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IC voltage mixing. |
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Jalex
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Joined: 08-Oct-17 Location: USA Online Status: Offline Posts: 414 |
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Topic: IC voltage mixing.Posted: 10-Jan-30 at 20:59 |
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I have two voltages. 0 - 5 and 5 - 0 from op-amps. I want to mix them together giving a unity gain ground (0 to 5) volts. How do I it? |
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Ivan
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Joined: 08-Aug-21 Location: Praha Online Status: Offline Posts: 396 |
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Posted: 10-Jan-31 at 11:18 |
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See e.g. http://devices.sapp.org/circuit/mixer/ Use equal values of resistors R1 = R2 =R3.
VBR from Ivan |
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Jalex
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Posted: 10-Feb-01 at 18:33 |
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Thanks Ivan Thanks Ivan Here is what I am playing with. I have a trimmer cap converted to voltage. I have two op-amps reading signal from a charge pump like a cap checker. One gives me 5 volts when the cap is fully meshed and the other gives me 5 volts when the cap is fully open. I want to use these voltages to make it read 0 to 5 volts on one complete revolution. That's an audio mixer and I am not sure that will do what I need. That will give me 0 volts when ever the polarity reversed. Maybe it's not a mixer I am looking for. I was thinking more like an instrument amp is what I want. I want 0 out when the polarity is (+ to -) and 5 volts out when the polarity is (- to +). |
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Jalex
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Posted: 10-Feb-01 at 19:31 |
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Hi Ivan I have thought about that some more and I don't think that will work either. I just can't picture in my mind how to do that but I still think it can be done.
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sm0vpo
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Joined: 08-Nov-02 Location: Lunda, Sweden Online Status: Offline Posts: 335 |
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Posted: 10-Feb-01 at 23:32 |
Question, Am I to understand that you have an input voltage that swings from -5v to +5v (total 10v swing) and you want the output to vary from 0v to +5v (total 5v swing)?
If this is the case then use a voltage divider (divide by 2, eg 10k + 10k) from +5v to your input voltage. The junction will vary from 0v to +5v.
If the input source or the output loads will affect the divider then use an OpAmp as a 1:1 voltage buffer where needed.
BR Harry
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Ivan
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Posted: 10-Feb-02 at 09:35 |
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Hi Jalex,
opamps work from 0 V to some limit (typically from 7 kHz to 100 MHz). If no coupling capacitor is used, the "audio" mixer will work from D.C., which seemed to be what you need. Of course, you must either use a true rail-to rail opamp, which will do with +5 V and 0 V, or use symmetrical supply at least +12 V and -12 V. Now I doubt what you really need. "That will give me 0 volts when ever the polarity reversed." You did not mention any polarity reversal before. Both input voltages should vary between 0 V and +5 V, so their sum can theoretically vary between 0 V and +10 V. " I want 0 out when the polarity is (+ to -) and 5 volts out when the polarity is (- to +)." Do you need a comparator ? It can be made of an opamp, too, or special ICs are on the market. Use a rail-to rail opamp powered by +5V and 0 V. Connect the first voltage via a protective resistor to the inverting input, the second one via a protective resistor to the non-inverting input, use no feedback. The output should give you what you need. Can you specify the purpose of the whole circuit, please ? Maybe there is a more effective solution. VBR from ivan |
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Jalex
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Posted: 10-Feb-02 at 18:01 |
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Thanks Harry and Ivan I am using two op-amps one puts out (0 to 5) the other (5 to 0), in other words one is inverted. The rail is 9 volts so I am ok there. There are no minus voltages as it's a single supply. Starting from the fully meshed point of the trimmer cap: Comparing the two voltages, when the polarity reverses the trimmer will be only at the half way point. (fully meshed to fully open, 180 degrees) . At this point the voltage needs to be 2.5 volts. For the continuation of the rotation (180 to 360 degrees) I need to voltage to climb from 2.5 to 5 volts and drop to 0v when it completes one revolution. This circuit monitors a rotating mechanical metal plate and it not really a trimmer but works in exactly the same way. It has 2 to 60 pf of capacitance. I hope this clears up what I am trying to do. |
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Ivan
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Posted: 10-Feb-03 at 09:00 |
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Hi Jalex,
now your problem seems to be much more clear. My remarks: 1/ The supply rails are +9V and 0V. The input/output swing is 0V - +5V. You still need a dual power source or a true rail-to-rail opamp, otherwise the voltages near 0V will be corrupt. Depending on the common opamp type, voltages in the 0V - +2V range may be imprecise ! 2/ Inverting the same signal in a separate step does not make much sense. Sum of a voltage and its invert should be always +5V in your case. Comparing these two voltages has the same effect as comparing the original voltage with +2,5V. (I assume linear change of the voltage with the angle.) 3/ Let me create a small table: rotation capacitor capacitance voltage deg pF V 0 meshed 60 0,00 90 half open 31 +1,25 180 fully open 2 +2,50 270 half open 31 +3,75 359 almost meshed 59,9 +5,00 360 meshed 60 0,00 Am I right ? If yes, the device can be hardly realized. Two different voltages should correspond to the same capacitance value. I imagine a RS-type flip-flop, which changes its state always at 0 deg and 180 deg (e.g. using two optocouplers) and switches the opamp between a follower (giving voltage from 0V to +2,5V) and an invertor with shifted output (giving from +2,5 V to +5V) ... BUT : You switch the device on with the plate in a random position. The capacitance will be say 30 pF. How will you set the flip-flop ? Should the output be +1,2V or 3,8 V ? Until the plate reaches 0 deg or 180 deg, the angle of rotation will be unjustified. If the direction of rotation is not uniform, the angle will be unjustified after each change of the direction. If this does not mind, you can use this solution. Othervise, a mechanical 1:2 gear between the axis and the capacitor solves all. VBR from Ivan OK1SIP |
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Jalex
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Posted: 10-Feb-03 at 19:18 |
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Yes you are correct. You chart is exactly what I am trying to come up with. Here is what I am going to try next. Let me know what you think. First of all the op-amps are biased at about 1 volt because the capacitance doesn't drop all the way to 0 pf. So the single power supply shouldn't be a problem with the 324 op-amp. Idea 1 My idea is this, using your chart I will use one op-amp in the 324 for a comparator looking at the output and set at 2.5 to switch in the output from the inverted op-amp that's biased at 2.5. I am not sure how clean the switching will be this way but it might be alright with a good regulated supply. Idea 2 I can put an IR detector across the rotor of the capacitor to read it's position and let that control the switching. When the led is covered it would turn on the inverted amp. This doesn't need to be a super accurate measurement. Just close would be fine. I guess I was just hoping to find a simple way to do this without the switches. LOL
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Ivan
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Posted: 10-Feb-04 at 15:07 |
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Hi Jalex,
let me continue. "all the op-amps are biased at about 1 volt because the capacitance doesn't drop all the way to 0 pf. So the single power supply shouldn't be a problem with the 324 op-amp." LM324 includes the lower rail both on inputs and or outputs. It may be "biased" by attaching the non-inverting input to ground. "Idea 1: My idea is this, using your chart I will use one op-amp in the 324 for a comparator looking at the output and set at 2.5 to switch in the output from the inverted op-amp that's biased at 2.5." This does not remove the ambiguity at least after start. "Idea 2 I can put an IR detector across the rotor of the capacitor to read it's position and let that control the switching. When the led is covered it would turn on the inverted amp." Yes ! The optodetector fixes the ambiguity by determining, whether the rotor is in the upper or lower part of his track. See pictures: http://ivanok1sip.atlasweb.cz/pics/Csensor.bmp Sorry, the schematics is a bit corrupt, but still readable. I had hard hours to upload it somewhere on the web. BR from Ivan |
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