Howland current pump question

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
Reply to
Phil Hobbs
Loading thread data ...

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.

formatting link

formatting link

formatting link

--

John Larkin         Highland Technology, Inc 
picosecond timing   precision measurement  

jlarkin att highlandtechnology dott com 
http://www.highlandtechnology.com
Reply to
John Larkin

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
Reply to
Winfield Hill

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.

Reply to
whit3rd

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
Reply to
John Larkin

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
Reply to
Tim Williams

Thanks, sorry for the delay in my reply. Any idea how to fix this? The circuit is copied directly from an application example...

--


----Android NewsGroup Reader---- 
http://usenet.sinaapp.com/
Reply to
bitrex

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...

Reply to
Frank Miles

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!

--


----Android NewsGroup Reader---- 
http://usenet.sinaapp.com/
Reply to
bitrex

ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.