and the refferenced output ground.
Man try saying that real fast! jer
"Your virtual ground should be the node at were r3 and r4 are connected. jer "
Or in your case the node at R3 and R4.
This ground system is a completely seperate enity and this becomes the virtual ground.
Any other inputs and outputs and sources that there may be must be tied to this virtual ground.
Better? jer
Or in your case the node at R3 and R4.
This ground system is a completely seperate enity and this becomes the virtual ground.
Any other inputs and outputs and sources that there may be must be tied to this virtual ground.
Better? jer
Might be a good time to show you the problem with this kind of virtual ground. a real ground should have a very low impedance. Yours will be around 2.3k. This will reduce the actual voltage across the load. The lower the resistance of r3 and r4 the better the virtual grnd will be, but at an increase in power. since your just simming it try R3=R4=1 ohm.
Ive designed circuits with this type of virtual ground, but instead of using a resistive divider I used a 7 volt regulator.
Ive designed circuits with this type of virtual ground, but instead of using a resistive divider I used a 7 volt regulator.
I used r3 and r4 just to keep it simple since it didn't work for me in the beginning. The TLE2426 looks like a good alternative to a resistor voltage divider.
Hmm,
It still doesn't work, atleast not in real life. I've connected the scope to the op-amp output and the scope's gnd to the virtual earth and got a straight line.
Connecting the scopes gnd to circuit gnd and everything works, but I still have the problem of dc at the op-amp output, don't I?
It still doesn't work, atleast not in real life. I've connected the scope to the op-amp output and the scope's gnd to the virtual earth and got a straight line.
Connecting the scopes gnd to circuit gnd and everything works, but I still have the problem of dc at the op-amp output, don't I?
If it is 0 ohms then the power supply and scope are not completely isolated and share a common ground system on the line side of the equipment.
Because of this you will find that you won't even need a ground connection from the scope to do measurements on your circuit.
When you hook the scope ground to your virtual ground you are essentaly shorting it back to 0 volts.
If you were to power the circuit from a battery the problem will (should) disappear.
Yes, you will always have an output d.c. voltage bias when operating from a single ended power supply in this configuration and it will be at what ever you set your virtual ground voltage too.
This is completely different from the d.c. offset voltage that the opamp create's.
Since's there is no negitive power supply voltage the opamp cannot swing past 0 volts to create the bottom half of the signal.
This is why capacitors are used is to block this d.c. voltage form the next stage(d.c. blocking capacitor).
This becomes a problem when trying to d.c. couple opamp stages.
Lets say that you have an output offset voltage of 10mv at the first stage and you have two stages after that with a gain of 10 each, your output offset voltage now becomes 10 X 10 x 10mv = 1v of d.c. output offset voltage at the last stage with 0v applied to the input to the first stage.
I did not take in consideration the d.c. output voltage of each sucsessive opamp stage in the above example.
Better? jer
Because of this you will find that you won't even need a ground connection from the scope to do measurements on your circuit.
When you hook the scope ground to your virtual ground you are essentaly shorting it back to 0 volts.
If you were to power the circuit from a battery the problem will (should) disappear.
Yes, you will always have an output d.c. voltage bias when operating from a single ended power supply in this configuration and it will be at what ever you set your virtual ground voltage too.
This is completely different from the d.c. offset voltage that the opamp create's.
Since's there is no negitive power supply voltage the opamp cannot swing past 0 volts to create the bottom half of the signal.
This is why capacitors are used is to block this d.c. voltage form the next stage(d.c. blocking capacitor).
This becomes a problem when trying to d.c. couple opamp stages.
Lets say that you have an output offset voltage of 10mv at the first stage and you have two stages after that with a gain of 10 each, your output offset voltage now becomes 10 X 10 x 10mv = 1v of d.c. output offset voltage at the last stage with 0v applied to the input to the first stage.
I did not take in consideration the d.c. output voltage of each sucsessive opamp stage in the above example.
Better? jer
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