Could some electronics guru...heh, no, I'm not that guy. But could some electronics guru explain how this circuit is supposed to stabilize the Vt dependence of a differential pair V->I exponential converter circuit? http://www.openmusiclabs.com/files/lm13700_expo2.jpg
exponential converter temperature stabilization
Started by ●August 17, 2016
Reply by ●August 17, 20162016-08-17
Doesn't look like it does anything useful. The op amp at top left is keeping the current through the top diode at zero, so if the PNP is passing any current, there'll be a full V_BE across the OTA's inputs, so it's output is always railed. Cheers Phil Hobbs
Reply by ●August 17, 20162016-08-17
On 08/17/2016 09:02 AM, Phil Hobbs wrote:> Doesn't look like it does anything useful. The op amp at top left is keeping the current through the top diode at zero, so if the PNP is passing any current, there'll be a full V_BE across the OTA's inputs, so it's output is always railed. > > Cheers > > Phil Hobbs >...its output,.... Grr, brain dead autocorrect on my phone! 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 17, 20162016-08-17
On Wed, 17 Aug 2016 09:47:57 -0400, Phil Hobbs <pcdhSpamMeSenseless@electrooptical.net> wrote:>On 08/17/2016 09:02 AM, Phil Hobbs wrote: >> Doesn't look like it does anything useful. The op amp at top left is keeping the current through the top diode at zero, so if the PNP is passing any current, there'll be a full V_BE across the OTA's inputs, so it's output is always railed. >> >> Cheers >> >> Phil Hobbs >> >...its output,.... > >Grr, brain dead autocorrect on my phone! > >Cheers > >Phil HobbsThere is a newsgroup... alt.possessive.its.has.no.apostrophe -- John Larkin Highland Technology, Inc lunatic fringe electronics
Reply by ●August 17, 20162016-08-17
On Tuesday, August 16, 2016 at 11:14:48 PM UTC-7, bitrex wrote:> But could some electronics guru explain how this circuit is supposed to > stabilize the Vt dependence of a differential pair V->I exponential > converter circuit? > > http://www.openmusiclabs.com/files/lm13700_expo2.jpgThe leftmost 1M resistor is to pseudo-ground, and is an input signal terminal. The current in the two OTA input diodes will mismatch according to the signal presented at the rightmost 1M resistor. It and the OTA output form a current summing junction, and the rightmost op amp keeps that current sum zero. The central 1M resistor is a bias current source, probably not an input. It appears that the multiplication is done by the matched input bias diodes of the OTA. Probably the temperature-sensitive gain of the OTA compensates it (but I'd want to scribble over a few sheets of paper to be sure).
Reply by ●August 19, 20162016-08-19
whit3rd <whit3rd@gmail.com> Wrote in message:> On Tuesday, August 16, 2016 at 11:14:48 PM UTC-7, bitrex wrote: > >> But could some electronics guru explain how this circuit is supposed to >> stabilize the Vt dependence of a differential pair V->I exponential >> converter circuit? >> >> http://www.openmusiclabs.com/files/lm13700_expo2.jpg > > The leftmost 1M resistor is to pseudo-ground, and is an input signal terminal. > > The current in the two OTA input diodes will mismatch according > to the signal presented at the rightmost 1M resistor. It and the OTA output > form a current summing junction, and the rightmost op amp keeps that current sum zero. > > The central 1M resistor is a bias current source, probably not an input. > > It appears that the multiplication is done by the matched input bias diodes of the OTA. > Probably the temperature-sensitive gain of the OTA compensates it (but I'd > want to scribble over a few sheets of paper to be sure). >On the forum where that came from there are a lot of wacky schemes to try to electronically compensate the Vt dependence of a differential pair. exponential converter. If it works this seems like one of the cleanest of the bunch. -- ----Android NewsGroup Reader---- http://usenet.sinaapp.com/
Reply by ●August 19, 20162016-08-19
On 08/19/2016 11:11 AM, bitrex wrote:> whit3rd <whit3rd@gmail.com> Wrote in message: >> On Tuesday, August 16, 2016 at 11:14:48 PM UTC-7, bitrex wrote: >> >>> But could some electronics guru explain how this circuit is supposed to >>> stabilize the Vt dependence of a differential pair V->I exponential >>> converter circuit? >>> >>> http://www.openmusiclabs.com/files/lm13700_expo2.jpg >> >> The leftmost 1M resistor is to pseudo-ground, and is an input signal terminal. >> >> The current in the two OTA input diodes will mismatch according >> to the signal presented at the rightmost 1M resistor. It and the OTA output >> form a current summing junction, and the rightmost op amp keeps that current sum zero. >> >> The central 1M resistor is a bias current source, probably not an input. >> >> It appears that the multiplication is done by the matched input bias diodes of the OTA. >> Probably the temperature-sensitive gain of the OTA compensates it (but I'd >> want to scribble over a few sheets of paper to be sure). >> > > On the forum where that came from there are a lot of wacky schemes > to try to electronically compensate the Vt dependence of a > differential pair. exponential converter. > > If it works this seems like one of the cleanest of the bunch. >But it doesn't. The top left op amp needs to run its noninverting input at some nonzero bias voltage, because otherwise there's no current through the top diode, as I said earlier. It can be fixed up, but it's broken as posted. 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 19, 20162016-08-19
On Friday, August 19, 2016 at 8:23:58 AM UTC-7, Phil Hobbs wrote: ...> > But it doesn't. The top left op amp needs to run its noninverting input > at some nonzero bias voltage, because otherwise there's no current > through the top diode, as I said earlier. It can be fixed up, but it's > broken as posted.... I think the 1M resistor at far left goes to a negative supply - not ground. The schematic is a bit confusing and ambiguous. kevin
Reply by ●August 19, 20162016-08-19
>> But it doesn't. The top left op amp needs to run its noninverting input >> at some nonzero bias voltage, because otherwise there's no current >> through the top diode, as I said earlier. It can be fixed up, but it's >> broken as posted....>I think the 1M resistor at far left goes to a negative supply - not >ground. The schematic is a bit confusing and ambiguous.Well, I think it's just wrong, meself. returning the 1M at left to +5 would make it do something more sensible. Cheers Phil Hobbs
Reply by ●August 19, 20162016-08-19
On Friday, August 19, 2016 at 9:54:08 AM UTC-7, Phil Hobbs wrote: ...> >I think the 1M resistor at far left goes to a negative supply - not > >ground. The schematic is a bit confusing and ambiguous. > > Well, I think it's just wrong, meself. returning the 1M at left to +5 would make it do something more sensible.... Surely not - returning it to a negative supply would forward bias the upper diode and cause the opamp output to go positive to maintain the inverting input at ground. In so doing it would make the OTA balanced when the current through the transistor is the same as the reference current through the 1M. The input current through the right 1M would unbalance the loop. The right opamp would then servo the base of the transistor to rebalance - the current through the transistor (which is also the output) would then be exponentially related to the current injected via the right 1M because of the diodes at the input of the OTA. The output (To VCO) would have to go to a VCO that is referenced to a negative supply to provide a suitable voltage across the transistor. kevin
