I am trying to understand the way a JFET op-amp works. I have a circuit like this:

enter image description here

You can see the LF347N chip I am using.

The IN+ terminal is connected to a TTL switch providing 0V or 5V. The VCC+ and VCC- is connecting to +15V and -15V. I just searched the manual of the LF347N, I don't understand what signal is at the OUTPUT terminal. Is it voltage or current? What's does the above circuit really do? My guess is that the TTL signal switches the output between 15V or -15V.


2 Answers 2


Any voltage feedback opamp works the same way. If operating it as an amplifier, negative feedback from the output to the inverting input attempts to do one thing: -

Keep both inputs at the same voltage level.

What I've just said is an approximation to reality but is a reliable and accurate approximation. So, in your circuit, the output produces a voltage that ensures the inverting input voltage equals the noninverting voltage and, because the inputs can be regarded as having infinite impedance, the voltage fed back equals the output voltage.

This makes your circuit a unity gain amplifier. Should you decide to add a resistor from the inverting input to signal 0 volt, the output will get bigger in order to satisfy the need to keep both inputs at the same voltage level. Now you have gain.

  • \$\begingroup\$ The circuit was found in a note and I don't know what's that for. But someone told me they might be used as a switch. Based on your reply, I think what you mean is above circuit is actually a voltage follower, if I send TTL HIGH (5V), it will send out some voltage about the same level if I choose the right feedback resistor. But if there is TTL LOW voltage (0V) fed, no voltage will be output. Is that right? \$\endgroup\$ Jan 25, 2014 at 22:17
  • 1
    \$\begingroup\$ You have described a voltage follower perfectly dude. \$\endgroup\$
    – Andy aka
    Jan 25, 2014 at 22:34
  • \$\begingroup\$ Thanks. I read a book on analysis of op-amp. One example is for the non-inverting amplifier, which have an input connect to V+, and there is a resistor (like R1 in the above diagram) connected the Vout and to V-, also, there is another resistor resistor (says R2) connecting V- to ground. After some math, it gives Vout = V+ * (1 + R1/R2). So I am going to use this result to analyze the diagram given above. In our case, there is no R2 connecting to the ground so I could chose R2 as infinity, which gives Vout = V+. So why we need R1 actually? \$\endgroup\$ Jan 25, 2014 at 23:13
  • \$\begingroup\$ The answer is you don't need that resistor for unity gain amplifiers. It's as simple as that dude. \$\endgroup\$
    – Andy aka
    Jan 26, 2014 at 1:04

I am trying to understand the way a JFET op-amp works.

The type of input transistor in the op-amp does have an effect, but not in a way that affects this answer. My answers below apply to any voltage feedback op-amp. (Current feedback op-amps are a different story.) JFET, bipolar, CMOS... it doesn't matter at the level of your question.

I don't understand what signal is at the OUTPUT terminal.

The voltage at OUT will equal the voltage at +IN because it's configured with OUT tied to -IN, which makes it a buffer. (Gain of 1.)

This is not specific to this one chip, it's a general property of op-amps.

Is it voltage or current?

Most op-amps are of the voltage feedback type, so they are best thought of as working on voltage. Since it is a simple G=1 buffer circuit, OUT equals +IN, but there are many possible configurations of feedback that give different results.

That said, op-amps also act something like buffers, meaning they can be used to convert a high-impedance source to a low-impedance. That is, they can put out more current than they take in on their input pins. In that sense, op-amps can be current sources, or current sinks.

Typical IC op-amps can source or sink 20 to 40 mA on their output while there is only microamps or picoamps of current flow on their inputs. The op-amp gets that current from the power rails.

My guess is that the TTL signal switches the output between 15V or -15V.

The schematic shows ±15 V supplies because most op-amps can't swing their output rail-to-rail. The LF347 is typical in this regard, with an output swing that can be as much as 3 V lower than its supply voltage. (See the "Output Voltage Swing" spec on page 3 of that datasheet.)

If you were to run this chip from the same 5 V single-ended TTL supply driving its input, the chip would basically not function. The ±3 V output-to-supply limit means you need more than 6 V across the power pins for this op-amp's output to be able to swing at all. ±15 V is a very common voltage to use for old op-amps like this, even when the full swing range isn't required. ±12 V is also very common.

I recommend that you read the free e-book Op Amps for Everyone. There is an updated paper version, but the online version is still quite good and useful.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.