• Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

12v to +-6V without changing main (12V) supply

gophert

Well-Known Member
Most Helpful Member
Apologies got busy last night, things fell apart.
Heres a schem I found, the diode keeps the op amp o/p a diode drop above ground so it always conducting, this circuit will only work reliably if the op amp i/p is kept above ground, some op amps dont like it if you try to drve the o/p below or above the rails, you could give this circuit gain by adding resistors.
Speaking of which in #7 I mentioned care is required in op amp selection, thats one of the reasons for the diodes in #8, they protect the op amp from the i/p going outside the rails.
There are if you look in suppliers blurb op amps that are meat for single rail supp;y such as the Lm358 (which would probably be fine wthout the diode below, you'd need that for something like the wonderful 741), its been with us a long time and is capable of simple applications.View attachment 125119

I thought the goal was an absolute value amplifier?
Your schematic is simply a half-wave rectifier that clips the peak of a 24v p-p input signal by one diode drop.

6258DCF7-885B-4CE0-B6A3-0BA9C4F0B0BE.jpeg
 
Last edited:

dr pepper

Well-Known Member
Most Helpful Member
For AC yes, like I said in the post this is for Dc voltages above ground where the diode would always be conducting.
If the o/p was close to ground the op amp would be a 0.6v.
Maybe I misunderstood the op, I thought this was a Dc amplifier.
 

v1.5

New Member
Super simple. The diodes only protect the op amp to keep input within allowable range. The diodes are not in the signal path - even though this is a precision rectifier circuit.

That is, the non-inverting amp needs the diode on the (+) input to keep input within 0.3V of negative rail (use schottky diode) during the bottom half cycle.

For the inverting amplifier, no diode is actually required if the op amps output can always properly drive the feedback resistor by reaching +12V when the AC source is at -12V (to achieve 0V at (-) input). Otherwise, the diode labelled "optional" should be used.

It's one dual op amp (rail to rail in/out) that can operate at 12VDC. Three resistors and a diode (second diode is optional).
I hope this works for you.

View attachment 125118
View attachment 125117
Hi
Many thanks for that very simple and usefull circuit.

But i need to understand one thing about this circuit. I didnt clearly understand positive cycle of supply.

For example let say input voltage is 5v. The unity gain amplifier’s output will be 5 v too. And this voltage will also be at inverting amplifier’s positive pin. Negative pin of inverting amp will be 5v too because opamp wants to keep 0v differantial voltage on input pins(if my knowladge is true!). On negative pin of inverting amp , there will be resistor divider on 10k and 1k. So the current must be 5/11k. On output of inverting opamp should be (5/11k)x1k. With my solve, which is wrong , it must be less than 5volts. But in simulation it seems to be 5 v ! How ? Can someone explaine how positive cycle works ? Which part of my solve is wrong ? Need to learn please. Thanks
 

gophert

Well-Known Member
Most Helpful Member
The 1k resistor is only there as a "load". As long as the op amp can drive a 12mA load (12V/1k = 12mA), the 1k resistor plays no role in determining the output of the op amp voltages.

For the positive half cycle, let's pick a voltage 5v.
-The non-inverting (Top) amps + input sees 5v and the follower circuit outputs 5v so the (-) pin also sees 5v. Easy.
- that means the (+) input to the Bottom op amp also sees 5v.

Now the hard part, if there were no feedback resistor on the bottom op amp, that op amp would output zero volts in the positive cycle because the (+) and (-) inputs are always the same voltage so the output doesn't measure any difference. However....

With the feedback resistor, a voltage divider is created. If the output is at zero volts when our source AC source is at 5v, the (-) input would see 2.5v. But that is not the same as the (+) input so the bottom op amps output changes until the (-) input matches the (+) input.

For the bottom op amp, During the positive cycle, the only way the (-) input voltage can be at the same voltage as the (+) input voltage is to set the output voltage to 5v. Pick any other positive voltage and it works the same.

Let me know if that makes sense to you.
 

v1.5

New Member
The 1k resistor is only there as a "load". As long as the op amp can drive a 12mA load (12V/1k = 12mA), the 1k resistor plays no role in determining the output of the op amp voltages.

For the positive half cycle, let's pick a voltage 5v.
-The non-inverting (Top) amps + input sees 5v and the follower circuit outputs 5v so the (-) pin also sees 5v. Easy.
- that means the (+) input to the Bottom op amp also sees 5v.

Now the hard part, if there were no feedback resistor on the bottom op amp, that op amp would output zero volts in the positive cycle because the (+) and (-) inputs are always the same voltage so the output doesn't measure any difference. However....

With the feedback resistor, a voltage divider is created. If the output is at zero volts when our source AC source is at 5v, the (-) input would see 2.5v. But that is not the same as the (+) input so the bottom op amps output changes until the (-) input matches the (+) input.

For the bottom op amp, During the positive cycle, the only way the (-) input voltage can be at the same voltage as the (+) input voltage is to set the output voltage to 5v. Pick any other positive voltage and it works the same.

Let me know if that makes sense to you.
Hi.
I understood clearly and that is a very good explanation. Compared to the first circuit, this circuit has a lot of advantages. I will note this circuit and use it in my future work.

I express my gratitude to you elders for sharing the circuit diagram and answering my questions gently.

Thanks. :happy:
 

gophert

Well-Known Member
Most Helpful Member
Hi.
I understood clearly and that is a very good explanation. Compared to the first circuit, this circuit has a lot of advantages. I will note this circuit and use it in my future work.

I express my gratitude to you elders for sharing the circuit diagram and answering my questions gently.

Thanks. :happy:

Note that getting a rail-to-rail output op amp to really get to the positive rail is only possible with a very light load - like a 20k to 100k resistor (or just connecting it directly to the input of an analog/digital converter (ADC) or a logic chip. The 1k load that I showed is not really possible if supplying only 12VDC to the op amp. Summary, ideally, an op amp acts like an ideal op amp. In reality, not so much.
 

Latest threads

EE World Online Articles

Loading

 
Top