BM235 vs BM786 reading issue.

[jknightandkarr]

Ok...... VERY ODD issue...
I have the BM235 and Bm786 from the EEVBlog's store, and have come across an interesting issue. Now, both meters seam ok, my BM235 I tried to get mA reading from the volt setting by mistake, didn't seam to hurt it, and since Dave in ep #1292 connected it to that fence controller and didn't blow that up, I can't imagine 5V & less then 200 mA sending it to multimeter heaven, and when testing the volts across my load, they match the BM786 perfectly, but on my load, when doing mA the BM235 gives me a reading of about 90.1 mA @ 2.13 v and the BM786 gives me like 77.01 mA @ 2.10 V, the bench top power supply set at 2.3V @ directly off the meter I get 2.400 or 2.3965 depending on which multimeter I measure the voltage out from the power supply. Now, at various attempts the mA varies, TekPower TP1803D, said lab grade, but cheap $50 supply so figuring thats the reason for the mA on the meters being different at different times, but this don't account for why one meter giving almost 15 mA difference.  Now another odd thing is the Amp meter on the power supply, even though it is 0.00 display, is matching the meter I hook up to the load, so at this point I am TOTALLY confused. Just for a test to see what happens when I hooked them both up to measure mA at the same time, in series, and now get the exact same reading, in parallel though I get almost 20 mA difference, the BM786 being the lower reading.... ANY clue what the heck is going on here???  Thanks.

Joe

[bdunham7]

If I understand how you are connecting things, it appears that the resistance of the meters are different in the mA range.  Any meter you are testing with becomes part of the circuit.  An ammeter is a low value (usually) resistor and the two meters just have different resistance values.  It appears that the BM235 is lower resistance since it allows more current in your load.  Another way of thinking about this--the normal way actually--is that the lower resistance meter will have a lower voltage drop across it at a given current.  This is called the burden voltage.  The difference between what your PSU is showing for output voltage and what is across the load is that burden voltage in each case.

[jancumps]

If you put the 2 meters in series, they should show a comparable current.

[Fungus]

Put both meters in series and see what happens... 

If you put the 2 meters in series, they should show a comparable current.

Edit: You beat me to it!

[jknightandkarr]

Interesting, since that is the case how can I tell what the actual current draw is?  I know what I expect but its so much lower then that value.  Expect around 150 mA, getting about 75 mA +/- like 15-20mA.  Guess I can use the volt readings on the leds and the circuit and add a small known value resistor and calculate it using Ohm's law.


Thanks for the replies.

Joe

[wraper]

Interesting, since that is the case how can I tell what the actual current draw is?  I know what I expect but its so much lower then that value.  Expect around 150 mA, getting about 75 mA +/- like 15-20mA.

Both meters show the actual current. You miss the fact that current changes depending on what meter you connect. If you connect both of them in series, current will be even lower but both meters will show the same figure. There is no issue with meters, the issue is with how you conduct the experiment.

[AVGresponding]

Can you show us the circuit you are testing, and how you connected the meter? We may be able to explain the difference between what you expected and what you measured.

[jancumps]

In each of the cases, you have measured the correct current draw. See bdunham7's earlier reply for the explanation.
The DMMs have a different sense resistor value, so the current each of them draws from the supply, just by themselves, is different.
When in series, the current through DMM A has to be the same as through DMM B (law).

[Fungus]

Interesting, since that is the case how can I tell what the actual current draw is?

In a purely resistive load? You can't.

Putting a meter inline will always alter the circuit in some small way.

[jknightandkarr]

     I understand the both in series part now, just now confused why both meters give me a different value when connected one at a time... but I've attached a quick diagram of the circuit, inc where I have the meter(s) connected.  The expected draw should be 150 mA, but maybe data sheet is wrong or something, in any case, just trying to find out what the mA being used is. having transistor issue, trying to figure out why.  2N3906 is what I am using, something is damaging the transistors, making the LEDs stay on even if base is connected to power.  Power it up, ground the base, and then damage occurs to the transistors, damaged 3 so far.  Expected to maybe blow some, but this seams to happen every time I try circuit.  I can smell that lovely burned electronics smell, no smoke or visible damage,  In the circuit the circuit is drawing about half of what I expect, no issues with that, but just confused whats going on?
     I just hooked led up via 27 ohm (26.6 measured), and set power supply to 2V, then adjusted to get 2v across the LEDs, so 3V on my power supply.  Got 33 mA instead of what I expected, still VERY far below rating on the datasheet of the transistor.

[bdunham7]

Grounding the base is not an appropriate way of turning the transistor on!  Try connecting the base directly to the collector and see if the smoke stays in.

In that circuit, you could easily just measure the current by putting the meter across the 27-ohm resistor and dividing the resulting voltage by 27.  It would save you the trouble of worrying about the burden voltages of the meters.

[Fungus]

I understand the both in series part now, just now confused why both meters give me a different value when connected one at a time...

Would you understand the difference between a 100 Ohm and a 200 Ohm resistor if you added one to your circuit?

It's the same with the meters. They have different resistances.

[AVGresponding]

Also bear in mind that a bipolar transistor is a current amplifier; you should normally (there are always exceptions) expect to see a larger value base resistor than the one on the emitter or collector. You could use a 470Ω base resistor in your circuit and it would reduce the chance of the magic smoke escaping.

Also as bdunham7 says, you're better off measuring the voltage drop across your resistor and calculating your current from that in this circuit. Ohm's Law ftw!

[jknightandkarr]

Grounding the base is not an appropriate way of turning the transistor on!  Try connecting the base directly to the collector and see if the smoke stays in.

In that circuit, you could easily just measure the current by putting the meter across the 27-ohm resistor and dividing the resulting voltage by 27.  It would save you the trouble of worrying about the burden voltages of the meters.

     I will try that, thanks for the tip.  You'd think in my DeVry classes, this info would been covered in basic electronics class but it wasn't...  I will add a resistor to the base pin to test a microcontroller on, this apperently is what I did wrong.  Bad time to forget things... bad holidays... lost my mother to cancer and stroke... 

Would you understand the difference between a 100 Ohm and a 200 Ohm resistor if you added one to your circuit?

It's the same with the meters. They have different resistances.

This is information, that I wasn't aware of.  I figured both would have had the same reading when checking, but I guess not.  Live and learn, so thanks for the info.  I REALLY wish my recent degree would have covered stuff like this.... Multisim isn't ALWAYS the best option sometimes....

Also bear in mind that a bipolar transistor is a current amplifier; you should normally (there are always exceptions) expect to see a larger value base resistor than the one on the emitter or collector. You could use a 470Ω base resistor in your circuit and it would reduce the chance of the magic smoke escaping.

Also as bdunham7 says, you're better off measuring the voltage drop across your resistor and calculating your current from that in this circuit. Ohm's Law ftw!

I will try that.  Thank you, and everyone for the help.

[bdunham7]

     I will try that, thanks for the tip.  You'd think in my DeVry classes, this info would been covered in basic electronics class but it wasn't...  I will add a resistor to the base pin to test a microcontroller on, this apperently is what I did wrong.  Bad time to forget things... bad holidays... lost my mother to cancer and stroke... 

I'm sorry to hear all that.  As for what you did and didn't learn in your school, there are plenty of people here who will help if you post fully developed questions.  That usually involves a schematic and so forth.  As for the burned transistors, remember that although a transistor is more complex than two back-to-back diodes, there are effectively two diode junctions in there.  When you short the base to ground, you are providing a short circuit through the E-B diode path.  If you connect B to C, the current goes through the load, not a short-circuit path--and of course the transistor turns on as well.  If you can't do that because your control signal is to ground, a proper resistor value--enough current to turn it on fully but not so much as to melt it--to ground will also work.  So what value resistor will you use? 

[jknightandkarr]

I'm sorry to hear all that.  As for what you did and didn't learn in your school, there are plenty of people here who will help if you post fully developed questions.  That usually involves a schematic and so forth.  As for the burned transistors, remember that although a transistor is more complex than two back-to-back diodes, there are effectively two diode junctions in there.  When you short the base to ground, you are providing a short circuit through the E-B diode path.  If you connect B to C, the current goes through the load, not a short-circuit path--and of course the transistor turns on as well.  If you can't do that because your control signal is to ground, a proper resistor value--enough current to turn it on fully but not so much as to melt it--to ground will also work.  So what value resistor will you use? 

Thanks for the sympathies. It helps out.  As for the resistor, since I am too tired to go over my school books at the moment, so I went over my "What is a microcontroller?" text from Parallax Inc., and for its NPN 2N3904 counter part, it says using 2-100k ohm resistors in parallel to get 50K ohms, so I will start there since the microcontroller in the text is a Basic Stamp 2, so another similar voltage circuit, just different transistor.

[Brumby]

The simple fact is that ANY meter inserted into a circuit is, essentially, a resistor.

For voltage measurements, this is easy.   The resistance needs to be as high as possible to minimise the loading of the circuit and affecting the reading.  With modern DMMs, this is typically in the order of 10M - which is quite high enough for the vast majority of use cases to be considered pretty much ideal.  There are, of course, some circuits where even 10M would be a problem, but that is something of which you need to be aware.

For current measurements, this is more challenging.  Ideally you would want a resistance of zero - but that is not practical (for the most part).  You need a voltage for the meter to measure and this is done by passing the current through a known resistor - known as a current sense or current shunt resistor - in order to create that voltage.  This voltage is known as the "burden voltage" - and the burden voltage is one of the greatest annoyances when measuring current.  This is especially annoying when you use different meters that have different burden voltage specifications - you get exactly the result you have discovered.  As far as your circuit is concerned, using meter X is like inserting a resistor of one value and using meter Y is like inserting a resistor of a different value.  Your circuit will respond differently - as you have experienced.

This exact issue is why Dave went that "extra mile" with the 121GW and had circuitry included to lower the burden voltage - thus minimising the problem (but not eliminating it).


Another option is to use a clamp meter that can detect DC current.  All you need to do is gain access for the clamp around one of the current carrying wires.  These are not all that common, but aren't too hard to find.  You just have to make sure you don't go for AC only clamp meters (which are everywhere and can be really cheap.)  The UT210E is an inexpensive unit that I and a fair number of other members here have in their kit.  Since these meters are sensitive to all magnetic fields - including that of the Earth - it does require a little bit of learning to use for consistent results, but you will soon get the feel of it.  I find the UT210E very effective, especially if you don't need super accurate figures.

[jknightandkarr]

The simple fact is that ANY meter inserted into a circuit is, essentially, a resistor.

For voltage measurements, this is easy.   The resistance needs to be as high as possible to minimise the loading of the circuit and affecting the reading.  With modern DMMs, this is typically in the order of 10M - which is quite high enough for the vast majority of use cases to be considered pretty much ideal.  There are, of course, some circuits where even 10M would be a problem, but that is something of which you need to be aware.

For current measurements, this is more challenging.  Ideally you would want a resistance of zero - but that is not practical (for the most part).  You need a voltage for the meter to measure and this is done by passing the current through a known resistor - known as a current sense or current shunt resistor - in order to create that voltage.  This voltage is known as the "burden voltage" - and the burden voltage is one of the greatest annoyances when measuring current.  This is especially annoying when you use different meters that have different burden voltage specifications - you get exactly the result you have discovered.  As far as your circuit is concerned, using meter X is like inserting a resistor of one value and using meter Y is like inserting a resistor of a different value.  Your circuit will respond differently - as you have experienced.

This exact issue is why Dave went that "extra mile" with the 121GW and had circuitry included to lower the burden voltage - thus minimising the problem (but not eliminating it).


Another option is to use a clamp meter that can detect DC current.  All you need to do is gain access for the clamp around one of the current carrying wires.  These are not all that common, but aren't too hard to find.  You just have to make sure you don't go for AC only clamp meters (which are everywhere and can be really cheap.)  The UT210E is an inexpensive unit that I and a fair number of other members here have in their kit.  Since these meters are sensitive to all magnetic fields - including that of the Earth - it does require a little bit of learning to use for consistent results, but you will soon get the feel of it.  I find the UT210E very effective, especially if you don't need super accurate figures.

I've been wanting the 121GW for quite a while now, hoping that now that got more hours, I can get one next couple of checks.  Had to recover from loosing my mom, so paying for that and the resulting overdue house hold bills that resulted....   I do appreciate everyone's help.  It's things like this about the meter's current resistance, I knew a known resistance was needed, but never figured it would change just because of different meters with different resistances.

[jknightandkarr]

Well, learned some things about the amp measuring and got the solution to my transistor issue fixed at the same time, so thank you all for the help!!  I intend on trying to get the 121GW meter soon, I hope, and lucky I forgot I have a AC/DC clamp meter, just not sure where I put it right now, but I tried a 56K resistor on the base pins and LEDs was hardly on, so tried 27K and hit the sweet spot with that right off the bat!  just 1.9599v across the LED of the 2v normal and 2.3v max, so next the red side to do, and a circuit that switches between the 2 when desired. I got a relay idea and one that uses a PNP and NPN to connect both base pins at the same time to go red when power on one state and yellow in the other state.  just gotta power all leds and measure current on each side to see which is the better option, but THANKS AGAIN!