Circuit challenge

What is "low impedance" ?

Can I use a transformer? ;)

Assuming the signal is symmetrical about zero volts, one approach is to run one phase through a RRIO inverting amplifier with a gain of -2 and a single positive supply. Sum its output with the other phase using equally weighted resistors of value less than

Cheers

Phil Hobbs

That actually only needs one phase, my error. ;)

Cheers

Phil

Ah, you're right, I'm sorry - I didn't qualify my request enough! ;)

Assume that the op amp in question is part of a section that requires both positive and negative supplies, and that it's a jellybean type (no RRIO).

No transformers allowed.

transistors/FETs allowed, so long as it's not more than a couple.

"Reasonable" amounts of other passives are allowed: diodes, full wave rectifiers, capacitors, resistors. Inductors not allowed!

Your solution is cute, but unfortunately it now does not meet the restrictions! ;(

Can you take each signal through a diode and sum them in an op amp?

Or would that not be "precision"?

tm

should be "bridge rectifiers" next to "diodes"

You wouldn't be compensating for the diode drop in any way, so unfortunately I don't think that would qualify.

Also, "low impedance" in this context means that hopefully, the rectified output can be taken from the output node of the op-amp, and not some other point in the circuit.

Be that way, see if I care. How about a quad transistor array?

Make a reference at -V_BE using a diode-connected BJT and a resistor. Attach a current mirror to that point (made out of two of the other transistors). Connect the mirror's input to one of the signals via a resistor 2R, and connect its output to the summing junction. Send the other input via a resistor R into the summing junction, with feedback resistor to taste.

It'll have a bit of crossover distortion for signals below about 50 mV.

Cheers

Phil Hobbs

Sounds cool. I'm having a little trouble visualizing - could you post an ASCII diagram?

Two equal signals 180 degrees out of phase sum to zero, yes? All you need is a wire to ground. And it's low impedance too. What am I missing? Also, the word "signal" leaves a lot of room for speculation. As does the word "precision".

On 11/27/2013 02:47 PM, bitrex wrote:

Posted an LTspice .asc file as an attachment--here it is inline. Still can't figure out a way to need that other phase! The 275k resistor takes out the offset due to the current mismatch between the mirror and the bias transistor near the zero crossing.

Cheers

Phil Hobbs

Version 4 SHEET 1 976 680 WIRE 768 -384 752 -384 WIRE 912 -384 896 -384 WIRE 768 -352 768 -384 WIRE 896 -352 896 -384 WIRE 768 -240 768 -272 WIRE 896 -240 896 -272 WIRE 480 -208 480 -256 WIRE 160 -96 32 -96 WIRE 480 -96 480 -128 WIRE 480 -96 240 -96 WIRE 608 -96 560 -96 WIRE 752 -96 688 -96 WIRE 480 0 480 -96 WIRE 560 0 560 -96 WIRE 560 0 480 0 WIRE 640 0 560 0 WIRE 752 16 752 -96 WIRE 752 16 704 16 WIRE 816 16 752 16 WIRE 32 32 32 -96 WIRE 32 32 -48 32 WIRE 64 32 32 32 WIRE 320 32 144 32 WIRE 640 32 624 32 WIRE -48 64 -48 32 WIRE 624 64 624 32 WIRE 160 128 64 128 WIRE 320 128 320 32 WIRE 400 128 320 128 WIRE 160 160 160 128 WIRE 320 160 320 128 WIRE 480 160 480 0 WIRE -48 176 -48 144 WIRE 64 208 64 128 WIRE 96 208 64 208 WIRE 400 208 400 128 WIRE 400 208 384 208 WIRE 416 208 400 208 WIRE 64 272 64 208 WIRE 160 288 160 256 WIRE 224 288 160 288 WIRE 320 288 320 256 WIRE 320 288 224 288 WIRE 480 288 480 256 WIRE 480 288 320 288 WIRE 224 352 224 288 WIRE 224 512 224 432 FLAG 64 272 0 FLAG 896 -240 0 FLAG 768 -240 0 FLAG 912 -384 +15 FLAG 752 -384 -15 FLAG 224 512 -15 FLAG 624 64 0 FLAG 816 16 Out FLAG 672 48 -15 FLAG 672 -16 +15 FLAG -48 176 0 FLAG 480 -256 +15 SYMBOL npn 96 160 R0 SYMATTR InstName Q1 SYMATTR Value 2N3904 SYMBOL npn 384 160 M0 SYMATTR InstName Q2 SYMATTR Value 2N3904 SYMBOL npn 416 160 R0 SYMATTR InstName Q3 SYMATTR Value 2N3904 SYMBOL res 160 16 R90 WINDOW 0 0 56 VBottom 2 WINDOW 3 32 56 VTop 2 SYMATTR InstName R1 SYMATTR Value 2k SYMBOL voltage -48 48 R0 WINDOW 3 -140 -39 Left 2 WINDOW 123 0 0 Left 2 WINDOW 39 0 0 Left 2 SYMATTR InstName V1 SYMATTR Value SINE(0 5 1000 1m) SYMBOL res 144 -80 R270 WINDOW 0 32 56 VTop 2 WINDOW 3 0 56 VBottom 2 SYMATTR InstName R2 SYMATTR Value 4k SYMBOL res 208 336 R0 SYMATTR InstName R3 SYMATTR Value 2k SYMBOL voltage 768 -368 R0 SYMATTR InstName V3 SYMATTR Value -15 SYMBOL voltage 896 -368 R0 SYMATTR InstName V4 SYMATTR Value +15 SYMBOL res 592 -80 R270 WINDOW 0 32 56 VTop 2 WINDOW 3 0 56 VBottom 2 SYMATTR InstName R4 SYMATTR Value 2k SYMBOL Opamps\\LT1037A 672 -48 R0 SYMATTR InstName U1 SYMBOL res 464 -224 R0 SYMATTR InstName R5 SYMATTR Value 275k TEXT 472 456 Left 2 !.tran 10m

Nice. So it looks like that when the input signal is negative-going, the op amp is operating in the standard inverting configuration. When it is positive-going, the op amp is acting as a TIA with current being pulled out, also leading to a positive going output?

I'm still pushing myself to think of a way to do it using both phases, with no active devices other than the op amp and diodes. It may not be possible.

Anybody doing this for real would use a dual op amp, and be done with it.

Two typical methods for that:

1. One amp connected as a unity gain inverter, with its noninverting input connected to Vin via a resistor. The other op amp section is an active diode clamp that prevents the noninverting input from going negative. That automagically switches the overall gain from +1 to -1 when the input crosses zero.

1. Unity gain inverter, with the feedback resistor inside a diode bridge, with the second stage used as a differential amp across the resistor.

If I could have had a transformer, I'd have used it to drive the op amp input differentially as a unity-gain inverter as above, but ground the - output of the diode bridge and let the inputs bounce around, and then dealt with the residual diode drop.

Cheers

Phil Hobbs

Assuming the two input signals count as "low impedance", another and more accurate one-amp method would be to connect the two phases to the output, one via a pair of NFETs and the other with a pair of PFETs, each pair tied G-G and S-S, with all the gates connected to the op amp output. Use the op amp as a comparator to pick whichever phase is positive.

Cheers

Phil Hobbs

(Who is finding it hard to focus on an urgent bit of lawyering due to a sinus infection--it's amazing how irritating people's patents are to read when you're in a foul mood and your head hurts.)

On 11/27/2013 04:33 PM, bitrex wrote:

It is with a transformer. The problem with the bridge rectifier approach is that the + end of the bridge is the high impedance (summing junction) end half of the time.

You can also use a bridge as a series clipper with pretty well zero offset, like this.

Cheers

Phil Hobbs

Version 4 SHEET 1 976 680 WIRE 768 -384 752 -384 WIRE 912 -384 896 -384 WIRE 768 -352 768 -384 WIRE 896 -352 896 -384 WIRE 368 -256 368 -304 WIRE 768 -240 768 -272 WIRE 896 -240 896 -272 WIRE 160 -96 32 -96 WIRE 560 -96 240 -96 WIRE 608 -96 560 -96 WIRE 752 -96 688 -96 WIRE 368 -64 368 -176 WIRE 368 -64 320 -64 WIRE 400 -64 368 -64 WIRE 560 0 560 -96 WIRE 640 0 560 0 WIRE 752 16 752 -96 WIRE 752 16 704 16 WIRE 816 16 752 16 WIRE 32 32 32 -96 WIRE 32 32 -48 32 WIRE 320 32 320 0 WIRE 320 32 32 32 WIRE 400 32 400 0 WIRE 560 32 400 32 WIRE 640 32 560 32 WIRE -48 48 -48 32 WIRE 320 48 320 32 WIRE 400 48 400 32 WIRE -48 64 -48 48 WIRE 560 80 560 32 WIRE 320 128 320 112 WIRE 368 128 320 128 WIRE 400 128 400 112 WIRE 400 128 368 128 WIRE -48 176 -48 144 WIRE 560 192 560 160 WIRE 256 208 224 208 WIRE 368 208 368 128 WIRE 368 208 336 208 WIRE 224 256 224 208 WIRE 368 384 368 208 WIRE 368 512 368 464 FLAG 896 -240 0 FLAG 768 -240 0 FLAG 912 -384 +15 FLAG 752 -384 -15 FLAG 368 512 -15 FLAG 816 16 Out FLAG 672 48 -15 FLAG 672 -16 +15 FLAG -48 176 0 FLAG -48 48 In FLAG 560 192 0 FLAG 368 -304 +15 FLAG 224 256 0 SYMBOL voltage -48 48 R0 WINDOW 3 -140 -39 Left 2 WINDOW 123 0 0 Left 2 WINDOW 39 0 0 Left 2 SYMATTR Value SINE(0 5 1000 1m) SYMATTR InstName V1 SYMBOL res 144 -80 R270 WINDOW 0 32 56 VTop 2 WINDOW 3 0 56 VBottom 2 SYMATTR InstName R2 SYMATTR Value 2k SYMBOL res 352 368 R0 SYMATTR InstName R3 SYMATTR Value 10k SYMBOL voltage 768 -368 R0 SYMATTR InstName V3 SYMATTR Value -15 SYMBOL voltage 896 -368 R0 SYMATTR InstName V4 SYMATTR Value +15 SYMBOL res 592 -80 R270 WINDOW 0 32 56 VTop 2 WINDOW 3 0 56 VBottom 2 SYMATTR InstName R4 SYMATTR Value 2k SYMBOL Opamps\\LT1037A 672 -48 R0 SYMATTR InstName U1 SYMBOL res 544 64 R0 SYMATTR InstName R7 SYMATTR Value 20k SYMBOL diode 304 48 R0 SYMATTR InstName D1 SYMBOL diode 384 48 R0 SYMATTR InstName D2 SYMBOL diode 304 -64 R0 SYMATTR InstName D3 SYMBOL diode 384 -64 R0 SYMATTR InstName D4 SYMBOL res 384 -160 R180 WINDOW 0 36 76 Left 2 WINDOW 3 36 40 Left 2 SYMATTR InstName R1 SYMATTR Value 10k SYMBOL res 352 192 R90 WINDOW 0 0 56 VBottom 2 WINDOW 3 32 56 VTop 2 SYMATTR InstName R5 SYMATTR Value 500 TEXT 472 456 Left 2 !.tran 10m

As in...

...Jim Thompson

Yup. Though I'm not the hugest fan of composite amps, because they're squirrelly at high speed, and usually have settling whoopdedoos at long times.

Cheers

Phil Hobbs