Author Topic: op amps using specific power supply configuration eg:+8v/-5v why?????????  (Read 506 times)

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Offline ifrenide

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hello, can somebody here tell me why they use specific power supply configuration +8v and -5v for op amps and comparators in some electronic devices instead of using +8v and ground. by the way I found this in a metal detector schematic (TGSL), and that make me confused also about the output of the compartor ,when an AC signal is fed to its inverting output why the out put is not +8vand - 5v Instead the output is +8v and 0v ?????????????????? |O
best regards
 

Online schmitt trigger

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*A probable explanation*

The LM358 input common mode range goes all the way to the negative supply, but it must be about 2.5 volts less than the positive supply.
Values from memory, check the datasheet for actual values.
And I believe its output swing behaves similarly.

Thus, to obtain a +/-5 v dynamic range, this trick was required.
 

Offline ifrenide

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ok my brother . lets consider that those op amps is powered with +8v and gnd what difference it make (understanding by using the oposite)  :D
and I i would like to mention that there are some comparators in the schematic what does this configuration with them ???
thank you
best regards
 

Offline StillTrying

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when an AC signal is fed to its inverting output why the out put is not +8vand - 5v Instead the output is +8v and 0v
best regards


The LM393 comparators will be outputting -5V to nearly +8V.
Many of the opamps have +IN connected to 0V and a diode in the feedback to -IN which will stop their output going below -0.7V even though their -Ve supply is -5V.

Edit:
Now I've looked again the diodes are the wrong way around. :-//
« Last Edit: April 27, 2019, 01:48:45 pm by StillTrying »
CML+  That took much longer than I thought it would.
 

Offline ifrenide

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but here in this schematic there is no diode to stop the output going into -5v.
if you don't talk about an integrated diode in the the copmarator .
 best regards
 

Online dmills

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The LM393 is an open collector part, so when the output transistor is off R21,24 will pull the output up to ground, and when on it will be pulled down to -5V.

This makes sense as the signals are used to switch a pair of jfets (TR4,5) in a kind of phase sensitive demodulator arrangement.
 
Not too sure what U107B is doing, looks like a low battery detector, but I would expect an LED somewhere to warn you.

Regards, Dan.
 

Offline reboots

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Sometimes an asymmetrical bipolar supply is just the simplest and cheapest solution. You start with a +8V power input, so that's a given. Then you need a to develop a negative voltage for the required signal range, so you have to build an extra bit of circuitry for that. If you only need -5V, there's no point in adding extra charge pump stages for -8V which would increase cost, complexity, and inefficiency.
 

Online dmills

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Yep, and -5V is quite sufficient to switch those jfets to high Z and to provide a little negative swing for those opamps.

You will notice that there is no actual regulation on the -5V rail, it is just what that charge pump happens to make with that load (8V - 4 diode drops approximately), which is good enough for the purpose.

Regards, Dan.
 

Offline ifrenide

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so your point of view is based on cost and simplicity not on technical reasons. here in this schematic jfets are bf245 wich is N channel switching jfet its gate needs +5v to switch the transistor the output of the Lm393 comparator gives +5v or little bit more and 0v to switch jfet on and of with a specific phase angle and frequency . here the question is why they use +8v and -5v to power most of op amps and comparators in the circuit not only the comparators the op amps also what ever the configuration of the op amp differential or with negative feedback as you see in the schematic .2nd does the voltage suply really affect its output or just to make them stable and avoid saturation or for some other reason???
best regards .
 

Offline ifrenide

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Online dmills

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BF245 is a deletion mode jfet, it is on with Vgs = 0, and cut off with negative Vgs.

You could build a single supply rail metal detector of course, but I doubt it would save you many parts and may actually be more expensive.

Lots of the doings are AC coupled so for a 'single supply' design you would need a reference somewhere above the negative rail to allow the AC amplifiers to work.

There are (as usual) plenty of ways to design such things, (I might have gone for a 'bus switch' as a synchronous detector for example in something resembling a H mode mixer, or maybe a cmos analogue switch), and often design with this sort of thing comes down to whatever the original designer had kicking around at the time.

Regards, Dan.
 

Offline ifrenide

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okmy brother . lets consider that bf245 will be active when vgs=0 . when it will be off so , the answer +5v or lets say +8v , then the output of the comparator wich drives bf245 must alternating between 0v and +8v, and then if the output of the comparator alters between these two valuses , its negative voltage supply -5v what it does  :-\ how to get 0v at its output while the negative volage supply is -5v, and this is a part of the problem without talking about the op amps of this detector .
best regards
 

Online dmills

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No, with the source of the BF245 at ~0V it will be off when the gate is at -5V, and on when the gate is at ~0V, the gate never goes positive because that would forward bias the intrinsic diode.

The comparator output stage is a NPN transistor with the emitter connected to the comparators negative supply pin (-5V), when this transistor is off the jfet gate is pulled to ground by a resistor.

Regards, Dan.
 

Offline ifrenide

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yes that's rigth .I cheked that thanks
 


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