> 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.
Reply by Phil Hobbs●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 whit3rd●August 19, 20162016-08-19
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.
Reply by Jim Thompson●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
>
>kevin
Yep. 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.
Reply by kevin93●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 kevin93●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 Hobbs
I'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 Jim Thompson●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 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
--
| 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 Phil Hobbs●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 Phil Hobbs●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 kevin93●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