I see circuits that add hysteresis to op-amp comparators by putting a positive feedback resistor network that acts as a voltage divider to the output.
Am I right in thinking that this will only work if you have a negative rail? If your negative rail is grounded this aint gonna give you hysteresis... is there a way this would work with a single supply opamp?
thanks!
itive feedback resistor network that acts as a voltage divider to the outpu= t.
il? If your negative rail is grounded this aint gonna give you hysteresis..= . is there a way this would work with a single supply opamp?
I think I have an idea.... I would have to figure out my thresholds and flo= at my opamp so that the positive feedback voltage divider gives me the thre= sholds I need...
So, if I want my high threshold at 2V and low at 1V then (assuming my divid= er splits my voltage in half) I'd need my high opamp rail at 4V and my low = rail at 2V... so I'd have to make sure my opamp works off of 2V and my inpu= t signal never goes beyond my rails...
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feedback resistor network that acts as a voltage divider to the output.
If your negative rail is grounded this aint gonna give you hysteresis... is there a way this would work with a single supply opamp?
If your threshold voltage is at zero, then you're likely to have other problems as well.
However, the resistor still provides hysteresis. Say you have a 10k resistor from the reference to the + input, and a 1M feedback resistor. With +5/0 outputs, the threshold will be:
V_LH = (Vref*1M + 5V*10k)/1.01M = 0.99*Vref + 50 mV V_LH = (Vref*1M + 0V*10k)/1.01M = 0.99*Vref
With a 5V swing and a 1% division ratio, you always get 50 mV hysteresis. If the reference voltage is halfway between the supplies, the hysteresis range is symmetrical, but not otherwise.
Cheers
Phil Hobbs
-- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 845-480-2058 hobbs at electrooptical dot net http://electrooptical.net
positive feedback resistor network that acts as a voltage divider to the output.
If your negative rail is grounded this aint gonna give you hysteresis... is there a way this would work with a single supply opamp?
my opamp so that the positive feedback voltage divider gives me the thresholds I need...
There you go.
splits my voltage in half) I'd need my high opamp rail at 4V and my low rail at
2V... so I'd have to make sure my opamp works off of 2V and my input signal never goes beyond my rails...But why are you using an OpAmp? That only guarantees slow transitions. ...Jim Thompson
--
| James E.Thompson, CTO | mens |
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I love to cook with wine. Sometimes I even put it in the food.
feedback resistor network that acts as a voltage divider to the output.
If your negative rail is grounded this aint gonna give you hysteresis... is there a way this would work with a single supply opamp?
Yes, it works the same way. You need to put a virtual common node at the (+) input when operating single rail. This feed back R is calculated working with this virtual network so that you can select the hysteresis window.
Jamie
You don't need a negative rail. Just having an output that swings from one rail to another is enough to provide hysteresis with one of those networks. Do the math, it'll jump out at you.
Note that using an op-amp for a comparator can lead to all sorts of subtle or not-so-subtle misbehaviors. In general, op-amps don't like operating in regimes where they aren't working in the linear region. Some op-amps show this with such mild misbehaviors as taking extra-long for the output to come off the rail, or not approaching the power rails in an easily predictable manner. Other's get nasty in various ways (I don't know the whole catalog, but oscillation, high input currents, and (I think) phase shift are all on the list).
Using a comparator for a comparator is often better.
-- My liberal friends think I'm a conservative kook. My conservative friends think I'm a liberal kook. Why am I not happy that they have found common ground? Tim Wescott, Communications, Control, Circuits & Software http://www.wescottdesign.com
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I went with an OpAmp out of convenience, and because what I'm looking at is= a really slow signal which, basically DC, it would rise and fall real slow= ly... I'm not sure if using a comparator buys anything when your not concer= ned about speed... maybe it does, I don't know, I'm all ears though, and I'= m gonna dig into it some more...
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Everyone should have a few LM393's in their parts box. I sometimes find a comparator to be too fast, but you can hang a capacitor on the output to slow it down.... nice linear ramps from one rail to the other.
George H.
(for a fast 'twichy' comparator I like the LT1016)
really slow signal which, basically DC, it would rise and fall real slowly... I'm not sure if using a comparator buys anything when your not concerned about speed... maybe it does, I don't know, I'm all ears though, and I'm gonna dig into it some more...- Hide quoted text -
Comparitors have outputs that are designed more like gates than linear devices. Opamps may not be well behaved when their outputs are driven hard into the rails. They're also not designed to have a differential voltage on the input. Using opamps as comparators can be done but it's not recommended.
If you drive a capacitor, watch the output current.
Your boss must be filthy rich. ..or perhaps dirt poor, after buying LT stuff. ;-)
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TNah, not much quantity so saving a few bucks is not worth my time. (the $3 comparator with a $3 opamp replaced a ~$100 (?) Amp-tek part. so my boss is happy.)
George H.
feedback resistor network that acts as a voltage divider to the output.
your negative rail is grounded this aint gonna give you hysteresis... is there a way this would work with a single supply opamp?
--- Depends on the opamp.
As others have suggested, instead of an opamp use a true comparator instead, like this: (View using a fixed-pitch font)
.Vcc>---------+--------+ . V2--+--|---+ | . | | | | . | | [R2] [R3] .Vin--[R1]-+-|+\ | | . | >-+----+--->Vout .Vref>-------|-/U1 . | .GND>---------+
Assume: Vcc = 5V Vref = 2.5V R1 = 10k R2 = 100K R3 = 1K Vout = 0V,
and Vin is ramping up from 0V. Then, with U1's output low, you have a voltage divider that looks like this:
Vin | [R1]10k | +--V2 2.5V | [R2]100k | Vout 0V
Now, since Vref is at 2.5V, as Vin ramps up eventually V2 will get to
For V2 to be driven to 2.5V, R2 must drop 2.5V and, if R2 is a 100k ohm resistor, the current through it must be:
E 2.5V I = --- = ------- = 25µA R 100kR
Then, since current in a series circuit is everywhere the same, the current through R1 must also be 25µA.
Since R1 is 10k ohms, with 25µA through it it'll drop:
Vin = IR = 25µA * 10kR = 0.25 volts.
So, since V2 is 2.5V and R1 must drop 0.25 volts, Vin must be either
2.75V or 2.25V.--- If anyone is interested, I'll go on.
-- JF
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If I specified a $3 comparator, it would be my last. In fact, anything LT will draw lightning in a design review (even a no-pop, BTDT).
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I know it's like I'm 'living in the past' selling circuits with two connectors, a rotary switch (or two) a bunch of 0.1% resistors, and an opamp (or two)... sprinkle in some C's.
The opamp is often the least expensive part of the equation, (we match the C's when it's important).
George H. (Hey, there must be someone besides me buying these IC's!)
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Hardly. A couple of high-end DSPs, a uC, and a couple of thousand other parts (BOM cost in the hundred$ - anticipated AOQ perhaps half million). Your contempt and cluelessness is noted, though.
When you're making a million of something, a $3 opamp doesn't make much sense when a $.10 opamp will do, now does it? Yes, I even get beat up for 1% resistors. Money matters.
Little guys with rich (and clueless) bosses, evidently. ;-)
really slow signal which, basically DC, it would rise and fall real slowly... I'm not sure if using a comparator buys anything when your not concerned about speed... maybe it does, I don't know, I'm all ears though, and I'm gonna dig into it some more...
-- Here's a good place to start: http://www.ti.com/lit/an/snoa654/snoa654.pdf
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Huh? I was describing what I do. No contempt was intended.
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Sure that's fine. I'm making tens of something. Next to the $3 opamp is a $10 switch, surrounded by $2 worth of R's. Different world.
You seem angry for some reason?
George (just trying to get along) H.
- Hide quoted text -
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