I'm looking to generate a differential signal to control a laser projector's galvos, and as I understand it needs to be +5V/-5V (10Vpp). I've found this circuit for a laser harp, but I'm confused about what this specific dual-opamp design does. It looks as if it's a pair of inverting and non-inverting amplifiers with a gain of 1, but they're being fed into each other. Here's a picture:

enter image description here

The original can be found here.

I'm curious if anyone could tell me what it's called, or how it works, because I've lookat at plenty of 'example circuits' and couldn't find anything that resembles it.

  • \$\begingroup\$ See the update to my answer for a link, with analysis, to essentially the same circuit. \$\endgroup\$ – Alfred Centauri Mar 20 '14 at 12:40
  • \$\begingroup\$ I'm finding a lot more info on Google when I look for 'balanced driver'. \$\endgroup\$ – Dave Van den Eynde Mar 20 '14 at 14:45
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    \$\begingroup\$ Necro comment, here, this is a discrete implementation of the documented internals of a Ti DRV134. It is unlikely to be balanced unless it is trimmed like the integrated circuit is. I made this one up with 1% components and the output is -3 +5, but at least accurately out of phase. \$\endgroup\$ – mianos Feb 17 '18 at 0:30

Looking at the topmost op-amp and ignoring the \$100 \Omega\$ resistors, write by inspection:

$$v_{X+} = v_{OUTX} + v_{X-}$$

For the bottommost op-amp, write

$$v_{X-} = v_{X+} - v_{OUTX}$$


$$v_{X+} - v_{X-} = v_{OUTX}$$

So, this circuit converts a single-ended input signal, \$v_{OUTX}\$ to a balanced output signal; it's an active 1:1 'transformer'.

An interesting 'feature' of this circuit is that, while the differential output voltage, \$v_{OD} = (v_{X+} - v_{X-}) = v_{OUTX}\$, is well defined, the singled-ended voltages \$v_{X+}\$ and \$v_{X-}\$ aren't.

For example, substituting the 2nd equation into the 1st yields

$$v_{X+} = v_{X+}$$

and similarly

$$v_{X-} = v_{X-}$$

So, in fact, the common mode output voltage output voltage

$$v_{OCM} = \frac{v_{X+} + v_{X-}}{2} = ?$$

is not determined without an additional equation (circuit constraint).

Update: I know I've seen and analyzed this type of circuit before but I haven't yet found my notes on it.

However, I did find this article at the Elliot Sound Products site for a "Balanced Line Driver with Floating Output" which appears to be essentially the same circuit except with a balanced input rather than single-ended input.

enter image description here

The whole amplifier, as it is dimensioned here, has a gain of 1. The same amount of voltage across the input terminals appears across the output terminals. This remains true if any output terminal is supplied with any voltage - like transformer coupled outputs do (provided both output voltages stay within the supply voltage area of course).

  • \$\begingroup\$ It occured to me last night, while I was pondering this, was that the whole idea of having the same signal buffered again sounded ridiculous (the input comes from another opamp that converts 0..2048mV into -10..+10V) but then it struck me: The two outputs need to be in perfect balance and phase, and having an opamp on one signal but not on the other would incur a (small) signal delay. For audio applications, this would be more critical than for positioning a laser mirror, but still. If the designer took inspriation there, it makes sense. \$\endgroup\$ – Dave Van den Eynde Mar 20 '14 at 14:41
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    \$\begingroup\$ This circuit has two interesting characteristics. The most important is the "floating" (within limits) differential output. The second is the 100R output impedance. I doubt either matter for laser mirrors, but I'd expect this might be a standard output circuit in pro audio. \$\endgroup\$ – Spehro Pefhany Mar 20 '14 at 15:21
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    \$\begingroup\$ The original source is AFAIK the HP 8903 audio analyzer. It is one of the "key circuits" the HP engineers published about in the HP Journal (Aug 1980, "Floating a Source Output, by George D. Pontis). \$\endgroup\$ – dom0 Apr 11 '15 at 20:34

At first I thought the circuit was a Differential Howland Current Pump.

Similar to this one here.

enter image description here

I thought maybe the cross-coupling makes the current sources share the available voltage.

But I did a simulation since the analysis didn't indicate that was possible..

With no load, the (-) output is a virtual ground and the (+) output equals the input voltage, which is not very exciting.

With a 1000 ohm load, the differential voltage is 90% of the input voltage (implying about a 100 ohm output impedance) but the (-) output is following the input by about +4%.

With a 100 ohm load, the waveforms look like this:

  • Green: input voltage

  • Purple: output +

  • Red: output -

  • Yellow: differential output voltage

enter image description here

I'm at a bit of a loss to understand the usefulness of this functionality if it's feeding coils directly.


As Alfred has pointed out, the circuit should have a high output impedance with respect to common, and as I said, the differential output impedance is low and matched to a twisted pair. So it would be a suitable driver for a balanced output feeding a twisted pair, going to receiver that might have a different (by as much as a few volts) ground potential from the transmitter. Very nice.

Here is a plot of the common mode impedance measured by applying a 1VAC signal to the center of a split load resistance of 100 ohms, and sweeping from 0.1Hz to 10MHz.

enter image description here

As you can see, it's 10K for low frequencies, crossing over at around 2.2kHz and dropping to 150 ohms or so at high frequencies. Perfect for situations where there is mains-frequency voltage between grounds, not so great for higher frequencies.

  • \$\begingroup\$ Looks to me like the inverted outputs feed back to the positive inputs, like in a "true differential op amp", where the negative out feeds to the positive input and the positive out feeds to the negative input, but implemented with two single ended op amps. \$\endgroup\$ – Scott Seidman Mar 19 '14 at 22:39
  • \$\begingroup\$ @ScottSeidman It's not very differential. See simulation. \$\endgroup\$ – Spehro Pefhany Mar 20 '14 at 3:39
  • \$\begingroup\$ Only the differential output voltage is well defined for this circuit, the single-ended output voltages are not without additional circuit constraints. \$\endgroup\$ – Alfred Centauri Mar 20 '14 at 12:16
  • \$\begingroup\$ So you would expect the outputs to have a high common-mode impedance wrt ground.. presumably expecting to see balanced loads to a common which might differ in voltage from the voltage at ground. This makes sense if the output is fed to a separate amplifier circuit. \$\endgroup\$ – Spehro Pefhany Mar 20 '14 at 13:01
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    \$\begingroup\$ I'm liking the post-edit explanation. \$\endgroup\$ – gwideman Mar 20 '14 at 23:51

Looking at the schematic you linked, evidently this op amp configuration is used to drive outputs that are part of the standard ILDA interface to laser projectors (as you alluded).


So the primary task is to create a differential signal from a single signal.

A differential signal is usually used to deliver an analog signal in an environment susceptible to noise, as laser shows might well be. Any noise will affect both the positive and negative copy of the signal approximately equally, and when the receiver recovers the signal by subtracting one from the other, the noise gets subtracted out.

Resistors R45 and R52 create some protection for the op amps in case the outputs are shorted, and possibly some impedance matching to the cable, though I'm not sure the need for that in this application (don't know the freqs involved).

But what about R48 and R49, and the apparent feedback they provide to the "opposite" amplifier? I think they might implement compensation for the attenuation introduced by R45 and R52, useful if the receiver input impedances are not balanced.

  • \$\begingroup\$ I know this, since I'm trying to accomplish the same thing. I just wanted to understand what this diagram is doing in order to understand how to build m own. \$\endgroup\$ – Dave Van den Eynde Mar 20 '14 at 14:36

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