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I am designing a circuit which means getting sinusoidal signal as as Input and pass it to an ADC. Since the ADC supports tensions from 0V to 5V and my input 5Vpp, then I need to escalate it properly (to 2.5Vpp)then shift it 2.5V

My first thought was the following: Circuit 1

The first AmpOp (IC1A) is to isolate the input from the circuit from the protection zeners. The second AmpOp (IC1D) is to escalate it to 2.5Vpp. The third AmpOp (IC1B) is to sum the input with 2.5V provided by the tension divider.

Which leads me to concern about (R5 series R6)//(R9 series R8) unintentionally; Problem solved adding a forth (IC1C) AmpOp between them.

But I got concerned about how the slew rate would be affected by associating so many AmpOps. I mean, my idea is to run it with ideally 1Mhz, and my SR is 2.9V/us. Any ideas?

(Actually I tried to reconsider each one of this ampOps, so things stay stable. The first one I could substitute by a really big resistor, and the second one for a tension divider, it'd be better maybe?)

I appreciate any thought about it

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    \$\begingroup\$ Are you really using an ADC with an input range of 2.7V to 5V? That sounds really really odd and more like the supply voltage range it will run with. Typically the ADCs inputs go from Vref(-) to Vref(+) or 0 to Vref(+). So maybe your whole circuit is based on a false assumption - can you tell us which ADC you are using? \$\endgroup\$
    – Arsenal
    Oct 8, 2015 at 16:28
  • \$\begingroup\$ Isolating the input from the protection zeners defeats the purpose of the protection zeners. IC1A is not going to produce an output outside its power rails, which are the same as the rails of IC1D, so the zeners serve no purpose. If IN is outside the rails it will (potentially) fry IC1A and break the circuit, and the zeners where you placed them provide no protection against that scenario. \$\endgroup\$
    – The Photon
    Oct 8, 2015 at 17:03
  • \$\begingroup\$ I think Arsenal is correct and you are probably making a mistake. Please provide 1) the model of ADC you are using, and 2) the exact specifications of your input signal. \$\endgroup\$
    – WhatRoughBeast
    Oct 8, 2015 at 17:26
  • \$\begingroup\$ @Arsenal, You are correct, thank you! It should be from 0 to 5V. I'm using TLV1572 from texas (ti.com.cn/cn/lit/ds/symlink/tlv1572.pdf) \$\endgroup\$
    – FernandoP
    Oct 8, 2015 at 17:30
  • \$\begingroup\$ @ThePhoton, You are correct either, that was pretty obvious but I didn't see it. Thank you! \$\endgroup\$
    – FernandoP
    Oct 8, 2015 at 17:32

1 Answer 1

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The first AmpOp (IC1A) is to isolate the input from the circuit from the protection zeners.

Isolating the input from the protection zeners defeats the purpose of the protection zeners. IC1A is not going to produce an output outside its power rails, which are the same as the rails of IC1D, so the zeners are not doing anything to protect IC1D.

If IN is outside the rails it will (potentially) fry IC1A and break the circuit, so the zeners where you placed them provide no protection against that scenario, either.

The third AmpOp (IC1B) is to sum the input with 2.85 provided by the tension divider.

If you want this to work, you need to add a series resistance between the output of IC1D and the input of IC1B.

But I got concerned about how the slew rate would be affected by associating so many AmpOps.

Slew rate will mainly affect whichever op-amp has the largest output amplitude.

If IC1D isn't slew rate limited, then IC1A won't be either.

Since IC1B has gain of 1, it will only slew rate limit if IC1D does too.

Possibly the distortion will be slightly worse from chaining several op-amps. The easy solution is either to find an op-amp with higher slew rate capability.

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  • \$\begingroup\$ Hum, I really didn't get how a resistor between IC1D e IC1B could work. The resistors from the IC1D and the divider wont be in parallel anymore but I still can have current flowing reversely from the divider when the input wave (let's say, a 500kHz sine, 2.5Vpp, 0 offset), goes to -2.5. And if current flows, tension divider has a charge, and its no more tension fixed. Could you explain a little bit? \$\endgroup\$
    – FernandoP
    Oct 8, 2015 at 17:48
  • \$\begingroup\$ @FernandoP, google "op-amp non-inverting summer". If one of the inputs to a non-inverting summer has zero (or near-zero) source impedance, the summer won't work. All other inputs will be ignored and you'll only get the output equal to the gain (1 in your case) multiplied by the input with low source impedance. \$\endgroup\$
    – The Photon
    Oct 8, 2015 at 18:36
  • \$\begingroup\$ Also a non-inverting summer is really an averager (a voltage divider connected between the different inputs) followed by a non-inverting gain amp. In your case you chose gain of 1 for the gain portion, so the overall gain will be less than one for either of the individual inputs being summed. \$\endgroup\$
    – The Photon
    Oct 8, 2015 at 18:38

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