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# Summing amplifier

#### Gryd3

Jun 25, 2014
4,098
So, I have a confession...
I have not built this circuit, or simulated it yet as I am at stuck at work for another few hours.
I had a quick question regarding the use of a summing amplifier.

If I connected Vout, to V1, would the summing amplifier instantly swing to it's upper/bottom-most rail? (Even if Vout was only a couple mV)
Would adding a series resistor then a parallel capacitor to Vout resolve this behavior and slow the rate of change to a defined amount? (Based on resistor/capacitor values)

#### Arouse1973

Dec 18, 2013
5,178
Depends on the value of the other two. Vout would be your new V1, so this would add to the others. So I think what you will get is V2+V3+V2+V3. But thats a strange question so I might be wrong. But I think it would soon saturate. As it kept adding.

#### LvW

Apr 12, 2014
604
Did I understand correctly: You intend to connect Vout with V1 ? Why?

#### Gryd3

Jun 25, 2014
4,098
Did I understand correctly: You intend to connect Vout with V1 ? Why?
The intention is to build a circuit that will continually increase its output when presented with a positive input, and continually decrease its output with a negative input.
Sort of like an integral function. I don't want the output to change when presented with 0V.

The idea here was to feed Vout to V1, so that when V2 = 0, then Vout will remain at it's last linear state.

#### Arouse1973

Dec 18, 2013
5,178
Did I understand correctly: You intend to connect Vout with V1 ? Why?

Hey Lutz
Yeah I wondered that.

#### hevans1944

##### Hop - AC8NS
Jun 21, 2012
4,889
Connecting Vout to V1 places R1 in parallel with Rin and shorts out V1 through the low (ideally zero) output impdance of A. The effect of that will depend on the source impedance of V1. In any case, V1 will then have little or no effect on the output.

If the source impedance of V1 is very low compared to the output impedance of A, it essentially "shorts out" the output and prevents summing of the other two inputs because there is no feedback current to maintain the virtual ground. If the source impedance of V1 is very high, the amplifier output shorts out V1 so. again V1 has no effect on the output. However, the other two inputs will sum correctly but with lower "gain" since R1 is paralleled with the Rin connection to V1.

I would not recommend connecting V1 to A output.

#### hevans1944

##### Hop - AC8NS
Jun 21, 2012
4,889
The intention is to build a circuit that will continually increase its output when presented with a positive input, and continually decrease its output with a negative input.
Sort of like an integral function. I don't want the output to change when presented with 0V.

The idea here was to feed Vout to V1, so that when V2 = 0, then Vout will remain at it's last linear state.
This is not "sort of like an integral function." It IS an integral function. Replace Rf with a low-leakage capacitor and the circuit will behave as you describe.

#### hevans1944

##### Hop - AC8NS
Jun 21, 2012
4,889
This is not "sort of like an integral function." It IS an integral function. Replace Rf with a low-leakage capacitor and the circuit will behave as you describe.
Well, more or less. The output will drift up or down with zero input, depending on the bias current and voltage offset at the input of the amplifier A as well as leakage current across the capacitor terminals. It is really difficult to build a analog integrator that doesn't drift. Better to do an ADC of the input and add successive digital outputs until overflow occurs. Add a DAC to the results register if you need analog output. Works fine, lasts a long time, costs a little (more), does not drift.

#### Gryd3

Jun 25, 2014
4,098
This is not "sort of like an integral function." It IS an integral function. Replace Rf with a low-leakage capacitor and the circuit will behave as you describe.
This sounds much easier than I was trying to draw out.
I should have clarified that V1 was an input label, and would only be connected to Vout. There would only be one external input placed on the V2 input.
Regardless, the information presented here helps me out.
I need to play with opamps more... I have a very general understanding but obviously not enough!

#### hevans1944

##### Hop - AC8NS
Jun 21, 2012
4,889
In my current line of work there is a requirement to "count" the number of high-energy positive ions implanted onto (and into) a target material. The time the ion beam is on can range from a few seconds (at low doses) to several days (at high doses). The target is mounted to an insulated aluminum carousel wheel (we can load up to 25 targets) mounted in vacuum. A rotary electrical vacuum feed-through connects the target and the wheel to the outside world.

To "count" the number of ions implanted, we measure the electron current that flows from ground to the wheel, each electron presumed to neutralize one positive ion. Now, if we could integrate that analog ion current signal (without drift!) the integrated result would be exactly proportional to the number of ions that landed on the target and were neutralized by electrons comprising the current we measure. An analog integrator, such as I described above, would not be accurate enough for our purpose when used for long intervals of time. Besides that, we need to know when a certain dose of ions has been completed, so as to turn off the ion beam and perhaps move on to the next target. If a perfect, drift-free, analog integrator existed, we could compare the integrator output against a set-point voltage, perhaps from a potentiometer connected to a stable precision voltage source, and turn off the ion beam when the integrator output voltage reached the set-point voltage.

Fortunately there is a better, more accurate way. Our accelerator uses a hybrid analog/digital device called a Model 1000C Current Integrator, manufactured by Brookhaven Instruments Corporation out of Austin TX. It is basically a selectable-gain current-to-voltage converter that drives a voltage-to-frequency converter. The latter has a full-scale frequency of 1000 Hz when the input is full scale. At less than full-scale, the frequency is a series of pulses with a repetition rate that varies from zero to 1000 Hz. The pulses are counted by now-obsolete Texas Instrument BCD counter/display ICs to a maximum of 9999 counts. The desired count is preset on four thumbwheel switches and the BIC stops the ion beam when that count is reached. The pulses can be divided down from 1000 Hz full scale to 100, 10, and 1 pulse per second to accommodate higher doses.

This was a garage-shop project a very clever engineer built about twenty years ago. I don't know how many are in existence, but not many. Their main application is to measure ion doses produced by particle accelerators, and there ain't a whole lot of them around. The instrument is almost bullet-proof. The input at the BNC is a "virtual ground" on any range where the output is less than full scale. It is important to select the correct range and not apply a voltage signal to the input. Voltage signals providing sufficient current will vaporize a small 100 Ω 1/8 watt resistor in the input path. The designer thoughtfully mounted this resistor on stand-off Teflon insulators, so it is easy to replace. I think I've smoked, or at least crisped, perhaps a half-dozen over the last eighteen years. Please don't ask how this happens; it's embarrassing.

So, if you need to integrate an analog signal, I unconditionally recommend this "V-to-F and count pulses" technique. Accuracy will depend mainly on your V-to-F circuit design, of which there are dozens of examples available elsewhere on the Web. The main thing, however, is to have FUN!

BIC 1000C Current Integrator:

#### LvW

Apr 12, 2014
604
The intention is to build a circuit that will continually increase its output when presented with a positive input, and continually decrease its output with a negative input.
Sort of like an integral function. I don't want the output to change when presented with 0V.
The idea here was to feed Vout to V1, so that when V2 = 0, then Vout will remain at it's last linear state.

Gryd3 - if you connect the node labelled as V1 to the output node you simply have Rin in parallel to Rf.
Hence, using the node labelled as V2 as an input signal you have an inverting amplifier with a closed-loop gain of (-Rf||Rin/Rin). That`s all. No integrating function at all.

#### Arouse1973

Dec 18, 2013
5,178
Yes that makes sense Lvw. I see it now.

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