Question about AD834 + AD811 as voltage controlled amplifier

Amandali

New Member
Aug 3, 2016
3
0
0
Hi,
Glad to be here.
According to page 15 of AD811 datasheet http://www.kynix.com/Detail/918337/AD811ARZ.html, 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=? 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+182? , 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.
kDPxg.png
 

SoundAndMotion

Well-Known Member
Mar 5, 2015
21
0
81
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.

Hi Amandali, it is not clear what you want to do and what your level of knowledge is, but I hope this helps. Do you know the difference between a differential and single-ended input? If not, we need to clear that up in another post.

1- The X input is the multiplier and it is differential in this diagram. X=(X1-X2) The red arrow points to the X1 (positive) input for X. The differential voltage X can vary from -1 to +1.
2- I think you mean page 12, and yes, the dB is a typo. It should read:
"Using feedback resistors R8 and R9 of 511 ?, the overall gain ranges from –70 dB, for VG = 0 V to +12 dB, (a numerical gain of four), when VG = +1 V. The overall transfer function of the VCA is:
VOUT = 4 (X1 – X2)(Y1 – Y2)"
In that diagram, Y2 is connected to ground (0 V), so Vin is single-ended.
You would never connect Vin to VG, unless you want to use the AD834 as a four-quadrant multiplier and you want to square the voltage: Vout=VG * Vin
 

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