On Friday, August 19, 2016 at 2:37:07 PM UTC-7, Phil Hobbs wrote:> >I disagree, I think it just needs a negative input to the 1M. > >The opamp output will then go positive to bias the input diodes > >and maintain the zero voltage at the diff input. > > Except that the bias current has to flow into both + and - inputs.Correct; the inputs are bases of NPN transistors, and the OTA saturates unless you bias both with net input current.>ISTM there are actually two mistakes there--the bias current polarity on I_diode and the lack of positive current into one input.The current into the (+) input comes from the 'compensation' diode, minus some PNP emitter current. Positive current from the (+) compensation diode, not from the (+) input pin. Net current positive. I think this is mainly confusing because we are accustomed to op amps, where input voltages balance; the input voltages on an OTA which is actually delivering current (to the circular terminal on the rightmost 1M resistor) are going to differ by few-to-dozens of millivolts. The multiplication output signal comes from that small imbalance.
exponential converter temperature stabilization
Started by ●August 17, 2016
Reply by ●August 19, 20162016-08-19
Reply by ●August 20, 20162016-08-20
On 08/19/2016 05:55 PM, Jim Thompson wrote:> On Fri, 19 Aug 2016 14:37:03 -0700 (PDT), Phil Hobbs > <pcdhobbs@gmail.com> wrote: > >> >>> I disagree, I think it just needs a negative input to the 1M. >>> The opamp output will then go positive to bias the input diodes >>> and maintain the zero voltage at the diff input. >> >> Except that the bias current has to flow into both + and - inputs. I'm on my phone, so I'm still working from memory, but ISTM there are actually two mistakes there--the bias current polarity on I_diode and the lack of positive current into one input. >> >> cheers >> >> Phil Hobbs > > Phil, It's an LM13700, so the diode polarity is bass-ackwards from > normal Norton Amplifier thought, so the 1Meg _does_ need to be > connected to some negative reference. > > But, all-in-all, the scheme is _not_ well thought out. > > ...Jim Thompson >Roight--the diodes are upside-down from the Norton approach. I was thinking of current mirrors, I expect. The 1M does need to go to a negative supply. 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 by ●August 20, 20162016-08-20
On 08/19/2016 06:39 PM, Jim Thompson wrote:> On Fri, 19 Aug 2016 15:17:37 -0700 (PDT), kevin93 > <kevin@whitedigs.com> wrote: > >> On Friday, August 19, 2016 at 2:56:09 PM UTC-7, Jim Thompson wrote: >> ... >>>> Phil Hobbs >>> >>> Phil, It's an LM13700, so the diode polarity is bass-ackwards from >>> normal Norton Amplifier thought, so the 1Meg _does_ need to be >>> connected to some negative reference. >>> >>> But, all-in-all, the scheme is _not_ well thought out. >>> >>> ...Jim Thompson >> ... >> >> Ahh, that's why Phil and me are talking at cross-purposes - we made different assumptions. >> >> thanks >> >> kevin > > Yep. I was likewise snagged until I looked up an LM13700 datasheet. > > ...Jim Thompson >There are a lot of goofy electronic "expo temperature stabilization" circuits on the forum where that came from. As you say, none of them seem particularly well-thought out or seem to offer much benefit over a simple tempco resistor in the input divider. Well, I'll keep thinking and looking.
