I need help designing an op amp with both a voltage gain and a variable current gain. My understand of op amps isn't perfect but I do know how to design an op amp circuit with a voltage gain and one that acts as a current source. That is relatively simple but my problem is combing them to have an op amp circuit with both a voltage gain and a variable current gain. More specifically, I need a voltage gain of 2 (5V to 10V) and a current gain of at least 300 (1mA to 300mA).
The goal of this circuit is to drive a piezoelectric actuator. I am using NI instruments that don't output enough current and that is why I need to design this circuit. My problems are as follows:
If I design a non-inverting op amp with a gain of 2 and then place my actuator across the output voltage, the actuator (correct me if I am wrong) will only draw as much current as the op amp can output. How do I design the circuit so I can drive a set amount of current to the actuator while maintaing a constant voltage gain? Could I put a potentiometer in series with the actuator to maintain the current but then wouldn't the voltage drop across the actuator? I hope I am being specific and not too confusing. I imagine this circuit will be pretty simple but I just can't put the pieces together. Any help would be greatly appreciated.
Electronic – Op amp design with both voltage and current gain
operational-amplifier
Related Solutions
The main issue I see in your schematic is that it looks like your op-amp power pins are not connected to anything. You label them with the VCC, and -12V nets, but you never connect these nets to any power supply that I can see.
For most situations where you use a linear regulator, you want to arrange for the amplifier/feedback circuit (the op-amp in your case) to be powered from the input voltage.
The second issue I see is that the way R4 and R5 are set, you are trying to get an output voltage of
\$ (6.2\ \mathrm{V})\times\dfrac{9000 + 240}{240}\$
which is 238 V, but your input voltage is only 12 V. In a linear regulator, the output voltage must always be less than the input voltage. In your scheme you're even more restricted, because the Darlington scheme of QZ and Q1 your maximum output voltage will be about 1.4 V below the input voltage.
Edit
I'll put my comments on the later versions here so the comments can eventually be cleaned up:
Second version
This is better, but I don't think you can get 10 V out. The AD841 datasheet isn't super clear on this, but I suspect its maximum output is about VS+ - 3 V (see Fig 2 in datasheet). Then subtract off 2 Vbe drops for Q2 and Q1, and you're not going to get more than 7.6 V out of this circuit. An op-amp that has "rail-to-rail" output would work better.
Third version
In this version your feedback resistors are set for a target voltage of 43.4 kV.
Also, be aware if you want to build this in reality, the max power supply of AD8542 is 5.5 V. Applying 12 V to V+ is likely to damage the device. This behavior will probably be ignored by the simulation models
Fourth version
Try reducing R1 to 1.2 kOhms. From the NXP datasheet for your zener, the zener voltage is specified for 5 mA current. Also make sure the simulator is seeing all the connections right around R1. With 12 V on one side, and 0 V on the other, there should be 1 mA through R1, not 24 pA. Are your probes somehow changing the operation of the circuit? Why is there a connection dot at the "bottom" end of R1?
How about this? You do give up very high impedance input, but the resistors do not have to all be 1K.
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Best Answer
Your description is not too bad but you have confused yourself by attempting to apply terminology which is inappropriate. The only way to force both current and voltage to both have fixed gains at the same time is to modulate the load impedance as well. (As i= V/R or V/z so they are all related.
What you want (and it's easyish) is an output current buffer and voltage gain. Effectively the current buffer is a power amplifier as it can provide more VI product = power at the same voltage swing as the input. Gain can be unity and in fact the buffer part of the circuit below has slightly less than unity gain. The opamp makes up for this overall.
You do not set the current - you just provide the ability to provide what current is needed when V and Z are set and the buffer provides whatever current is needed to maintain the voltage required.
The easy starting point is an emitter follower at the opamp output if you want only +ve drive (Q1 below) and an equivalent inverted follower if you want bidirectional drive (Q2 below). Effectively a current buffer. The op amp includes the buffer in its feedback loop in the final circuit.
Note carefully that both the buffer transistors are "emitter followers" - unity voltage gain (or less) stages with the output taken from the emitter of both the transistors. You will see numerous circuits where the transistors are both NPN (usually) or both PNP (sometimes) with the output taken at the midpoint between them. This arrangement is sometimes referred to as "totempole" output - it has advantages for some applications but is inferior when current dive capabilities are the aim.
The circuit below is a starting point. In practice it MAY be slightly more complex, but not much - and in many cases this circuit will work OK as-is.
Note that they do not show feedback connections - the output becomes the current buffered amplifier output and you take feedback from there, not from the opamp output.
Satisfy yourself as to how this works and then ask more questions.
Circuit from here
Note that there is an about 2 x Vbe = 1.2V dead spot as the opamp crosses the V+/2 area. This is not seen in practice as the opamp output 'slews' top compensate but it can cause problems at higher frequencies. There are ways of removing this which are not shown here.
They say:
Because almost all monolithic amplifiers use class-B output stages, they have good loaded output voltage swings, delivering ±10V at 5 mA with ±15V supplies. Demanding much more current from the integrated circuit would require, for one, that the output transistors be made considerably larger. In addition, the increased dissipation could give rise to troublesome thermal gradients on the chip as well as excessive package heating in high-temperature applications. It is therefore advisable to use an external buffer when large output currents are needed.
A simple way of accomplishing this is shown above. A pair of complementary transistors are used on the output of the opamp to get the increased current swing. Although this circuit does have a dead zone, it can be neglected at frequencies below 100 Hz because of the high gain of the amplifier. R1 is included to eliminate parasitic oscillations from the output transistors. In addition, adequate bypassing should be used on the collectors of the output transistors to insure that the output signal is not coupled back into the amplifier. This circuit does not have current limiting, but it can be added by putting 50Ω resistors in series with the collectors of Q1 and Q2.
The circuit below shows the same principle used in a complete circuit with negative feedback applied from buffer output back to op-amp inverting input.
Circuit from here
c:\in\AN LED strip high side drive.jpg