What's wrong with my DMM???

[miqbal]

Recently I built the Rocket (that's not working well, squealing, but sound great).


Take a look at the schematic above. The buffer section (Q1) is biased by a voltage divider, 1Meg and 470k. According to voltage divider formula, if we use  a 9Volt power source (i.e., Vcc=9Volt), the voltage at the junction should be at 9V*1M/(1M+0.47M) or about 6.12 Volt.

I simulate this voltage divider on a breadboard, no transistor, no caps, just 1Meg and 470k resistor and an 8.65 Volt battery.
STRANGE,.... I always get a reading of 4.31 Volt at the junction, I've tried a few different resistor (same value, different rating), the result is the same. Another weird thing, if I measure the voltage drop at 470k resistor (i.e., voltage between junction and Vcc), it read only 2.3 Volt.
If we sum all the reading, it would be 4.31+2.3=6.6 Volt. According to Kirchoff's formula, the correct value for the summation should be 8.65V, same with Vcc. Where did the other 2.05Volt go (8.65-6.6=2.05V)

So, what's wrong with my DMM? Well, I know it is a cheap one.


IQB
 

[Sock Puppet]

Hi,
Because the circuit is high impedance, the loading effect of your multimeter needs to be taken into account.
This is often in the range of 20k ohm per Volt so if your meter is on a 20V range then the parallel impedance of the meter is 20x20k = 400k.
Selecting a higher voltage range of say 200V will provide less loading (4M) but is still considerable in this case.

S.

[gez]

The way to take accurate low-voltage readings with a DMM is to wire up a FET-input op-amp as a follower (wire from output to -input) and connect its +input to the point on the circuit that you wish to measure.  Use the same power rails as your circuit and take readings with your meter from the output of the op-amp.  I usually wire up the circuit on a breadboard and use croc-clips to connect everything.

PS  bear in mind the limitations of op-amps re input/output swing.  If you use an amp from the ICL76*1 range (replace * with 1 for single; 2 for dual etc) both inputs can be pulled to the rails and it has rail-to-rail output swing.

[George Giblet]

When you have a voltage divider you *assume* the voltage there is what you usually expect from the divider equation.   The assumption for this based on the fact the divider isn't loaded down by something, say the multimeter, or, other parts of the circuit.

When you see funny results you need to check that your assumptions are OK.

If the divider is unloaded you would expect 8.65 * 1M / (470k + 1M) = 5.88V

Step 1:  Is it the multimeter loading it down?

Most digital multimeters are 10MEG ohm input impedance so 1M//10M = 909k; you need to find out what you multimeter really is.
Then, V = 8.65 * 909 / (909 + 470) = 5.7V

So the multimeter has a small impact but it's not the answer.

Step 2:  Is the circuit doing this?

The divide feeds a transistor.  The transistor has base current, usually the base current is small enough to ignore but is it valid here?

The quick but imprecise way to check is.  Assume the voltage at the voltage divider is the unloaded voltage, then calulate an *estimate* for the base current.  The voltage at the emitter will be about 5.7V - 0.6 = 5.1V, so the current through the 10k emitter resistor is about 5.1/10k = 510uA.  Suppose the transistor gain is 200 then the base current is about 510uA / 200 = 2.55uA.

Now, the current through the 1M resistor in parallel with the 10Meg meter resistance is about 5.7 / 909k = 6.3uA.

So you have an extra 2.55uA going into the base and there's only 6.5uA going down the divider  so immediately you know that the base current is not insignificant in comparison to the bias circuit current - that appears to be the answer.

You can calculate a rough estimate for the expected voltage.   The 470k will be carrying about 2.55uA + 6.5uA = 9.05uA, so the voltage drop across it will be 470k * 9.05uA = 4.25V.  The voltage at the divide is then estimated to be 8.65 - 4.25 = 4.4V.    That's getting into the order of what you measured and it's likely to be the answer.  The only thing is I'd expect the voltage to be a little higher than 4.4V.

The way I've calculated the voltages is here not exact is only a quick/rough way to see if the base current is a feasible explanation.  You can of course calculate an exact solution but the math is messier - it is easier to just plug the circuit into a circuit simulator and get it to calculate the voltage!

A second way you can make an estimate is to simply say the transistor multiplies the 10k emitter resistor by the gain, so 200*10k = 2MEG.  Since 2M is within the same order as 1M you would expect the to see loading of the divider voltage.  If you plug 2M//1M//10M = 625k into the divider you get V = 8.65 * 625 / (625 + 470) = 4.9V, again this is a rough calculation but the calculation is good enough to make you suspicious the base current is the cause.

You can further confirm this by pulling the base leg of and measuring the voltage you should get the meter loaded answer of 5.7V.

[gez]

So the multimeter has a small impact but it's not the answer.

Step 2:  Is the circuit doing this?

I simulate this voltage divider on a breadboard, no transistor, no caps, just 1Meg and 470k resistor and an 8.65 Volt battery.
STRANGE,.... I always get a reading of 4.31 Volt at the junction

It's not the circuit, as he isolated the divider.

Cheap DMMs can have relatively low impedance (definitely not 10M!), as I well know (hence the fix I outlined).  An alternative would be to measure the voltage at the emitter of Q1, then add .4V for Ge transistors and about .6V for Si.  That would get you pretty close to what the voltage is at the base.

[miqbal]

Hi all, thank you very much for great replies and analysis.

I have tried another way to see if the circuit (Step:2) is responsible in this situation, please correct me if I'm wrong.

I change the 1M and 470k to 100k and 47k. All are in circuit (with transistor and caps). Voila..., I got 5.8V at the junction. Looks like my DMM is really loading the original circuit. I think Step:1 still applied, as Mr.Gerry and Sock Puppet mention before.

Based on my previous voltage reading (4.3V at junction), I do some calculations for the total resistance between 1Meg and my DMM in parallel. If the reading is 4.31V, battery voltage is 8.65 V, and total resistance is designated as RT, than 4.31=(8.65*RT)/(RT+470k). If I solve this equation for RT, than I got 470k for the total resistance (1Meg in parallel with DMM). If we further solve this result for DMM resistance only (using parallel resistor formula), I got 887k Ohm!!. Pretty low input impedance

If I measured the voltage drop at 470k with this DMM (i.e.,DMM in parallel with 470k), the total resistance for DMM+470k would be 307.2k. And the reading should be (8.65*307.2)k/(307.2k+1M)=2.03V, quite close to my previous reading at 470k resistor(2.3V)!!

Again, please correct me if I've done something wrong.

Bytheway, here is my DMM:
http://www.constantinstruments.com/Pages/50.htm

I can't find any information about its input impedance. The price is so low, its only Rp 80,000 (Indonesian rupiahs; my country), or about $8.9. For such price, I should only allowed be for a very low input impedance DMM

Looks like I got as much as I paid for

The most important things from you all guys: I know that my pcb trace is OK, and I know I should be really really more careful in using this multimeter. Considering a more expensive one...

IQB

[miqbal]

The way to take accurate low-voltage readings with a DMM is to wire up a FET-input op-amp as a follower (wire from output to -input) and connect its +input to the point on the circuit that you wish to measure.  Use the same power rails as your circuit and take readings with your meter from the output of the op-amp.  I usually wire up the circuit on a breadboard and use croc-clips to connect everything.

Mr.Gerry,

Do you mean like this:




1. Is is correct?
1. Are there any additional components that should be used to run the circuit?
2. As you can see at schematic, I've used an external power supply for the OPAMP+DMM circuit. Since 9v - 18V voltage is  very common on effect pedal, can I use the same power source for the OPAMP+DMM circuitry?

Thank you very much in advance.

IQB

[gez]

Yes, that's right, Muhammad.  I tend to wire up the circuit on a breadboard whenever I need it and use the effect's power supply, though I only apply power (including to the effect) once all connections are made.  Also bear in mind that many op-amps don't like having their inputs being pulled close to the power rails (not an issue in this case).

[miqbal]

...........  An alternative would be to measure the voltage at the emitter of Q1, then add .4V for Ge transistors and about .6V for Si.  That would get you pretty close to what the voltage is at the base.

Mr. Gerry, you got it again!! My voltage reading at the emitter is 4.97V. For Si transistor, the actual voltage should be near to 4.97+0.6=5.57V. This gives me the explanation why I always get signal/sound at the emitter, although my base voltage is lower than the emitter.

Haven't try with the higher impedance circuit above, but soon I will.

Thank you all guys, once again, for a lot of knowledge and great tips.

IQB