# Fix a node voltage

Discussion in 'General Electronics Chat' started by beedees, May 28, 2015.

1. ### beedees Thread Starter Member

Mar 29, 2015
37
0
Hi!
I'm trying to obtain a voltage of circa 0.4 V in node A, as you can see in the attached circuit.

So I decided to use an adjustable 3-terminal positive voltage regulator (LM317LZ) and two resistors in order to have an output of 0.4V. This voltage is applied to the non-inverting terminal of op-amp and the inverting terminal is connected to node A. To accomplish my goal a current has to flow from VDD towards the BJT, so the BJT must be on!
I'm trying in every way! But I don't understand why I am mistaken...
Maybe I have to pay attention to the input common-mode voltage range of the op-amp. That in the circuit is a rail-to-rail op-amp, but it is only an example. I don't know if it could be a possible solution.
Maybe I need a resistor between the output of op-amp and the base terminal of the BJT.
Have you any ideas? I don't know what is wrong.
Thanks!

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
Did you try swapping the inputs to the op-amp? The feedback, as you currently have it, is of the wrong polarity.

Last edited: May 28, 2015
3. ### AnalogKid AAC Fanatic!

Aug 1, 2013
5,655
1,606
In the common-emitter configuration you have, the NPN transistor is an inverter. Think of it as an inverting opamp in series with the first one. For the overall negative feedback to work, the nature of the inverting and non-inverting inputs of the first opamp must be viewed with respect to the output of the second amp (your NPN). This means that the + input is acting as the - input, and vice versa.

ak

4. ### beedees Thread Starter Member

Mar 29, 2015
37
0
Yes, now it seems to work!

Effectively I haven't understood very well how the feedback works in this particular case. Why is the + input acting as the - input and vice versa? I don't understand this passage..
I think I realize that the NPN transistor is an inverter because when it is ON the collector is at ground and when it is OFF the collector is high (the collector is the output). But I don't realize how I can consider it as an inverter connected to the first op-amp, and then how I can calculate the overall feedback.

5. ### AnalogKid AAC Fanatic!

Aug 1, 2013
5,655
1,606
Although logic inversion and signal inversion are not exactly the same thing, they are similar enough in this case. In the same way that running a logic signal through two inverters inverts it twice, acting essentially as a non-inverting buffer and returning the output signal to the logic polarity (or phase) of the input signal, opamps can do the same thing. For negative feedback to work, the output signal, wherever it comes from, is fed back to the inverting input; that is, the input that is out of phase with the output. But if you put an extra inverter in the signal path, even a linear amplifier inverter, then it is the + input that is out of phase with the overall output and the - input that is in phase. This effectively swaps the actions of the two inputs.

ak

6. ### beedees Thread Starter Member

Mar 29, 2015
37
0
Ok! Now it's clearer. I have understood! Thanks.
Since the voltage in input to the voltage buffer amplifier is small I suppose I have to choose either a rail-to-rail op-amp or an op-amp characterized by a low input common-mode voltage range. Otherwise I risk the op-amp isn't able to provide the desired voltage, right?

Last edited: May 28, 2015

Aug 1, 2013
5,655
1,606
Correct.

ak

8. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
Notwithstanding AK's agreement, I am puzzled by your statement concerning the op-amp requirements - at least in your current application.
You stated that the op-amp should have a low input common mode range or have rail-to-rail output capability. Surely these are distinctly different issues in general and (in particular) with respect to your requirements. Why would it be an either / or situation? Perhaps I have mistaken your meaning.
I'm not sure what your circuit is really doing other than setting a "node" voltage. Where does the Wheatstone bridge concept play its part in the overall design?
In any event, the criteria that your op-amp must meet are a low input voltage on the inputs and the ability to drive the base of the transistor Q1. Perhaps if you are seeking accuracy for the controlled node voltage with respect to the input reference, you may require an amplifier with a low input offset voltage. Although your initial preamble (post #1) seems to indicate otherwise.
If the op-amp common mode input range includes ground with single supply operation, then you have no issues on that score. Also, driving Q1 base via a suitable resistor should present no problems, since the base is going to be much less than 1V above ground. It's unlikely you would have output headroom problems so long as the base resistance is suitably chosen. There are probably more suitable op-amps than the one you have chosen, particularly if precision & stability in the node voltage control are important factors - otherwise any general purpose single supply op-amp with an input common-mode range including ground would suffice. You could probably even get by with a vintage LM324 (wash out my mouth with soap). If accuracy and stability are important you might go to an amplifier with characteristics typical of the LT1006 for instance.

Last edited: May 28, 2015
9. ### AnalogKid AAC Fanatic!

Aug 1, 2013
5,655
1,606
Don't be picking on the LM324. Along with the LM339, they changed the world of analog and interface circuit design. Despite all of the raving about gigundous production numbers for CPU, uC, and RAM devices, very few part numbers rack up nine or ten digits per year. 324, 358, 339, 555...

ak

t_n_k likes this.
10. ### Bordodynov Well-Known Member

May 20, 2015
1,114
335
Hello. I would like to remind you that if even change the input pins of the operational amplifier, the system may not work (the question of stability). And most likely require additional elements of correction. In any case it is necessary to use modeling.