Op Amp Question. Can this be done?

Can we give positive voltage to V- pin of an Op-Amp? I am asking this because in the book "Introduction to Electric Circuits" by Dorf and Svoboda 9th Edition page 220. The Figure shows +V being applied to V- pin of Op-Amp.

I did not see the book but i think you can apply positive voltage to V- pin,as we know that V- pin is connected to emitter of npn & collector of pnp in IC 741 op-amp. But that would limit the voltage swing at the output if a simultaneous increase is not present at V+ pin(u can't raise Voltage at V+ probably beyond 22/25V).

Yes. We can give. It'll work fine.
It matters only the potential difference.

It should work even if u give +18V at +V and +5V at -V. (But u should plan for the I/P and O/P accordingly)

An op-amp allows any kind of input on the input pins. The V- pin is called the inverting input, and V+ pin is called the non-inverting input.
Considering the case when the gain ratio (Vo/Vi) is 1 (no amplification nor attenuation):
* The inverting input (V-) will take any input, and will create a 180° phase shifting (will flip the signal upside down) on the output. So, if the input is a triangular signal like /\/\/ the output of the op-amp will be \/\/\.
* The non-inverting input (V+) will take any input, but as opposed to the inverting input, it will not create any phase shifting: what you get in will come out. So, if the input is a triangular signal like /\/\/ the output of the op-amp will be exactly the same, /\/\/.

I am not talking about the inverting (-) or non-inverting (+) pins of Op-Amp. I am talking about the V+ and V- power pins of the Op-Amp.

Hello!

Can we give positive voltage to V- pin of an Op-Amp? I am asking this because in the book
"Introduction to Electric Circuits" by Dorf and Svoboda 9th Edition page 220. The Figure shows
+V being applied to V- pin of Op-Amp.

Click to expand...

It depends on what you call V+ and V-, I don't have this book.

For instance in this tutorial:
https://gadgetgangster.com/tutorials/428
V+ is the positive supply and V- the negative supply.
In this case, of course, you should not put more in V- than in V+.

Examples: -5V in V- and 5V in V+, or 0 in V- and 20V in V+, etc...
Note: verify in the specs (absolute maximum ratings) which power supply you can feed.
There are small opamps that will stand only 5V, single supply (or -2.5, 2.5).

Now let's call the power supply +Vcc / -Vcc and the inputs I+ and I- so that everything is clear.
In this case, both I- and I+ can be anywhere between -Vcc and +Vcc.

Dora.

You should have specified that you're talking about the power supply, not the inputs.
In this case, as of any polarized device, it will not work if the power supply is not properly polarized. Inverse polarization will break the device.
But, you can put +45V to the V+ pin and +15V to the V- pin. Their voltage differential is 30, and it's polarized V- to V+ (lower to higher voltage), as same as it would happen when you're using +15V on V+ and -15V on V-. But you can also put -50V at the V+ pin and -80V at the V- pin. Their differential is still 30 V polarized V- to V+, the same as in the previous two power conditions.

Applying a positive voltage on the V- is possible and allowed under certain circumstances. But you should consider that if the V- voltage is positive (or 0, GND), a negative voltage input will be unable to be output as negative, and instead, it will be output as the lowest voltage available on V-. For example, an op-amp powered at +30V on V+ and GND on V- (non-inverting configuration, gain ratio: 1) , with an input signal ranging from -3V to +3V, the op-amp will only be able to output the 0 to 3V signal while the -3V to 0 signal will be lost (actually output as 0 because it is the lowest voltage available on V-).

https://www.google.co.in/imgres?sa=X&biw=1366&bih=667&tbm=isch&tbnid=VgPug9XwDMtguM:&imgrefurl=http://datasheetoo.com/tag/philips-ic-datasheet&docid=uID8HSwy_RKUvM&imgurl=http://datasheetoo.com/wp-content/uploads/2009/08/philips-ic741-equivalent-circuit-schematic.jpg&w=447&h=336&ei=WMGkUZq6NK2ziQfSroGgAw&zoom=1&iact=rc&dur=1381&page=1&tbnh=187&tbnw=248&start=0&ndsp=15&ved=1t:429,r:4,s:0&tx=104&ty=96
If u see schematic of IC 741 u can see that V- supply pin is connected to npn & pnp transitor. So u can see that u can give positive voltage at V- pin.

@rahdirs: please, tell us which transistors are you referring to, because I can see many PNP transistors, and only 1 NPN transistor (Q14) connected to V+.

Hi all

npn and pnp transistors eh....

Current flows from positive to negative everywhere including inside a (741) OPA. If the negative supply becomes more positive than the positive supply the OPA will not work and will most likely latch-up (i.e. present a short circuit), get very hot and die shortly thereafter.

The input and output terminals must (generally) remain between the rails.

So,

positive supply most positive

input and output

negative supply least positive

There are a few OPAs which will tolerate the input (i.e. the voltage at the input pin itself) going a few hundred millivolts below the negative rail (not the 741 though).

The supply voltages are not required to be symmetric about 0V; both may lie above or below zero, but often do lie either side. The OPA supply voltage is limited to (typically) 30V, again this means that the difference between the two voltages must not exceed 30V but the voltage wrt 0V may be irrelevent. So +/-15V is OK, +/-20V is not, but +220 and +200 is OK - provided the inputs and outputs lie up there also.

Many OPAs get upset if the inputs get within a couple of volts of either rail and are incapable of driving the output nearer than 2V to either rail.

If you look at the circuit posted you will see the current flows from positive to negative, in the direction of the arrows, from top to bottom, npn and pnp not withstanding.

grahamedriver said:

The supply voltages are not required to be symmetric about 0V;

Click to expand...

That's right, the GND point of the op-amp should be either half the voltage differential of V+ and V- (symmetric) or V- >= GND <= V+ (asymmetric). In the explanation I gave above I supposed the case where V- = GND.

About the input not tolerated above a couple of volts, that's incorrect. It depends on the configuration the op-amp is used. If it's used as amplifier and the power supply cannot afford to reach a very high value it will output the maximum value reachable. Suppose gain=10, supply ±15V, input 5V, op-am configuration non-inverting: the output should be +50V, but the maximum it can reach will be V+ = 15V. Unfortunately, this can damage the op-amp on long operations as it is kind of forced to output more than available, and it may not support such situation. But if you configure it to a gain ratio of 0.5, then you can even input 30V because the 0.5 gain of 30 is +15V, still less or equal of V+.

@grahamedriver: From what i've posted about supplying positive voltage to V- supply pin,I meant there should be a minimum potential difference of 6 V to a maximum of 36 V between V+ and V- supply pins.Regarding +220 v and +200 v i don't know maybe you should take into consideration the break down voltage of the bjt transistors.

@T3STY: I meant transistors Q11,Q22,Q20 would operate fine even if V- pin is positive but at a lower potential than V+ pin.

I wonder if we should have a warning sign at edaboard:
"This thread is containing many erroneous, confusing or misleading statements and not suitable for electronic beginners"

Can you provide the right answer?

rahdirs said:

@grahamedriver: From what i've posted about supplying positive voltage to V- supply pin,I meant there should be a minimum potential difference of 6 V to a maximum of 36 V between V+ and V- supply pins.Regarding +220 v and +200 v i don't know maybe you should take into consideration the break down voltage of the bjt transistors.

@T3STY: I meant transistors Q11,Q22,Q20 would operate fine even if V- pin is positive but at a lower potential than V+ pin.

Click to expand...

The OPA will not suffer operating at any voltage wrt GND so long as all voltages are in the same region. You could, for instance mount it on top of a supply grid pylon and subject it to 200kV wrt to GND and it will be fine provided everything around is at 220kV. It is the voltages on the pins with respect to each other that matter.....

Can you provide the right answer?

Click to expand...

What's the exact question? What do you mean with "give a positive voltage to V- pin"?

As explained in the few good contributions to this thread, only the voltage difference matters. V- must be always the most negative and V+ the most positive voltage, the voltage at the other pins must not exceed this range. Otherwise the device might be damaged. The same happens if the maximum supply voltage (difference between V+ and V-) is exceeded. Input common mode range has to be kept for correct operation in addition.

FvM said:

As explained in the few good contributions to this thread, only the voltage difference matters. V- must be always the most negative and V+ the most positive voltage

Click to expand...

@FvM: Does this mean i can use all Dual Supply op-amps with single supply also,as long as i do not exceed maximum range

I said input common range has to be kept for correct operation.

E.g. TL071 has minimum guranteed common mode range of +/- 11 V when supplied with +/- 15 V. Or OP07 is specified with +/- 13 V minimum CM range. Typical common mode range is larger, but never includes the negative rail. This means that all single supply circuits with an OP input voltage near the negative rail don't work.

@FvM: No,i wasn't trying to say that.There is no worry about input.I'll adjust it to stay away from both supply rails,in the mid-region near two supply rails.So,no issue about input.
Say, i have AD 8672 op-amp.It was given in it's datasheet that it needs Dual Supply Operation:±5 V to ±15 V.So can i use AD 8672 with supply rails V+ at 10 V & V- at 0 V.

Usually the term single supply implies operation of in- and outputs near a supply rail, mostly the negative rail. Otherwise it would be meaningless (at least for OP operation) if the supply is e.g. +/-5 or +10/0 V.