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Hello again,
We would all like to know why the current mirrors are being used. It could be as simple as extending the usable output voltage range (because it will do that), but i dont want to guess at this point what the original designer had in mind.
To clarify again, when i say the first stage i mean both the first op amp on the left and the (now) four transistors. The output of the first stage then appears across R1 because the right side of R1 is at virtual ground, and virtual ground is very close to zero volts.
As i am sure you know, a current mirror takes an input current of some level and outputs a current at that same level in the output branch. The two current mirrors are connected to the supply lines of the left op amp, so they mirror the positive supply current and negative supply current. Since their output junctions are connected together and to R1 that has the other end connected to virtual ground, the current through R1 is the difference in currents that the op amp draws from the positive and negative supply rails. So if the op amp draws 1.1ma from the positive supply rail and the op amp draws 1ma from the negative supply rail then the current through R1 is the difference 1.1ma-1.0ma which equals 100ua. So the current through R1 is:
iR1=Ip-In
where Ip is the positive rail current and In is the negative supply rail current.
Now when the left op amp has zero input, there is zero output (assuming zero offset for now). Zero output voltage produced zero current in R2, so the positive rail current is the same as the negative supply rail current which would be the quiescent current draw of the op amp. If this was 1ma then that would appear in both rails, so iR1=0 also.
When a positive voltage is applied to the left op amp, the output goes higher by the same amount. So for 1v input the output of the left op amp goes to 1v also. This produces a current equal to:
Iout1=Vout1/R2
This current must come from the positive rail, so now the positive rail current is higher than the negative rail current so we get a net output current from the two current mirrors of:
iR1=Ip-In
and since the current in R2 is now 1/10000=100ua we have iR1=100ua also.
We could now state the transconductance of the first stage, which would be:
gm=100ua/1v=0.0001/1=100uS
but it's also a simple matter to state the voltage output of the first stage because the current iR1 produces a voltage Vout1=iR1*R1 so the voltage is:
Vout1=Vin/R2*R1
so the voltage gain of the first stage is:
Vout1/Vin=R1/R2
The second stage (the right side op amp) has a gain R3/R1, so the overall voltage gain of the whole circuit is:
Av=(R1/R2)*(R3/R1)=R3/R2
We might be able to call the two current mirrors used that way a "Norton Amplifier" because it works on the input difference of two currents rather than the difference of two voltages, for what it is worth. More strictly speaking though we could probably call the two current mirrors combined with the output op amp stage a Norton Amplifier because it takes the difference of two input currents and amplifies that difference into a voltage output. We could look into this more.
But the negative battery is shown as a positive battery.Hi,
Yes but one is specified as -15v rather than just plain old 15v. That also makes one of them negative.
If you built the circuit on a solderless breadboard then the capacitance between the many rows of contacts and the many long jumper wires all over the place might cause oscillation.When I try to regulate offset voltage of op amp, the output voltage starts oscillating.
If you built the circuit on a solderless breadboard then the capacitance between the many rows of contacts and the many long jumper wires all over the place might cause oscillation.
The positive and negative supplies need a bypass capacitor very close to pin 4 and pin 11 of the opamp and close to ground to prevent oscillation.
The unused opamps should be properly disabled to prevent them from oscillating.
But the negative battery is shown as a positive battery.
Sorry, I forgot that the power supply pins of the opamp have current mirrors that also might make them oscillate.You mean I should add capacitor to negative and positive supplies of both 2 op amps?
When I try to regulate offset voltage of op amp, the output voltage starts oscillating.
I want to see this kind of oscillation in pspice simulation but I have two problems,
First, I don't know how can i find an op amp with offset pins because almost all of these
sort of ports are unmodeled. Second, I don't know how to apply noise to the circuit in
pspice to see the oscillation. I ran into a big problem with circuit. Please Help :-((
Hi,
An interesting trial would be to connect the output of the current mirror to the inverting terminal rather than the output of the op amp.
Hi,
I'm sorry, i worded that wrong. I meant this:
An interesting trial would be to connect the output of the current mirror to the inverting terminal of the first op amp, rather than having the output of the first op amp connect to the inverting terminal. The resistor 10k would still go on the output of the first op amp, but the feedback would now come from the output of the current mirror rather than the output of the op amp.
Just in case that is still not clear, another way of saying this is to do this:
Disconnect the output of the first op amp from the inverting terminal but not from the 10k resistor. That leaves then inverting terminal open.
Next connect the output of the current mirror to the inverting terminal, then try running it up and see what happens. Leave the output of the current mirror connected to the 1k too and the second op amp can stay in place also. It's just that now the feedback for the first op amp comes from the output of the current mirror rather than the output of the first op amp. That's the only difference.
This is just something to try to see how it works that way.
To simulate this more exactly and maybe get more real world results with either circuit, build up the op amp circuit itself using the data sheet internal diagram for the op amp which uses transistors. You can use very low power signal type transistors for the parts they show in the diagram.
Hi,
That looks right but it looks like it is causing a latch up so something else would have to be added to prevent that. The idea was to feed back the output of the current mirror in order to ensure it was corrected, if needed. We'd have to investigate this more to see why it latches up.
In the mean time, are you saying that you get 1.04v out before this change? So it doesnt oscillate anymore then?
Hi,
What is the frequency of oscillation? That could tell us what is causing this.
Frequency of oscillation is 2Mhz, frequency is high for this circuit I don't know what should I do to reduce it :-((
In your new circuit, dont you have to have one end of that 1k resistor (on the output of the lower current mirror) connected to ground? Then the output of the current mirror would also connect to the non inverting terminal of the right side op amp.
Hi,
In the lower section, if you apply a voltage the output current mirrors generate a current, and that current should run through a resistance so that it produces a voltage. In the top circuit this is no problem because the 1k goes to a virtual ground, but in the lower section the 1k goes to a very high impedance of 500k. That might not work. Remember the current mirrors generate a CURRENT not a VOLTAGE by their nature.