Re: Hysteresis needed in 741 Op-Amp circuit
From: CF (form_at_web.site)
Date: 06/10/04
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Date: Thu, 10 Jun 2004 07:27:26 GMT
Oh wow Bob! What a fantastically helpful, educational and thorough reply. Even
a circuit diagram no less.
Unless anyone raises an objection in the meantime, I'll give your suggestions
a try either tomorrow or on the weekend, and post a follow-up.
Very gratefully,
Craig
In article <p8Txc.4297$2i5.3115@attbi_s52>, "Robert C Monsen"
<rcsurname@comcast.net> wrote:
>Nice problem description!
>
>The way hysteresis works for these kinds of circuits is that you
>influence the + input by tying it to the output through a resistor,
>and use another resistor from that node to the reference. Thus, the +
>input is basically set by a voltage divider which goes from the output
>to the comparison input.
>
>Say your input is set at 5V, and your output swings from 0 to 10V. If
>the resistor from output to + and from + to 5V are equal, then its
>clear that when the output is high, the + input will be at 7.5V. Thus,
>the - input will have to get up to 7.5V to cause the output to go low.
>
>When the output is low, the + input is at 2.5V. Thus, the - input will
>have to go down to 2.5V to cause the output to go high.
>
>Because of this, you have a 5V 'guard' region. When the output swings
>from low to high because the - input has gone below 2.5V, the trigger
>voltage will immediately go up to 7.5V, meaning that fluctuations
>around 2.5 won't cause it to turn on and off repeatedly.
>
>Now, for your heater circuit, you are varying the + input using the
>thermistor, rather than the -. Thus, by putting a resistor from 6 to
>3, and another one from 3 to where the old 3 was connected, at the
>node connecting the 10k and the thermistor, you'll add some
>hysteresis.
>
>Adding it to 2 won't be good, because that means that when the output
>goes low, the negative input will go lower, meaning the + will have to
>go down even further, thus turning it back on, etc... Not good. Also,
>you really want your opamp acting like a comparator, not an opamp, so
>negative feedback is not a good thing.
>
>I'd use a 330k potentiometer between pin 6 and the junction of the 10k
>and the thermistor in your diagram (the node that used to be attached
>to pin 3). Then, attach pin 3 to the wiper of the pot, and set it for
>the proper hysteresis.
>
> 12V
> +------------------------------+
> | |
> | |
> .-. |
> 10k | | |
> | | |
> '-' 330k |
> | ___ |
> +----|___|--------+ |
> | A | |
> | | | |
> | | |\ | 3.3k |
> | +----|+\ | ___ |<
> | | >--+--|___|---|
> .--------+--. Vref-|-/ |\
> | | |/ |
> | Sensor | .--------.
> | 10k | | |
> | | | Relay |
> '--------+--' | Etc |
> | | |
> === | |
> GND '--------'
>created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de
>
>
>Regarding the value of that resistor from the opamp to the PNP
>transistor, its value should be such that the current through it is
>about 1/10 of the current you want through the relay when its on. If
>you want 30mA, then with about a 10V drop, you want 10V/3mA = 3.3k.
>That should fully saturate it. If not (if the collector isn't near
>Vcc), then use a smaller resistor.
>
>Regards,
> Bob Monsen
>
>
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