Wide input range Amp

[camerart]

Hi,

I am trying to make an Amplifier circuit, with an input that can detect signals between 1mV to 10V AC, and convert this to 0V to 5V DC for a PIC chip A/D input.

I think it is a Logarithmic amp, and will need to amplify the lower input signals more and squeeze the higher ones, while the output should be linear.

Cheers.

 

[Nigel Goodwin]

It's called AGC (Automatic Gain Control), it works to some extent, depending on your exact purpose - but a switched gain system might be of more use, under PIC control.

Exactly what are you trying to do?.

 

[camerart]

It's called AGC (Automatic Gain Control), it works to some extent, depending on your exact purpose - but a switched gain system might be of more use, under PIC control.

Exactly what are you trying to do?.

Hi Nigel,

I have just started this new thread, because the last one 'small signal to 5V output amp' got too long because of my explanations, so is it possible to simply think about the circuit?

I am detecting signals, with different set-ups, that can vary a lot. Once set-up the signals normally don't vary too much (But can do). So I need to be able to detect very low signals, this is the main problem, and not be overloaded if the signals go high.

Cheers.

 

[Nigel Goodwin]

Like I mentioned above - use a switched gain amplifier, and use the PIC to switch the gain if you like.

 

[camerart]

Like I mentioned above - use a switched gain amplifier, and use the PIC to switch the gain if you like.

Hi Nigel,

The PIC chip on the main circuit board that uses the A/D signals has no spare pins left. So can a Switched gain Amplifier use it's own PIC chip, and output the A/D signals needed on the main circuit? I am not that good at programming, so will it need clever programming?

Why do you prefer this type of program instead of an analogue circuit?

Can you show me a Circuit please?

Cheers.

 

[Nigel Goodwin]

Hi Nigel,

The PIC chip on the main circuit board that uses the A/D signals has no spare pins left. So can a Switched gain Amplifier use it's own PIC chip, and output the A/D signals needed on the main circuit? I am not that good at programming, so will it need clever programming?

No, it wouldn't need clever programming.

Why do you prefer this type of program instead of an analogue circuit?

Both are analogue - but AGC introduces lot's of noise, and causes all kinds of weird problems as the gain goes up and down, it's ALWAYS a compromise.

Can you show me a Circuit please?

Just use CMOS switches to switch in different resistor values in the opamp feedback.

 

[ericgibbs]

Hi,

I am trying to make an Amplifier circuit, with an input that can detect signals between 1mV to 10V AC, and convert this to 0V to 5V DC for a PIC chip A/D input.

I think it is a Logarithmic amp, and will need to amplify the lower input signals more and squeeze the higher ones, while the output should be linear.

Cheers.

hi,
Question, do you want the 1mV thru 10V AC to be linear over the compression range required, ie: 0mVac = 0Vdc thru 10Vac = 5Vdc...??

 

[camerart]

No, it wouldn't need clever programming.



Both are analogue - but AGC introduces lot's of noise, and causes all kinds of weird problems as the gain goes up and down, it's ALWAYS a compromise.



Just use CMOS switches to switch in different resistor values in the opamp feedback.

Ok, Nigel

I imagine that my application might suffer noise ok, so it depends on how the weird problems affect it. If the signal is fairly steady, this would be fine.

So before I embark on this type of circuit, I'll see if there are more simple AGC circuits suggested first, although it does sound interesting.

Cheers.

 

[camerart]

hi,
Question, do you want the 1mV thru 10V AC to be linear over the compression range required, ie: 0mVac = 0Vdc thru 10Vac = 5Vdc...??

Hi Eric,

Changes at the lower end 0V to a few mV need to be detected, and I think the 1V to 10V could be much coarser. So stretch the bottom end and shrink the top end. To give you an idea of input (Guesswork): 1.2.4.7.10.13.16.20.25.30.40.50.70.100.150.200.250.500mV.1.2.4.6.10V AC. Only 256 steps.

Output would be linear 0-5V DC.

Cheers.

 

[ericgibbs]

Hi Eric,

Changes at the lower end 0V to a few mV need to be detected, and I think the 1V to 10V could be much coarser. So stretch the bottom end and shrink the top end. To give you an idea of input (Guesswork): 1.2.4.7.10.13.16.20.25.30.40.50.70.100.150.200.250.500mV.1.2.4.6.10V AC. Only 256 steps.

Output would be linear 0-5V DC.

Cheers.

Ouch!

I know you have probably been asked a thousand times, but what exactly are you trying to do.

More questions:
1. What frequency range is the AC signal input.?
2. Does it pass thru 0V, ie +/-V
3. How fast do you want the system to respond to changes in freq and amplitude.?

 

[camerart]

Ouch!

I know you have probably been asked a thousand times, but what exactly are you trying to do.

More questions:
1. What frequency range is the AC signal input.?
2. Does it pass thru 0V, ie +/-V
3. How fast do you want the system to respond to changes in freq and amplitude.?

Hi Eric,

Please allow me to save you, going round in circles, be the long explanation into what I am doing, that has caused more problems than helped.

1/ 55930hZ.

2/No. minimum = zero.

3/ Say, 10times/second, but I'm sure faster is best.

Cheers.

 

[ronsimpson]

AD8310 Linear input log output
It input range is 0 to 1 volt so it needs your 0 to 10V devided down.
Output range 1 to 2.5V
input output
1V 2.5
0.1V 2.1
0.01 1.6
0.001 1.1
0 1.0
See fig 39 in data sheet.

 

[crutschow]

If you have no spare pins on the PIC, how will you know the gain setting on the AGC circuit? Or don't you care about that?

If you want linear response than I suggest considering an external 16-bit, serial output, A/D converter. That would give you a minimum (1-LSB) resolution of 76µV for 5V full scale, which will more than cover your minimum requirement of 1mV. No gain switching or AGC required.

 

[camerart]

AD8310 Linear input log output
It input range is 0 to 1 volt so it needs your 0 to 10V devided down.
Output range 1 to 2.5V
input output
1V 2.5
0.1V 2.1
0.01 1.6
0.001 1.1
0 1.0
See fig 39 in data sheet.

Hi Ron,

Thanks, I'll investigate.

Cheers.

 

[camerart]

Hi Crutschow,

I meant, It has A/D inputs for this application, then no more.

16 bit A/D converter, I'll look into it.

Cheers.

 

[camerart]

AD8310 Linear input log output
It input range is 0 to 1 volt so it needs your 0 to 10V devided down.
Output range 1 to 2.5V
input output
1V 2.5
0.1V 2.1
0.01 1.6
0.001 1.1
0 1.0
See fig 39 in data sheet.

Hi Ron,

This Chip is nearly ok, but for me is a bit expensive as I need a few, and it needs altering for it to do my job, but i haven't discounted it.

I have extended the Fig39 Graph for clarity to understand Logarithms better (I'm not very mathematical). At 10V input Logarithms would produce 3.1 Volt. It would be better, if the line was steeper so that 10V AC input gives an output of just under 5V DC. So I hope for some mathematical genius to be able to get op amps to perform in a similar way.

Cheers.

 

[crutschow]

Is that 10Vrms?

 

[camerart]

Is that 10Vrms?

Hi Crutschow,

I don't know what rms means exactly, but 10V is approximate.

 

[crutschow]

RMS is the DC power equivalent voltage of a sine wave. It's approximately 0.707 of the peak voltage (not peak-to-peak).

So is your value RMS, peak, or peak-to-peak?

 

[4pyros]

don't know what rms means exactly, but 10V is approximate.

RMS = Root Mean Square
Google it!
Andy