Programable Gain Amplifier, using cd4051

[garg29]

potensiometer digital using cd4051

Hi friends,

I'm trying to design a simple PGA (Programable Gain Amplifier), using cd4051 (8-channel Analog Multiplexer) with OP07 opamp. i'm controlling the channel with microcontroller 89c51. (Please see the diagram, attachment).

The problem:-
now let say i send 000 on address line to select X0 line, ie. X-X0 are now connected, ideally i should be having the feedback resistor of 47K on opamp, but i dont get that, instead i get very low resistance. If i remove all other resistors except 47K on X0 i get proper resistance on opamp.

Please help me.Thanks

With best regards,
Garg

 

[Sudhir.K.A]

Hi garg29,

I have got two queries regarding your ciruit.

1. What is the output Voltage of OP07?
(Is it more than CD4051 supply voltage?)

2. What is the measured supply voltage for CD4051 in your system?

If the OP07 output voltage is > supply voltage, the input CMOS protection circuit
i.e diode connected to supply will start conducting. If conduction happens for all channels, then resistors connected to different channels come in parallel to each other.

Before reaching to this conclusion i want to know more about the OP07 output leveland supply voltage level :|

 

[garg29]

thanks mr.sudhir for replying.the supply voltage on cd4051 is +5-0 volts and on op07 is +15v to -15volts . the output of op07 is 5 volts.

 

[Sudhir.K.A]

hi garg,

now let say i send 000 on address line to select X0 line, ie. X-X0 are now connected, ideally i should be having the feedback resistor of 47K on opamp, but i dont get that, instead i get very low resistance. If i remove all other resistors except 47K on X0 i get proper resistance on opamp.

Now when you connect resistors in for other channel my guess it is coming in parallel due to some reason.

I still dont know the reason.But what u can do to conform the claim is connect some known resistance in other arm and then measure the resistance my guess would 47K parallel with new resistance.

the output of op07 is 5 volts.

Is the output always camped to 5V irrespective of input?

In your circuit variable resistance is in the feedback loop.
I think it is better to put the variable resistance in the -ve input of the opamp.

ALL THE BEST

bye

 

[polymath]

garg29
In the configuration you are using for the opamp it is usual to place a resistor from +input node to 0V,
the value is equivalent to the feedback resistor, which you have made variable.
This helps CMRR, balances input differential amp and references +input to 0V which you have floating.

I would suggest that you change the opamp to inverting configuration.

You will need to keep opamp supply within range of 4051 Vdd-Vee and Vdd-Vss limits and your µC supply Vdd-Vss.

Vdd = +5v , Vee = -5v , Vss = 0v

So opamp supply is ±5v
µC supply is +5v-0v

You could use ±7v5 but you will need to driver logic control inputs
with Open Collector or level shifting buffers.

Anything outside the Vdd-Vee Vdd-Vss limits and you will turn on the substrate diodes of the 4051 as Sudhir.K.A said or exceed output limits of µC.

regards Polymath

 

[garg29]

Thanks Mr. Sudhir and "polymath" for replying i was able to solve out the problem. I applied -5 on Vee instead of 0 and it started working.

I've cliked on helped me for both of you.

 

[happy2005]

There are integrated adjustable gain instrumentation amplifier chips, for example, MAX 4460, MAX4461 or MAX4462. I think these chips will be more robust and stable than 4051.
Regards,

 

[atferrari]

And certainly much much more expensive!!:!:

 

[kender]

There are integrated adjustable gain instrumentation amplifier chips, for example, MAX 4460, MAX4461 or MAX4462. I think these chips will be more robust and stable than 4051.

I skimmed through the datasheets and as far as I can tell, MAX4462 is the only adjustable in-amp in that family. It's adjusted with feedback resistors, which means that a digital pot or a mux (such as 4051) with ressitors will still be required to adjust the gain digitally.

 

[polymath]

Programmable Gain Amplifiers for Microchip are controlled via SPI port:
https://ww1.microchip.com/downloads/en/DeviceDoc/21861b_.pdf

Or you can try a novel approach using a Dual DAC chip as shown by Maxim
**broken link removed**

It has a transfer function of:

Vout/Vin = - 4096/12Bit DAC code ......... 12bit resolution serially programmed

data sheet:
**broken link removed**


hope you find this interesting ..... Polymath

 

[kender]

In your circuit variable resistance is in the feedback loop.
I think it is better to put the variable resistance in the -ve input of the opamp.

Out of curiosity, why the -ve input is preferred over the feedback loop for the variable resistance?

At the moment, I'm designing my own programmable gain peak detector. More discussion about it can be found here:

- Nick