# ACS758 + LTC1968

#### geo0rpo

Joined Sep 9, 2020
9
Hi,
I am using an ACS758LCB-050B-PFF-T to measure chopped AC current and I feed the output to an LTC1968 to get DC output.
The ACS has an output voltage of 2.5V for 0A current and swings from 0.5V-4.5V.
The LTC1968 has input 1V p-p so I need a way to decrease the ACS voltage from 0.5V-4.5V to 1v p-p and keep the 2.5v as zero output voltage.
As simple resistor divider cannot work because it will decrease all the range including the 2.5V.

So, I need a circuit (op-amp?) with 2 inputs.
Input 1: 2.5v fixed reference voltage.
Input 2: 0.5-4.5v
The output should be:
1.5V for 0.5v on input 2
2.5V for 2.5v on input 2
3.5V for 4.5v on input 2

Any ideas?

#### Papabravo

Joined Feb 24, 2006
15,518
Since those three points all lie on a straight line. You can solve the problem with a gain of 1/2 and an offset of 1.25
The equation of the line is:

$$y\;=\;\frac{1}{2}x\;+\;1.25$$

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#### geo0rpo

Joined Sep 9, 2020
9
Papabravo,
thanks for the reply. Do you have an example circuit?
Here is what I want to do. I am missing the limiting circuit.

#### Papabravo

Joined Feb 24, 2006
15,518
Not off the top of my head. What voltages do I have to work with?

#### geo0rpo

Joined Sep 9, 2020
9
Exactly as it is in the circuit attached, up to the left side of the LTC1968. Ignore the right side.
The supply would be 5 volts and 0 volts and half the supply voltage (2.5 volts) is applied to the IN2 input.
The ACS758LCB-050B naturally produces a quiescent DC level of 2.5 volts. But I have to limit the input voltage range to a peak voltage of 1 volt. So I need to reduce the output swing of 0.5 volts to 4.5 volts from the ACS758LCB-050B to 1.5 volts to 3.5 volts for the LTC1968.

#### DickCappels

Joined Aug 21, 2008
7,171
Just curious, what is the point of chopping the input?

#### Papabravo

Joined Feb 24, 2006
15,518
That makes it more difficult. I'll let you know if I have a solution.
Here you go:

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#### geo0rpo

Joined Sep 9, 2020
9
Just curious, what is the point of chopping the input?
Constant current on heating element.

#### geo0rpo

Joined Sep 9, 2020
9
That makes it more difficult. I'll let you know if I have a solution.

#### Papabravo

Joined Feb 24, 2006
15,518

#### DickCappels

Joined Aug 21, 2008
7,171
Constant current on heating element.
From memory, the '1968 can measure DC as well as AC, so I don't understand the need for chopping. Just looking to see if your problem can be simplified.

#### geo0rpo

Joined Sep 9, 2020
9
Wow! Thanks!
I see the 1/2 fain but where is the 1.25v offset?

#### Papabravo

Joined Feb 24, 2006
15,518
Wow! Thanks!
I see the 1/2 fain but where is the 1.25v offset?
It is the y intercept associated with the chosen slope and the fixed point at (2.5, 2.5) It does not have to show up explicitly. Both opamps are biased to Vcc/2 because you have a unipolar supply so that is the "virtual" Ground. We need to use two inverting stages because you cannot realize a gain of less than 1 with a non-inverting stage. The two stage gains are -1/2 and -1 for a total gain of +1/2. Depending on the actual opamp you choose you may or may not get all the way to 0V and +5V

#### geo0rpo

Joined Sep 9, 2020
9
From memory, the '1968 can measure DC as well as AC, so I don't understand the need for chopping. Just looking to see if your problem can be simplified.
The project demands constant current in a 220v AC heating element. In order to do this you have to sync with the power line zero crossing and trigger the triac some time after the zero crossing. So you chop the original waveform.
Then I have to measure the current in order to keep a constant current
The ACS758 works really nice as long as the waveform is pure sinusoidal but it turned out that it does not like the "chopped" waveform and the reading are wrong.
That is where the LTC1968 comes in and it really takes the chopped waveform and gives DC output. Verified!
But the ACS758 outputs 0.5V - 4.5V for -50 to 50A and 2.5v for 0A and the LTC1968 wants 1vp-p to the input.
That is why I want to scale down the voltage but keep the 2.5v in the middle.

#### geo0rpo

Joined Sep 9, 2020
9
It is the y intercept associated with the chosen slope and the fixed point at (2.5, 2.5) It does not have to show up explicitly. Both opamps are biased to Vcc/2 because you have a unipolar supply so that is the "virtual" Ground. We need to use two inverting stages because you cannot realize a gain of less than 1 with a non-inverting stage. The two stage gains are -1/2 and -1 for a total gain of +1/2. Depending on the actual opamp you choose you may or may not get all the way to 0V and +5V
OK, I get it all except for the 1.25v offset. I understand the 2 stages but not the 1.25v offset.
In any case thanks a million for your time. I will try it tomorrow and update here

#### Papabravo

Joined Feb 24, 2006
15,518
OK, I get it all except for the 1.25v offset. I understand the 2 stages but not the 1.25v offset.
In any case thanks a million for your time. I will try it tomorrow and update here
You have to write the node equation for U1 to see it. If Va is the voltage at the summing junction of U1, and Vin = 0, then:

$\frac{(V_a-0)}{20\text K}\;+\;\frac{(V_o-V_a)}{10\text K}\;=\;0$

The result is Vo = 3.75 volts which is 1.25 Volts above the bias point. U2 will perform an inversion about the bias point of 2.5 volts to produce an output of 1.25 V

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#### geo0rpo

Joined Sep 9, 2020
9
Even though I did not get it... It worked perfectly!
Sir you are a magician. I thought I knew things...
Thank you very much for your time and knowledge.

#### Papabravo

Joined Feb 24, 2006
15,518
Even though I did not get it... It worked perfectly!
Sir you are a magician. I thought I knew things...
Thank you very much for your time and knowledge.
No magic here just basic principles and a bit of algebra. I think the thing that is confusing you is the need to work with a single supply voltage and the nature of what "inversion" means when your "zero" level is 2.5 volts.