Apparently, when used in the continuous mode, (as opposed to the switched mode) the local oscillator input to a needs to be distorted with a diode circuit. What is exactly this diode circuit -- a single diode or something more complex. Thanks in advance.
The LM13700 OTA has them built-in. The datasheet shows how they are connected.
The HFA3101 solves the linearity issue a different way - the emitter coupled pair emitters aren't connected together, allowing external resistors to be used.
Regards, Allan
It's another transistor, connected so as to effectively form a current mirror with each input transistor. The LM13700 and NE5517 (faster and higher current) show them as being diodes, but they're really transistors with C&B shorted.
Cheers
Phil Hobbs
-- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC / Hobbs ElectroOptics Optics, Electro-optics, Photonics, Analog Electronics Briarcliff Manor NY 10510 http://electrooptical.net http://hobbs-eo.com
There's an alternative, to gain-control the input so that only a small signal is present on the difference inputs. At circa 20 mV, it's nearly linear without the 'linearizing diode' addons.
Yes, but at the potential expense of increased noise.
In my limited experience, diode connected transistors are more 'ideal' than diodes. (I,V,T wise)
George H.
That is because the voltage and current paths are separated.
In a transdiode, the current goes via collector and the voltage is sensed via the base-emitter diode. The effect of the base resistance is diminished by the beta, as most of the current is diverted off the resistance.
-- -TV
Base resistance is an important effect, but even more important is avoiding high-level injection, which makes the recombination rate increase with bias current. That's what makes real diodes not follow the 'diode equation'. Even with the exponential constant increased to ~35-50 mV vs. kT/e ~26 mV they're still not nearly as accurate.
A diode-connected transistor is running in normal bias, so the depletion zone accelerates carriers through the base, so they have little chance to recombine and spoil the exponential character.
Cheers
Phil Hobbs
-- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC / Hobbs ElectroOptics Optics, Electro-optics, Photonics, Analog Electronics Briarcliff Manor NY 10510 http://electrooptical.net http://hobbs-eo.com
The diodes allow for, essentially, (almost) perfect multiplication of two currents. To wit:
A transistor has a nominal small signal trannsconductance, gm=40.Ic. i.e. Iout =40.Ic.Vin.
So, if Ic=k.Vx, then
iout=40.k.Vin.Vx
However, this is based on an approximation to an exponential transfer function of input voltage to output current. For large input volts swings, the exp() bit starts biting.
Essentially, the input transistors Log() the input current. This results, effectively in implementing Log(xy) = Log(x) + Log(y) and exp(Log(xy))
If you write the nodal equation, one derives Ic1/Id1 = Ic2/is2, then using the fact that the differential input current is held constant at IK, one gets
DeltaIC = IK. (Ic1-Ic2)/(Ic1+Ic2)
I leave the rest to the reader.
-- Kevin Aylward
How's the old glam rock tribute band going lately, Kev? You still getting your 6" platforms and gold Lurex suit on?
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Well, G.S.... mellowed out to MOR
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-- Kevin Aylward
Right, it's a diode with gain. (I think Phil H said that.)
George H.
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