How to scale down voltage to 5V? Will it effect external ckt

kender · Joined: 09 Aug 2004 Posts: 768 Location: Silicon Valley

Your circuit should work as you drew it. However, you are not showing ground explicitly.

EDIT: PLEASE DISREGARD THIS POST FOR NOW

leesing2k3 · Joined: 08 Nov 2005 Posts: 28

But will it effect what the voltmeter will measure? originally and after resistors are added into the circuit?

kender · Joined: 09 Aug 2004 Posts: 768 Location: Silicon Valley

You have clarified you drawing, and I have to retract my previous post for now. Could you plese specify where you ground is? Also, what kind of diode are you testing (silicon, zener, schottky)?

Since you already have most of your test rig, here's a way for you to obtain the answers to your question experimentally. The A/D of a PIC has a high impedance (a couple iof megaohms). You can simply add the 2K resistors without the PIC and see if you're satisfied with voltmeter readings.

leesing2k3 · Joined: 08 Nov 2005 Posts: 28

Modified

asmallri · Joined: 12 Aug 2004 Posts: 1634 Location: Perth, Australia

Your original schematic shows a voltmeter that is floating (not referenced to any system ground). The comment on the circuit "20mA clamped at 10V" is ambiguous. Does it refer to the voltage source applied to the anode of the diode under test or does it apply to some unknown circuitry connected between the cathode of the diode to some common ground?

If the cathode of the diode ALWAYS connects directly to ground then the circuit you have proposed will work however, to be safe, I would add a 4.7V Zener diode between the resistor junction and ground. This will protect the PIC in the event that the right hand resistor were to go open circuit. Another safeguard (some would use it as an alternative but it is complementary) would be to add a series resistor between the junction of the two resistors and the A/D input on the PIC.

There is a subtle difference in behaviour of the circuit wrt the original implementation and that is that in the original circuit current flow would be negligable until the forward diode volt drop appraoched 0.7v (silicon general purpose diode) or 0.2v (schottky diode) when noticable current will start to flow. With the voltage divider approach current will alway flow through the divider. This is really only an issue if you are trying to profile the tranfer characterisitics of the diode.
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kender · Joined: 09 Aug 2004 Posts: 768 Location: Silicon Valley

Thanks! From the drawing alone it seems that you are driving 20mA directly into the ground. Did you actually mean this?

Guest · Posted: Mon Feb 27, 2006 11:56 pm

leesing2k3 · Joined: 08 Nov 2005 Posts: 28

Ya, I actually do mean that. If I use smaller resistor like only 2k, this will effect the reading of the meter.
The resistor will effect the reading less if i use bigger resistors like this

kender · Joined: 09 Aug 2004 Posts: 768 Location: Silicon Valley

kender · Joined: 09 Aug 2004 Posts: 768 Location: Silicon Valley

leesing2k3 · Joined: 08 Nov 2005 Posts: 28

You mean just use a Zener diode without any resistor?

leesing2k3 · Joined: 08 Nov 2005 Posts: 28

kender · Joined: 09 Aug 2004 Posts: 768 Location: Silicon Valley

rnielsen · Joined: 23 Sep 2003 Posts: 852 Location: Utah

I believe the zener idea will work. Since the LED requires a current limiting resistor, that same resistor will limit the current through the zener in case the LED opens up. The zener will not conduct any current, in normal operation, since the voltage drop across an LED is, tipically, 1.2V. This circuit could detect if the LED opens up or even shorts out. If you want to add a little redundancy you could add two zeners, say a 3.3V and a 4.7V, into your circuit. The 3.3V zener would take the initial load in case of the LED failing. Then if the 3.3V zener were to open up the 4.7V zener would be able to protect the input and your program could watch for the different voltages and act accordingly.

One problem, though, if any of the components were to become shorted you would not be able to tell which one it was since any one of them would pull the input low and the LED would not light up. This is not very likely though but possible.

Ronald