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I'm trying to discover a circuit that will produce a voltage which is some factor of the square root of the input voltage. I.e. \$V_{out}(t) = K\sqrt{V_{in}(t)}\$. The factor K is irrelevant.

I looked at the circuit at the bottom of this page. The problem is it uses a MOSFET and the formula predicting the output requires various parameters \$\mu_n, C_{ox}, V_{th}\$ (some of which I imagine vary greatly even among devices of the same model, and some of which I wouldn't know how to find from the data sheets)

I'd like to find an alternative circuit which has a consistent and predictable output, before I purchase the necessary components.

When I say K is irrelevant, I just meant that I can later amplify the output by a constant factor if needed. It needs to be consistent and predictable though.

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  • \$\begingroup\$ If the factor K is irrelevant, then the factors C, mu, W,L are irrelevant. \$\endgroup\$ – david Aug 8 '13 at 8:22
  • \$\begingroup\$ Those factors vary between transistors of the same model #. Irrelevant in the sense that I can amplify to normalize K to a certain value, but I can't tune each circuit individually based on the transistors properties. \$\endgroup\$ – Keegan Jay Aug 9 '13 at 19:27
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An easy approach would be to use an analog multiplier (MC1495 was an early one, Analog Devices AD633 or Burr-Brown(oops, Texas Instruments!) MPY534 are better newer ones) as a squaring circuit, in the feedback loop of an op-amp.

To use a multiplier to square a voltage simply connect that voltage to both inputs. Connect your input voltage to opamp's non-inverting input, the opamp output to the multiply inputs and the multiply output to opamp's inverting input.

If \$ V_{out}^2 = V_{in} \$ then \$ V_{out} = \sqrt{V_{in}} \$.

schematic

simulate this circuit – Schematic created using CircuitLab

Details such as DC biasing left as an exercise...

(Side note : the analog multipliers rely heavily on "matched pairs" of transistors; it is relatively easy to match 2 transistors if you make both at once in the same area on the same chip!)

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  • \$\begingroup\$ Nice idea! Very elegant approach. Might be a hair on the expensive side, but should work nicely. \$\endgroup\$ – Scott Seidman Aug 7 '13 at 20:44
  • \$\begingroup\$ Hang the expence, nicely done. \$\endgroup\$ – Andy aka Aug 7 '13 at 20:51
  • \$\begingroup\$ 1495s aren't that expensive ... the better ones, yes - you get what you pay for. \$\endgroup\$ – Brian Drummond Aug 7 '13 at 20:55
  • \$\begingroup\$ This is actually kind of amazing as the formula for the square root simply falls out of the nodal analysis of the op-amp. Thanks! \$\endgroup\$ – Keegan Jay Aug 7 '13 at 21:02
  • \$\begingroup\$ I'm wayyyy late to this party, but this answer is the non-inverting (read: less stable) version of the solution given by Application Note AN489, Analysis and Basic Operation of the MC1595 by Ed Renschler. It's a little hard to find now, but it explains in great detail just how the circuit works and have several sample circuits, of which this is one. There is currently a scan of it here. \$\endgroup\$ – TrivialCase Aug 21 at 17:33
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if you have some BJTs plus an op-amp lying around, a quick translinear BJT analog square root is all yours! V(OUT) = SQRT(V(IN))/10 in this case:

(Open & run the DC Sweep simulation in CircuitLab.)

As far as "matched transistors", in this case:

  • mismatch in Q1/Q2/Q3/Q4 or in Q6/Q7 will produce a slight scale factor error (which you've stated you don't care about much anyway)
  • Q5 is not match dependent
  • variation in temperature between different transistors may produce scale error
  • you can simulate mismatch by adjusting I_S of one transistor. See this LED example for something similar in the LED case. (You can also have B_F "beta" mismatch, but in this particular circuit, it's less of a factor.)

I added some notes on the schematic. I'm sure others can help simplify or make this more robust, but I hope it's a good start using parts you probably already have on your bench!

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From TI application note 31:

enter image description here

May work with other op-amps. See the application note for details on running the LM101A off a single-ended supply.

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  • \$\begingroup\$ What is meant by "matched pairs"? Also, the output of this circuit isn't described other than "root extractor." Should I assume it computes the exact square root or are there some factors or constants? Sorry, my EE experience is limited. \$\endgroup\$ – Keegan Jay Aug 7 '13 at 20:37
  • \$\begingroup\$ Transistors are said to be "matched" when their various characteristics are very close to the same value (since transistors of the same model can vary due to inconsistencies in the wafer). As for what it does, it comes from the mathematical operations section; I haven't actually built it before so I can't attest to its accuracy. \$\endgroup\$ – Ignacio Vazquez-Abrams Aug 7 '13 at 20:40
  • \$\begingroup\$ So does that mean in order to build this circuit, I have to purchase a bulk quantity of 2N3728s and find pairs with close properties? Thanks for your help! \$\endgroup\$ – Keegan Jay Aug 7 '13 at 20:41
  • \$\begingroup\$ You may be able to substitute a IC already containing a matched pair as long as the other characteristics are similar to that of the 2N3728; this AN was written a long time ago and some things have changed or advanced since then. \$\endgroup\$ – Ignacio Vazquez-Abrams Aug 7 '13 at 20:42
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This is not intended to be an answer or an explicitic solution, rather an expanation why there aren't any single chip integrated solutions. Perhaps the demand is too low, when you can use a digital solution now with quantization using 12 or 16 ADC's with log codecs or log algorithms and divide by 2 in binary as the log of the exponent ^(0.5) has a 0.5 multiplier in the result.

Square root designs come in many analog variations from 1 to 16 integrated parts with complexities of precision matching, current mirrors, bias mirrors to use the quadratic non-linear behavior of FETs. They have been a perpetual research topic of EE Profs with getting controlled results spanning from 3 to 7+ decades. Problems result from variations in RgsON , Vgs threshold and self heating.

Few of these research topic experiments have ever made their way into production, perhaps because of the difficulty of controlling the process of doping and fabrication contros to get the required consistency, which are orders of magnitude more difficult than CMOS logic. The zero reference is the most critical for errors and a differential output offers more linearity in the Sq Rt result. Considering negative feedback is inverting is used, it is academic that sq. rt amps tend to take a negative input to give a positive ouput, yet this is not an maginary number. Ha.

Have fun.

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I'd recommend a log amplifier, followed by a 0.5 gain linear amplifier, followed by an antilog amplifier. You might be able to buy the log and antilog amps as single purpose IC's. The Burr-Brown 4127 would handle log and antilog, but is obsolete. AD8307 might be another choice

Another approach, depending on your bandwidth requirements and some other things, would be to hand the problem to a microcontroller and a DAC.

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Works in Spice, took about a day to figure out. second stage resistor network removes offsets that caused the third op amp section to overdrive Accuracy within 1db for RF/MW Schottky Diode Detector enter image description here. .0005vdc - 1.000vdc

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    \$\begingroup\$ Can you explain how this circuit works? It looks like there may be a problem with some of the resistor values. \$\endgroup\$ – Dave Tweed♦ Jun 17 '15 at 15:02
  • \$\begingroup\$ I didn't actually build and test it but simulated it on spice. If simulated the original way the simulation would not work. This is the actual schematic from the spice simulation. Don't really have an explanation as to why. It just works. Tried other op amps in the sim such as MAX410 LM124N and a few others. The LM318 works the best. \$\endgroup\$ – Dan Klonoski Jun 17 '15 at 16:27
  • \$\begingroup\$ I actually started off with a 10K/10K voltage divider and the circuit would not work at all and would overdrive the third stage. I removed the divider Network and added a small gain network and then started manipulating values and added a third resistor to control current to the second transistor to get the results I wanted. This is what I ended up with. \$\endgroup\$ – Dan Klonoski Jun 17 '15 at 16:36
  • \$\begingroup\$ The only thing that I can think of is the original schematic called for a 2n5428 transistor and I couldn't really find any information on that type since it's 30 years obsolete so I substituted with a common 2n3904. \$\endgroup\$ – Dan Klonoski Jun 17 '15 at 16:49
  • \$\begingroup\$ In simulation your transistors are perfectly matched. This will not be the case in the real world. \$\endgroup\$ – mng Jun 17 '15 at 17:42

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