giovanni
Well-known member
A while ago I saw an op-amp based buffer circuit with optional polarity reversing switch that uses a single op-amp, so I started searching for it, but I only found this (and the articles that inspired it). Digging a bit more in the darkest corners of the interwebs, I finally found the circuit I remembered, obviously from R.G.Keen. Now call me old fashioned, but I always thought the "less-is-more" philosophy is a really good fit for electronics design, so let's compare the two.
Here's the first circuit, using 2 op-amps:
As you can see, it's pretty much two textbook op-amp buffers, one inverting, one not, with a switch that lets you choose the polarity. Nothing more to see here, move along.
Next, the more clever and spartan design:
This time, the design is a bit more complicated and kind of an amalgam of the two circuits above. The input is connected to both op-amp inputs, but with the switch open the positive input has gain of exactly 2 (modulo component tolerances) and the negative side exactly -1 (ditto), for a grand total of 2-1=1 (ditto), i.e., a non inverting buffer. When closing the switch, the positive input is essentially grounded (Vbias is equivalent to ground for AC) so the gain automatically becomes -1, i.e., an inverting buffer. That comes at the cost of one extra resistor but one fewer op-amp.
Is the first circuit easier to design (at least for a first year EE student)? You bet it is. Is it more expensive to make? Maybe a bit given that an op-amp (or a dual op-amp) is probably more expensive than a resistor. Does it consume more power? Most likely although I'd have to do the math (I assume the op-amps consume the lion's share of the power). Is it less efficient? Definitely. And don't get me started on the applications of the second circuit as rectifier/octave and ring-mod. Also I'm not sure why the pulldown is on the input in the first circuit but on the output in the second. Does anybody have any idea?
Anyhoo, I thought this was a fun pair of circuits to share.
Here's the first circuit, using 2 op-amps:
As you can see, it's pretty much two textbook op-amp buffers, one inverting, one not, with a switch that lets you choose the polarity. Nothing more to see here, move along.
Next, the more clever and spartan design:
This time, the design is a bit more complicated and kind of an amalgam of the two circuits above. The input is connected to both op-amp inputs, but with the switch open the positive input has gain of exactly 2 (modulo component tolerances) and the negative side exactly -1 (ditto), for a grand total of 2-1=1 (ditto), i.e., a non inverting buffer. When closing the switch, the positive input is essentially grounded (Vbias is equivalent to ground for AC) so the gain automatically becomes -1, i.e., an inverting buffer. That comes at the cost of one extra resistor but one fewer op-amp.
Is the first circuit easier to design (at least for a first year EE student)? You bet it is. Is it more expensive to make? Maybe a bit given that an op-amp (or a dual op-amp) is probably more expensive than a resistor. Does it consume more power? Most likely although I'd have to do the math (I assume the op-amps consume the lion's share of the power). Is it less efficient? Definitely. And don't get me started on the applications of the second circuit as rectifier/octave and ring-mod. Also I'm not sure why the pulldown is on the input in the first circuit but on the output in the second. Does anybody have any idea?
Anyhoo, I thought this was a fun pair of circuits to share.
