According to page 15 of AD811 datasheet , it can be used as a voltage controlled amplifier in conjunction with AD834 . In the schematic below, I am a bit confused about the configuration.
If we assume that W1 and W2 are in phase shift=π and supplied by AD834 internal Open collector transistors which are configured as common emitters by external resistors, it easy to understand that AD811 is working in a differential mode. VG stands for gain control voltage ( between 0-1v ?) and I think Vin must be the input signal. but what I can not understand is the + point in the schematic (red arrow in the schematic below).
What is this point connected to? it can not be V+ because 12v is well above the maximum ratings for inputs (page 4 of AD834 datasheet: with 11mA in Vs+ and R6+R4= 294+182Ω , we have 5.2v drop in VS pin but the maximum allowed Vin is equal to Vs). On the other hand AD834 is designed for working in +/-1.3 clipping voltage between X and Y and a positive voltage above 1v on this pin simply goes to its clipping region.
My questions are:
1- To where the indicated point is connected?
2- On page 15, line 10 of AD811 datasheet, it says that the gain is -70dB for VG=0 dB and 4 for VG=1 volt. I think this dB is a typographic error. If it is really dB and not volt, it may mean that for gain=-70dB we need to connect Vin to VG. Is it correct?
Any suggestion appreciated in advance.
Answer
Vg can be a differential input - if you look carefully they also suggest grounding Vg-, then Vg+ becomes a dc level relative to 0V that controls the gain of the device.
I also think it is a typo on page 15 - with zero control voltage (Vg- at 0V and Vg+ at 0V) there is a big attenuation (70dB) of the real input (Vin). When Vg+ is 1V greater than Vg- the voltage gain of the circuit is +12dB.
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