A/D signal lowering 8.5v to 5v

[RobertD]

Hi folks, I need to lower a voltage from 8.5v to 5v for the A/D pin. I was told to do a bridge with two resistors, like a trimpot. But that would mean there is always a drain on the input. Is that correct? In order to have zero current drawn while not using the A/D I was thinking to put a single resistor between the 8.5v and the pin to reduce the voltage to 5v. Would that work? That way, when the pin in not in use, there would be zero current drawn, which is what I want, and not a bridge where there would always be a leakage to ground through the resistors.

Something like this:
**broken link removed**

 

[ericgibbs]

hi,
The first cct in your diagram is way that its normally done, a simple resistive potential divider.
I wouldnt try the second circuit, thats the resistor in series with the input.

The ADC inputs draw very small currents from the source.

Why are you concerned about the current in the divider.?

 

[RobertD]

Because the voltage is used for another reading, and if my circuit draws, it will change the voltage and skew the other reading. So I don't want any change in the 8.5v circuit. I could use a trimpot to lower the voltage, but I don't want the AD to draw when it's not being used.

 

[ericgibbs]

Because the voltage is used for another reading, and if my circuit draws, it will change the voltage and skew the other reading. So I don't want any change in the 8.5v circuit. I could use a trimpot to lower the voltage, but I don't want the AD to draw when it's not being used.

hi Robert,
In that case use a CA3140 OPA as an non inverting isolating buffer.
You could also make the gain slightly less than unity in order to drop the +8V down to +5V.
Its connected between the source and the potential divider.

The CA3140 has a VERY hi input impedance so it will not load the source.

 

[RobertD]

Thank you sir, duly noted... I'm on my way to Creatron this morning, and will look for that.

Can I ask why you wouldn't use a resistor in serie with the input to lower the voltage...? I only need to sample the voltage one microsecond, then the resistor is off line.

 

[Boomslang]

The second pic with the resistor in series will not drop the voltage, it will only limit current.... it's that stupid ohms law that always comes into play .

 

[Nigel Goodwin]

Can I ask why you wouldn't use a resistor in serie with the input to lower the voltage...?

Not only wouldn't, you couldn't - such a scheme relies on a second resistor inside your chip (so it's the same as the first anyway) - but a PIC analogue input doesn't act like a resistor.

 

[blueroomelectronics]

The A/D has a tiny sample & hold cap built in. Just a simple voltage divider is all that's needed. Keep the total resistance under about 2.5K for maximum sample rate.
(A pair of 1K resistors in series should work nicely)

If you want very low power and are just looking for a threshold trigger use the comparator mode instead.

 

[ericgibbs]

The A/D has a tiny sample & hold cap built in. Just a simple voltage divider is all that's needed. Keep the total resistance under about 2.5K for maximum sample rate.
(A pair of 1K resistors in series should work nicely)

If you want very low power and are just looking for a threshold trigger use the comparator mode instead.

Hi Bill,
The OP is concerned about loading his sensor voltage source, wants to keep the load impedance high.
So I have suggested a CA3140 HiZ unity OPA setup.

He could tag your divider on the output of the OPA.

Regards

 

[Nigel Goodwin]

What if I did something like this, use a large slow charging cap to act as a ground for a few uS...? When the AD closed the pin, the cap discharges through the 1M resistor. The size of the cap is calculated to draw 3/8th of the charge while the AD is reading, how big exactly, I'd have to calculate it, large anyway. The other cap is a decoupling cap.

Sorry, but it's a silly idea - use a simple voltage divider and a buffer opamp, as already suggested.

 

[RobertD]

The A/D has a tiny sample & hold cap built in. Just a simple voltage divider is all that's needed. Keep the total resistance under about 2.5K for maximum sample rate.
(A pair of 1K resistors in series should work nicely)

If you want very low power and are just looking for a threshold trigger use the comparator mode instead.

Hi Bill, I went to Creatron Saturday, and got a CA3140 opamp, as well as decoupling caps. I thought of using a trimpot for the voltage divider, but indeed, as Eric mentionned, I don't want to load the source.

 

[ericgibbs]

Hi Bill, I went to Creatron Saturday, and got a CA3140 opamp, as well as decoupling caps. I thought of using a trimpot for the voltage divider, but indeed, as Eric mentionned, I don't want to load the source.

Hi Robert,
As a follow up on that OPA, if you want to reach the +8V output from the OPA, you must use a supply rail of at least 2 to 3 volts more.
This applies to most common OPA, they need that overhead voltage.

So about +12Vsupply would be OK.

The divider could be a lowish resistance pair on the output of CA3140, that give a down step of +8V to +5V.
The PIC's ADC requires to 'see' a source resistance of 10K or less.
OR
you could use a high resistance divider, megohms range on the NONINV input, to give an input of +5V from +8V.
That means the max output swing would be +5V, so you could use a +9Vsupply to the OPA.

With regard to the minimum voltage output from CA3140, you can get within a few +mV of zero,
when a single rail voltage is used.

Hope you follow that.

 

[RobertD]

Yes 12v on the OPA and a low voltage divider resistance, 2k i.e. I'm thinking of using a trimpot.

I'm taking notes...

 

[ericgibbs]

Yes 12v on the OPA and a low voltage divider resistance, 2k i.e. I'm thinking of using a trimpot.

I'm taking notes...

hi,
A trim pot will be fine.

If the OPA and the 'bridge source' are powered from different psu's, I would add a 47K resistor between the source and the NI input, just in case the CA3140 gets powered down and the source isnt.

Lets now how it goes.

 

[RobertD]

Not ready to build the circuit yet, I'm trying to load the program into the chip and get that to work first. Once the chip is functional, then I'll build the circuit. That shouldn't take too long I hope, unless I run into unforeseen problems. The main problem was the A/D input, and how to connect it, and that's it right here. The rest of the circuit is powering a DC motor, off the pin, so that's no problem.