On 08/19/2016 01:41 PM, kevin93 wrote:> 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 >I'm going from memory here, but I thought the 1M went to the OTA input and the op amp was driving the I_diode pin negative. The input 'diodes' are actually diode-connected NPN transistors nominally identical with the diff pair, so you have to dump current into the inputs and suck it back out of the I_diode pin. So if the 1M is connected to the input, the other end has to be positive for the circuit to work. 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
exponential converter temperature stabilization
Started by ●August 17, 2016
Reply by ●August 19, 20162016-08-19
Reply by ●August 19, 20162016-08-19
On 08/19/2016 03:03 PM, Phil Hobbs wrote:> On 08/19/2016 01:41 PM, kevin93 wrote: >> 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 >> > I'm going from memory here, but I thought the 1M went to the OTA input > and the op amp was driving the I_diode pin negative. The input 'diodes' > are actually diode-connected NPN transistors nominally identical with > the diff pair, so you have to dump current into the inputs and suck it > back out of the I_diode pin. > > So if the 1M is connected to the input, the other end has to be positive > for the circuit to work. > > Cheers > > Phil Hobbs >I've probably got I_diode upside down--I vaguely recall that there's another current mirror on I_diode so you don't need a negative supply, but don't have time to look up the LM13700 datasheet and see. One way or another, the op amp is trying to put a constant bias of zero on one of the input diodes, which needs to be fixed. Cheers Phil Hobbs (scrambling madly to get a BOM done by COB today) -- 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 12:03:11 PM UTC-7, Phil Hobbs wrote: ...> I'm going from memory here, but I thought the 1M went to the OTA input > and the op amp was driving the I_diode pin negative. The input 'diodes' > are actually diode-connected NPN transistors nominally identical with > the diff pair, so you have to dump current into the inputs and suck it > back out of the I_diode pin. > > So if the 1M is connected to the input, the other end has to be positive > for the circuit to work.... No, it's the other way round. The diode bias pin needs to be positive wrt the diff inputs. They could be NPN transistors diode connected but they are in series with diff inputs with their collectors joined and brought out to the bias pin. kevin
Reply by ●August 19, 20162016-08-19
On Friday, August 19, 2016 at 12:14:09 PM UTC-7, Phil Hobbs wrote: ...> > > I've probably got I_diode upside down--I vaguely recall that there's > another current mirror on I_diode so you don't need a negative supply, > but don't have time to look up the LM13700 datasheet and see. One way > or another, the op amp is trying to put a constant bias of zero on one > of the input diodes, which needs to be fixed.... 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. kevin
Reply by ●August 19, 20162016-08-19
>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
Reply by ●August 19, 20162016-08-19
>No, it's the other way round. The diode bias pin needs to be positive wrt the diff inputs.because there's a current mirror in there. they aren't PNPs.>They could be NPN transistors diode connected but they are in series >with diff inputs with their collectors joined and brought out to the bias pin.They aren't in series, they're in shunt. Have a look at the schematic in the datasheet. cheers Phil Hobbs
Reply by ●August 19, 20162016-08-19
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 HobbsPhil, 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 -- | James E.Thompson | mens | | Analog Innovations | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | San Tan Valley, AZ 85142 Skype: Contacts Only | | | Voice:(480)460-2350 Fax: Available upon request | Brass Rat | | E-mail Icon at http://www.analog-innovations.com | 1962 | I'm looking for work... see my website.
Reply by ●August 19, 20162016-08-19
On Friday, August 19, 2016 at 2:39:56 PM UTC-7, Phil Hobbs wrote:> >No, it's the other way round. The diode bias pin needs to be positive wrt the diff inputs. > > because there's a current mirror in there. they aren't PNPs. > > >They could be NPN transistors diode connected but they are in series > >with diff inputs with their collectors joined and brought out to the bias pin. > > They aren't in series, they're in shunt. Have a look at the schematic in the datasheet. > > cheers > > Phil HobbsI'll wait until you have a computer available. I agree I was being inaccurate in saying they were in series, I meant that bias current could flow through the diodes and then into the bases of the input transistors - from the AC point of view they are in parallel as you say. Here is the schematic for the LM13700 from www.idea2ic.com where there is a description by the designer. http://www.idea2ic.com/LM13600/SpiceSubcircuit/LM13700_SpiceModel.html cheers kevin * ^ VCC * /_\ * _____|____________________________________________ * | | | | | | * -> <- -> <- ___ _| | * QP1 `|___|' QP2 QP4 `|___|'QP5 |BIN|_|' QN11 | * _ '| | |`_ _ '| | |`_ |___| |`-> _| * ___ | |____| | |____| VN12B |___|'QN13 * |LIN| | VP2B | | VP5B | ____| |`-> * |___| | <- | <- | _| | * QN6 | |______|'QP3 |______|'QP6 |_|' QN12 | * _________|_ |VP3B |`_ |VP6B |`_ |`-> | * | _| | _| | | ______| | |______| * |_|' |_|' QN7 | | | | | * |`-> |`-> | |_/|\_____ | _|_ * | | | | | | |BUF| * ___ | | _| |_ | | ___ |___| * |INN|_|_____/|\__|' QN4 QN5 `|_ | |_|OUT| * |___| | |`-> <-'| | | | |___| * | |____________| | | | * ___ | VN3C | | | | * |INP|________|__________________/|\___| | | * |___| | | | * ___ _| |VN10B _| * |IBA|___________|' QN3 |______|'QN10 * |___| | |`-> | |`-> * | VN2B____| | VN9B____| * |_ | _| |_ | _| * QN1 `|_|_|' QN2 QN8 `|_|_|' QN9 * <-'| |`-> <-'| |`-> * | | | | * |_________|______|_________| * _|_ * ///
Reply by ●August 19, 20162016-08-19
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
Reply by ●August 19, 20162016-08-19
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 > >kevinYep. I was likewise snagged until I looked up an LM13700 datasheet. ...Jim Thompson -- | James E.Thompson | mens | | Analog Innovations | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | San Tan Valley, AZ 85142 Skype: Contacts Only | | | Voice:(480)460-2350 Fax: Available upon request | Brass Rat | | E-mail Icon at http://www.analog-innovations.com | 1962 | I'm looking for work... see my website.
