Well, that's sort of how a sampling bridge works.
Hopefully it'll repay the effort.
Cheers
Phil Hobbs
Well, that's sort of how a sampling bridge works.
Hopefully it'll repay the effort.
Cheers
Phil Hobbs
-- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 hobbs at electrooptical dot net http://electrooptical.net
Sure, but one corner of the sampling bridge sees about 25 ohms to ground, and the sampling pulses are really short. They're not really current sources.
-- John Larkin Highland Technology, Inc picosecond timing precision measurement jlarkin att highlandtechnology dott com http://www.highlandtechnology.com
That's an "update" of the 2008 NSC app note, and a significant change was that TI removed Bob Pease's name as the author.
-- Thanks, - Win
In this example, transformer T1 couples the sink/source pair of currents, which ensures (for short sample times) source and sink current equality. It effectively decouples the symmetry problem from the trim resistors and associated wiring.
T1 is a balun, but it does balance the short sampling pulses.
-- John Larkin Highland Technology, Inc picosecond timing precision measurement jlarkin att highlandtechnology dott com http://www.highlandtechnology.com
Just use an op-amp and transistor per output, with supply side current sense (emitter / source resistor). This makes a transconductance amp (gain = Iout / Vin). Now all you need to do is supply equal voltages to all the stages, and you get N accurate current sources.
To get a +V referenced voltage, hang a sense resistor from +V and drive it with a sink from -V. The input and sense resistors all have to be matched/trimmed within the desired accuracy to keep unity gain, as well as the op-amp Vos's. Or mismatched to add offset or change the gain.
Tim
-- Seven Transistor Labs, LLC Electrical Engineering Consultation and Contract Design Website: http://seventransistorlabs.com
Thanks, sorry for the delay in my reply. Any idea how to fix this? The circuit is copied directly from an application example...
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Without checking your 'schematic' : if the device is saturated you may not have a low enough load impedance, and the positive feedback is 'winning'.
HTH...
I think I found the problem. The non inverting input has to come from the top of the output load, and the inverting has to come from the other side of the sense resistor. I had things reversed...doh!
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