4

I want to make a +/- 2.5 V rail from a 5 V supply using an op-amp. I was wondering if I could use a totem pole for driving the output like this:

opamp_totem_pole

Image from here

But while I was looking at it, I noticed that since both transistors need to be (almost/close to) half way open, they will act like a resistor and will just sink current to GND (the original GND) even if I do not use the new "Virtual GND)

Am I right? None of the tutorials I watched mentions this.

Also, which is my return path for current for the devices I power from the new "Virtual GND"? Is it the Virtual GND or the original GND?

I need to understand this in order to know where to connect the copper pour.

Null
  • 7,448
  • 17
  • 36
  • 48
Christianidis Vasileios
  • 2,645
  • 2
  • 9
  • 32

3 Answers3

5

But while I was looking at it, I noticed that since both transistors need to be (almost/close to) half way open, they will act like a resistor and will just sink current to GND (the original GND) even if I do not use the new "Virtual GND)

In your circuit (an op-amp follower buffered by a complementary emitter follower), always one of the two transistors will be off and the other will operate in active mode (as a single emitter follower) as follows:

If the return current tries to enter the buffer (the device "blows" current through the load into the buffer output), the virtual ground will try to increase. The op-amp will sense this increase and will decrease its output voltage with approximately 0.7 V so the virtual ground will stay exactly in the middle. The return current will pass through Q2.

If the return current tries to exit the buffer (the device "sucks" current through the load from the buffer output), the virtual ground will try to decrease. The op-amp will sense this decrease and will increase its output voltage with approximately 0.7 V so again the virtual ground will stay exactly in the middle. The return current will pass through Q1.

As you noted, it is preferable that always "the two transistors act like a resistor" connected between the supply rails or, more precisely speaking, like a voltage divider. For this purpose, you have to insert a biasing network between the op-amp output and bases consisting of diodes and resistors like in the similar circuit diagrams below:

Push-pull stage at negative input voltage

Fig. 1. Push-pull stage at negative input voltage

Push-pull stage at zero input voltage

Fig. 2. Push-pull stage at zero input voltage (OP's case)

Push-pull stage at positive input voltage

Fig. 3. Push-pull stage at positive input voltage

Also, which is my return path for current for the devices I power from the new "Virtual GND"? Is it the Virtual GND or the original GND?

I think the answer is already clear - the virtual ground.

It would be extremely useful for you to draw the full paths (loops) of the currents in all cases like in the pictures above.

Circuit fantasist
  • 13,593
  • 1
  • 17
  • 48
3

That splitter circuit seems good in principle. Problem with it is that it won't work with a supply as low as +5v (+/-2.5V) with a 741 op amp.

The opamp works by driving its output positive or negative by about 1.2V to turn on the appropriate transistor to source or sink current depending on which of the two loads is largest.

The 741 can only drive its output to within about 2V of either supply rail before it saturates. With a +5V supply, 0.7V dropped by the polarity diode and +/-2V saturation limits it only leaves about 0.3V available for opamp output swing, far less than the +/- 1.2V required.

The circuit should work if you replace the 741 with a rail to rail output op amp such as a MCP601 op amp which will work with a +4.3V voltage supply.

2

As far as I know, in order to generate a negative voltage you need:

  1. A charge pump made of 2 diodes, 2 capacitors and a square wave input signal

enter image description here or

  1. A DC/DC converter

See this answer:

How to create a negative voltage supply?

Enrico Migliore
  • 4,161
  • 1
  • 9
  • 12
  • I did not know that. Cool! But I think this will generate switching noise to my supply, which I do not want. I will also need filtering afterwards for the ripple so probably not. – Christianidis Vasileios Mar 30 '21 at 15:42
  • 3
    It's just a matter of what you define as ground. A rail splitter as proposed in the question is a totally fine concept, as long as the current is quite small. If you define ground as half of your supply voltage, you automatically have a negative supply. You just need to keep your new ground stable. – jusaca Mar 30 '21 at 15:44
  • You will have to filter out the residual ripple on both negative and psitivie supply rails. It won't be difficult. A capacitor is more than enough because operational amplifiers have excellent PSRR's (Power Supply Rejection Ratio). – Enrico Migliore Mar 30 '21 at 15:46