Voltage problem

Cenlasoft

Hello,
I have been trying to divide a voltage from an LC circuit. The attached circuit shows a working circuit. At test point T2, It measure 5 V AC. My question is I am want the diode to rectify the AC into DC so I can send it to a ADC0831. If I put the diode before the 10 M resistor it does, but my voltage is almost zero at the divider junction. Any suggestions would be appreciated.
Thank you,
Cenlasoft

Philldapill

I'd say use an opamp to buffer the voltage since you'll probably have loading issues otherwise. A voltage divider works by sinking current from what you are trying to "divide". If you are using 11M ohms, total, you will have a very tiny current to begin with. Using something like a multimeter or the ADC0831, you may be loading this divider. After all, these devices need to sink a little bit of current to measure a voltage.

An opamp at the initial stage should allow you to source more current to a voltage divider so that you can measure it.

Leon

You won't get any output with that voltage divider!

I'd use a parallel resonant circuit with the diode connected to a tap on the inductor, or a link winding, to prevent it loading the circuit too much.

Leon

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Peter Jakacki

The diode is not at all suitable as it has an inherent voltage drop which also varies with temperature and device etc. At first glance I would probably use an opamp to buffer the signal and bias it to around half supply. An opamp is a good choice because your ADC requires a fairly low imput impedance anyway and you can't get accurate results with a 10M resistor.

*Peter*

P.S. darn, I see some other replies now, same thing, that's what happens when work interrupts you!

Cenlasoft

Thanks for all the help. It seems that I'll have loading problems at each step. I going to try to find some way of measuring AC volts instead of converting to DC and then measuring. I want to use a coil as a sensor with the propeller. I want the coupling with metals and I want to measure the voltage drop and show it digitally. I was thinking of using ViewPort and its digital scope with XY?
Thanks,
Cenlasoft

kwinn

Try replacing R3 with a small capacitor first, and make sure you are using a signal diode for D1. The capacitor value depends on how fast a response you need. Start with 0.01uF.

Cenlasoft

Thanks, I'll try that and then I'll post the results.
Cenlasoft

Cenlasoft

Hello,

I tried replacing R3 with various small caps connected to ground. No change. I going to have take a different approach. Maybe I can analyze the AC signal instead of converting to DC.

Thanks,

Cenlasoft

Tracy Allen

I think I'd try this. Make R2 = 0 Ohms instead of 10 MOhms. And make R3 at least 10 Mohms in parallel with a small cap like 0.01 uF or 0.001 uF. And D1 a signal diode like 1N4148, not a rectifier diode.

The loading at T2 lookiing into the ADC0831 should be a very high resistance.

Why? As the voltage on the capacitor charges up to the peak level of the sine wave, the diode impedance increases exponentially and ceases to load the resonant circuit, except for the leakage through R3 and the input impedance of the ADC0831. The forward voltage drop of a 1N4148 at 0.5 microAmp average current at equilibrium would be around 0.3 volt. The response time will be set roughly by the parallel combination of R3 and its parallel capacitor. Say 10 MOhm || 0.01 uF => 0.1 second. That circuit should not substantially load down the resonant circuit, except briefly when the input level makes a step increase.

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Tracy Allen
www.emesystems.com

Cenlasoft

Hello,
Thanks Tracy. I'll try that tonight. Your explainations are great. My students will be happy if it works.
Cenlasoft