# How to make a symmetrical DC power source?

I have been tasked with creating a logic probe that reads High/Low/High Impedance, but what works in simulations, isn't working in practice. This was my idea:

I made a symmetrical power source using two resistors and a virtual ground, and fed the 741 OP AMP using these +VCC/2 and -VCC/2. The idea was for the probe voltage to be compared to half VCC, that way we'd get a -VCC/2 to +VCC/2 voltage output, and if there was high impedance, both LEDs would be off.

What ends up happening however, is no voltage output. I think the problem is in the symmetrical power source, because whenever the OP AMP is connected to the circuit, the perfect VCC/2 equilibrium between the two resistors is gone, and it goes to like 4v or whatever when feeding it with 5v.

Any ideas on why this circuit doesn't work? I have tried searching for ways to create a symmetrical power source, but all of them give me the same simulation results, so i figured it wouldn't change on practice.

• Shouldn't your + terminal of the opamp be connected to the virtual ground between the resistors? – pjc50 Dec 8 '17 at 20:07
• What are the voltage levels you need to probe ? – altai Dec 8 '17 at 20:08
• Use a "rail-to-rail" opamp. The 741 is not guaranteed to work if the inputs or output are within 3 volts of the supply rails - tough to use with a total 5 volt supply! – Peter Bennett Dec 8 '17 at 20:27
• @altai I can choose it, but i decided 2.5v and above is High. – Alex RD Dec 8 '17 at 20:44
• @pjc50 Not really, because that would give me 0 volts. – Alex RD Dec 8 '17 at 20:49

Any ideas on why this circuit doesn't work?

Oh yeah. In one word (or at least one number): 741.

Never, ever, try to use a 741 with power supplies much different from +/- 15 volts. +/-12 is probably OK, and I've seen +/- 9 used. +/- 2.5? Hahahahaha.

Sorry. I got carried away.

But seriously, that's your problem (or at least the most obvious one). If you really must operate from a 5 volt supply, you need to use a comparator (not an op amp) rated for that voltage.

And your LED circuit is asking for trouble. If you use an LED with Vf less than about 2.5 volts and a rail-to-rail drive, you risk killing the LED. If you use something like a white LED, with a 3 - 3.5 volt threshold, you will get little or no output from the LED.

Umm. And let's talk about your 5 volt supply. You realize, I hope, that you will be tempted to use the same 5 volt supply for your logic as your probe. And if you do you cannot possibly establish your virtual ground. Look at your schematic again, and put a ground connection on the - side of the 5 volts, and think about what will happen.

Finally, be aware that the LED current will affect the midpoint ground set by the two 1k resistors. In fact, if you have (let's say) 10 mA through the LED, this is 4 times the nominal current through the resistors and will swamp them.

EDIT - In comment, an alternative (LM339) was suggested.

That won't work either, and in some respects is an even worse choice than a 741. Regardless of supply voltages, it cannot drive the High LED at all. The output of the LM339 is essentially a one-way switch between the output and the - supply (called ground on a 339). When active (LOW output) it will turn on the Low LED. When not (HIGH output) it will do nothing at all, so the High LED will never turn on. See page 10 of the data sheet.

It also has considerably higher current capacity than a 741, so is almost guaranteed to kill the Low LED if it does turn it on. LEDs need current limiting.

• I'm sorry, i should have explained it better. In trying to make the post as succinct as possible, i left out some important information. First, i did try different OP AMPs other than the 741 like the LM339, and they yielded the same results. Second, this is the circuit for the probe. What i meant by the voltage source in the non-inverting input is the voltage it's going to analyze if it's low or high. But the forward voltage of the LED is something i completely overlooked, and just assumed it to be the same as a normal Diode like 0.7v. – Alex RD Dec 9 '17 at 17:07
• @AlexRD - See edit. – WhatRoughBeast Dec 9 '17 at 17:24
• I was thinking that the output resistance would be enough to limit the current, especially since the maximum output voltage would be 2.5v - 2v VF, giving me a current of 0.5v/RS. I still can't think of a solution to this problem. Say i find an amplifier that can be fed with low VCC like 2.5v, i still have the problem of the output current affecting my resistors current. Is there a way to create a symmetrical power source out of one voltage source that doesn't get affected by the output? – Alex RD Dec 9 '17 at 17:58
• @AlexRD - Yes, essentially you make the imedance of the virtual ground much less than loads. In your case, you would buffer the virtual ground with something fast and beefy. – WhatRoughBeast Dec 10 '17 at 2:53

Ok, it sounds like what you're trying to do is basically amplify the signal from the probe using your op amp, so it's like you have the leds connected to your test circuit, except they don't draw any current.

One issue with this circuit is that the op amp gate is VERY high impedance. Since you're not pulling any real current from your probe, you can't tell whether it's high impedance or not. You'll need to add a small amount of impedance by connecting a resistor between the probe and "GND". 50k aught to do it.

The second issue is the op amp configuration. The op amp goes high when + > -, since + is tied directly to V-, the op amp will always be low. Set it up like a regular non inverting amplifier, with feedback resistors and all that, then just make the gain something like 10 or 100. Don't forget the v- is NOT your ground.

3rd issue. The virtual ground will swing around a lot due to those LEDs. An easy fix is to keep your voltage divider, but add a second op amp to act as a buffer for its output.

4th issue. This is a small one, but now that your virtual ground isn't so current limited, you'll need to put a current limiting resistor between the leds and ground. I'd recommend keeping the max current low, like 10ma maybe, unless you're sure your op amps can handle more than that.

5th issue, I just noticed, your supply voltage might not be enough for that 741 op amp. Also don't forget that many op amps (like this one) are not rail to rail, so the max output wont be -2.5 to 2.5 (for 5v), but more like -2 to 2. You could try to find a lower voltage op amp, or you could increase the supply voltage.

I made some (very ugly) edits to your schematic to illustrate these changes. It may have errors due to me rushing, but it should get you moving in the right direction.

• The outputs of the 741 have to stay over two volts away from the rails. So, no useful output range when operating from 5V. The inputs on the 741 have to stay over 2V away from the rails, so no useful input range when operated from 5V. Did I mention that the 741 is an ancient piece of crap that you shouldn't be using anymore? – JRE Dec 8 '17 at 21:58
• I'm sorry, i explained it very poorly. The voltage source called probe is the signal i'm trying to analyze. So that could vary from 0 to 5v, so when i compare it to the to half VCC it either goes to High or Low depending on the source. The problem i'm having is the output current affecting the virtual ground current, making the voltage division in those resistors uneven. I have tried your buffer idea, but even in simulation the resistor voltage is still uneven. Also, yes i do absolutely need to use an amp op other than the 741 heh. I really appreciate the help, thank you. – Alex RD Dec 9 '17 at 18:21
• Ok I guess I did misunderstand. I know the probe is the signal, but I thought you wanted -2.5v = low light on, +2.5v = high light on, and nothing connected = no lights. You should only have to change to op amp configuration (and the loose the 741) to get 0=low, 5=high, and high z=both off. In fact, I think all you'll have to do is change the probe voltage reference to v- instead of vGND. – Drew Dec 10 '17 at 0:18