maybe it's best explain with an example:
This is my circuit:
The way I read the graph is that if I require Iout=2amps then difference between Vsupply and Vout must be at around 3V. (though I'm unsure what Vsupply means in the case since it's a +/- supply)
Here's the graph (There's a high chance I'm looking at the wrong graph but I cannot find anything else suitable in the datasheet)
Here's the full datasheet: https://www.ti.com/lit/ds/symlink/opa548.pdf
This is my circuit:
- Vin is connected to a 5V reference (via a zener diode or a chip like a REF02)
- Vsupply is +/- 16-18V (controlled via an LM317 or equivalent chip so it can be anything within the LM317 limits)
- Vout is calculated at around 12V (Rf=24Kohms | Rin=10Kohms ) (5V * 24/10 = 12V)
The way I read the graph is that if I require Iout=2amps then difference between Vsupply and Vout must be at around 3V. (though I'm unsure what Vsupply means in the case since it's a +/- supply)
Here's the graph (There's a high chance I'm looking at the wrong graph but I cannot find anything else suitable in the datasheet)
Here's the full datasheet: https://www.ti.com/lit/ds/symlink/opa548.pdf
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The graph is in effect showing how close to the supply rails the OPA548 output terminal will swing at various output currents. If you need 2A (for example) then you can only swing to 2.2V above the negative rail (bottom line) and 3V below the positive rail (top line).
Are you planning to use OPA548 as a voltage regulator? If you are using it as a poweramp then your included graph will tell you the maximum swing you'll be able to get into various loads, up to the current limit (3A continuous, 5A peak)
Are you planning to use OPA548 as a voltage regulator? If you are using it as a poweramp then your included graph will tell you the maximum swing you'll be able to get into various loads, up to the current limit (3A continuous, 5A peak)
1) "it can be anything" does not cut it; for any answer including numbers you MUST choose a rail supply voltage and stick to that.
2) then suppose we have +/-18V rails
3) graph shows how close can output get to a rail depending or rail voltage and current.
4) it does show values for both rails, not sure why your doubt.
5) your gain calculation is wrong, in this case it´s (Rf/Rin)+1 so 3.4X
So output will try to reach 5*3.4=17V
But it can´t, best it can do at 2A is 18V-3V=15V
And those are bad/wrong 15V, since you achieve them by clipping, not because of normal operation.
If rails less than +/- 18V, output will be worse.
Not analyzing negative output since you only mentioned +5V reference.
I suggest you study Electronics from the beginning, up to and including Op Amps.
Reading without a previous solid base and asking at every rock you stumble upon is the long and tortuous path, you´ll amass a Ton of individual disconnected answers, it´s better yo follow the proper path.
That said, congratulations on being motivated to learn Electronics.
2) then suppose we have +/-18V rails
3) graph shows how close can output get to a rail depending or rail voltage and current.
4) it does show values for both rails, not sure why your doubt.
5) your gain calculation is wrong, in this case it´s (Rf/Rin)+1 so 3.4X
So output will try to reach 5*3.4=17V
But it can´t, best it can do at 2A is 18V-3V=15V
And those are bad/wrong 15V, since you achieve them by clipping, not because of normal operation.
If rails less than +/- 18V, output will be worse.
Not analyzing negative output since you only mentioned +5V reference.
I suggest you study Electronics from the beginning, up to and including Op Amps.
Reading without a previous solid base and asking at every rock you stumble upon is the long and tortuous path, you´ll amass a Ton of individual disconnected answers, it´s better yo follow the proper path.
That said, congratulations on being motivated to learn Electronics.
Let us analyze what is happening to this circuit.
If we put 1 volt into the non-inverting (+) input, the output voltage will rise until there is 1 volt on the inverting (-) input. At that time there would be 1 volt across the 10K resistor (labeled Rin) so the the feedback and input cancel. If there is 1 volt across the 10K resistor common sense would tell you there must be 2.4 volts across a 24K resistor. Thus there must be 2.4 + 1 volt from the output, in total 3.4 volts.
Of course if you don’t have common sense you can use Ohm’s law. To get 1 volt across a 10,000 ohm resistor for V=I*R, I (current not me) must be V/R or 1/10,000 or .0001 Amps also known as .1 milliamps. If we use Ohm’s law again for Rf the the voltage across it must be .0001 * 24,000 = 2.4 volts.
Now for the really confused V stands for a DC voltage, I for the steady current in amperes (shortened to amps because we write it so much) and R for resistance.
Now if we are not talking about DC but Alternating current V becomes v and I becomes i. However as math folks use i to mean imaginary numbers we do not and instead use j.
However R always means resistance. As it is in theory the same value for both DC and AC circuits.
If we have not just resistance but also an element that contains inductance and or capacitance we call that impedance and is represented by Z.
If we put 1 volt into the non-inverting (+) input, the output voltage will rise until there is 1 volt on the inverting (-) input. At that time there would be 1 volt across the 10K resistor (labeled Rin) so the the feedback and input cancel. If there is 1 volt across the 10K resistor common sense would tell you there must be 2.4 volts across a 24K resistor. Thus there must be 2.4 + 1 volt from the output, in total 3.4 volts.
Of course if you don’t have common sense you can use Ohm’s law. To get 1 volt across a 10,000 ohm resistor for V=I*R, I (current not me) must be V/R or 1/10,000 or .0001 Amps also known as .1 milliamps. If we use Ohm’s law again for Rf the the voltage across it must be .0001 * 24,000 = 2.4 volts.
Now for the really confused V stands for a DC voltage, I for the steady current in amperes (shortened to amps because we write it so much) and R for resistance.
Now if we are not talking about DC but Alternating current V becomes v and I becomes i. However as math folks use i to mean imaginary numbers we do not and instead use j.
However R always means resistance. As it is in theory the same value for both DC and AC circuits.
If we have not just resistance but also an element that contains inductance and or capacitance we call that impedance and is represented by Z.
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It all falls into place easily if you realise that in normal operation, the inverting and non-inverting inputs are basically the same.
After that it is just reasoning as simon7k showed.
Jan
After that it is just reasoning as simon7k showed.
Jan
Ja..
If you can’t even spell my full handle…
Actually the non-inverting input is assumed to be a high input impedance and the inverting input is treated as a short to common or a zero ohm input impedance!
ES
If you can’t even spell my full handle…
Actually the non-inverting input is assumed to be a high input impedance and the inverting input is treated as a short to common or a zero ohm input impedance!
ES
For clarity I´d slightly extend that to:
"in normal operation, the inverting and non-inverting inputs are basically the same. voltage"
meaning:
IF Op Amp has any NFB, output will take any voltage possible so inverting input stays at same voltage as non inverting one.
In this case: non inverting input gets +5 Volts; we have a 24k/10k voltage divider NFB, so output NEEDS to reach +17V for inverting input to reach +5V and exactly match what´s present at the non inverting input.
Again, the inverting input will always try to be at same voltage as the non inverting one.
And what if it can´t?
Then Op Amp will clip.
And if we have NO NFB?
Then Op Amp will NOT be able to keep both inputs same voltage, output will try anyway, and the slightest voltage difference (we are talking a couple mV or less) will slam output against a rail.
In this case it will be working as a comparator.
Or, to be more precise: a difference in the two input voltages will drive the output towards nulling out that difference.
(There will always be some difference between the inputs of course, otherwise there will be no output. But it will be very small - assume an opamp gain of 100,000 and an output voltage of 10V, then the two inputs have a difference of 10/100,000 = 100uV
Jan
Next we can discuss input offset voltage and offset current. A practical limit on the inputs matching. One trick of design is to make the input resistors sized so that the resistance time the offset current matches the offset voltage. That is because as temperature changes that positively affect the offset voltage tend to negatively affect the current! One limit on this practice for audio and other uses is the added noise from high value resistors. The other limit on low value feedback resistors is the limited output current.
Data sheet values are fine for this calculation. Of course if offsets are critical as in some measurement applications there are special opamps and ways to reduce the errors.
Data sheet values are fine for this calculation. Of course if offsets are critical as in some measurement applications there are special opamps and ways to reduce the errors.
It would work if we knew beforehand what the offset (current and voltag) would be and at what polarity ...
Jan
Jan
The graph indicates that the positive output, the top of the totem pole has a minimum voltage drop of (about) 1.75 Volts and a resistance of about 0.66 Ohms. The bottom side has a resistance of about 1 Ohm and a minimum voltage drop about 0.5Volts.
So if you want 2 Amps at 12 Volts, the supply has to be at least 12+3=15Volts. If you want -2Amps at -12 Volts, the negative supply has to be at least -12+ -2.25= -14.25 Volts.
These values imply that the output circuit on the positive side is a Darlington and current source, ie two VBE + Vce, while the bottom is a CFP with some gain, ie the op-amp has a "quasi complimentary" output.
So if you want 2 Amps at 12 Volts, the supply has to be at least 12+3=15Volts. If you want -2Amps at -12 Volts, the negative supply has to be at least -12+ -2.25= -14.25 Volts.
These values imply that the output circuit on the positive side is a Darlington and current source, ie two VBE + Vce, while the bottom is a CFP with some gain, ie the op-amp has a "quasi complimentary" output.
Usually as a range around zero. If you can predict it will be + or -, you should instead play El Gordo.