# It=sum of all current branches...not with transistors??

Discussion in 'The Projects Forum' started by rougie, Oct 12, 2012.

1. ### rougie Thread Starter AAC Fanatic!

Dec 11, 2006
410
2
Hello,

Everyday that I play around with transistors, they never cease to amaze me.

Today I wanted to take down some current measurements, so I started measuring currents with my multi meter... here's what I found:

In the attachment below I have the following currents coming into the base circuit:

18.5ua + 8.1ua = 26ua

And I have the following currents leaving the base circuit:

18.9ua + 26ua = 44.9

Why is there more current leaving then coming in ???
Where is all this extra current coming from?

All feedback very appreciated!

Thnak you
r

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2. ### Audioguru Expert

Dec 20, 2007
10,668
1,190
Your current meter causes a voltage drop which affects the very low base-emitter voltage which affects the currents.

Instead of measuring currents, measure the base voltage and calculate the currents in the resistors using ohm's Law.

3. ### wayneh Expert

Sep 9, 2010
15,235
5,573
Or use two meters at the same time. They're cheap!

4. ### WBahn Moderator

Mar 31, 2012
23,396
7,101
Any time you measure something, you affect the quantity that you are measuring. As already noted, when measuring small currents in high impedance circuits, typical ammeters cause significant errors. You either need better measurement tools (ammeters designed specifically for this type of application or using voltmeters to make less intrusive measurements and then calculating the currrents from there) or you need to understand your mesurement technique well enough to compensate for the errors and back out the actual results that you would have measured had you been able to make an ideal measurement.

5. ### rougie Thread Starter AAC Fanatic!

Dec 11, 2006
410
2
So then, I can do this for the 33k which would be 0.7/33k = around 20ua
and i can also do the same for the 75k resistors... right?

But what if I wanted to know the current entering the base????
if we don't know the resistance of the Be junction, then we
are missing the ohmic value!!! so to get the actual ib, instead
of measuring it I would just assume that ib = ic(beta) right?

it's funny though how the measured base current is so much
higher than it should be due to my ohm meter's internal
resistance. 26ua that's way off... and also one more thing,
shouldn't the effect of the internal resistance in my amp
meter make my ib much smaller instead of bigger???

thanks Audioguru !!!!
thanks all
r

6. ### WBahn Moderator

Mar 31, 2012
23,396
7,101
Don't make the mistake of assuming that the current your meter is reading is the correct value. You have to analyze the circuit to determine the effect of having your meter in the circuit and to determine what the current would probably be if the meter weren't there.

As for determining how much current is entering the base, you can either do it by carefully measuring the current in series with the base ('carefully' means taking into account the disruptions) or measuring all of the other currents carefully and using KCL to determine the base current.

7. ### Audioguru Expert

Dec 20, 2007
10,668
1,190
No.
You must measure the base-emitter voltage, not guess it. The datasheet shows a typical base-emitter voltage of 0.62V at a collector current of 1mA but every transistor is different. Maybe the current in the 33k resistor is 18.8uA.

Yes, if you accurately measure the resistances and voltages.

The add and subtract the accurate currents.

It sets its own voltage, not resistance.

But you do not know the beta that is different for every transistor and it changes with current and with temperature.

Your meter messed up the circuit.

It messes up the entire circuit. Measure the resistances and voltages accurately then calculate accurate currents.

8. ### rougie Thread Starter AAC Fanatic!

Dec 11, 2006
410
2
Audioguru, I apologize because I assigned the wrong feedback resistor value. So here it is with the new values... see attachment below.

I measured evrey ohmic value and every voltage across every resistor.
Please view if my calculation is correct... Ib seems to be 10ua... please confirm.

Another question is why doesn't VR2 + VR3 = 1.110 VDC ???? KVL no?

PS. Note t1 is a pn2222.

OOOOPS I made an error in the attachment... VR2 = 0.331VDC (VR2 voltage varies so much!!! )

So my currents now become:
IR1 = 18.787
IR2 = 0.331/62500 = 5,296ua <***
IR3 = 19.41 ua

Calculated Ib now become:
Ib = (18.7 + 5.29)-19.4 = 4.59ua

Beta now seems to be 239 ?? is this possible??
HFE = 1.0977ma/4.59ua = 239!!!

r

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Last edited: Oct 14, 2012
9. ### wayneh Expert

Sep 9, 2010
15,235
5,573
You have a second voltage source at the base, so the currents are not equal on both sides and don't have to add up to Vc.

Anyway, looks like you've got a current gain of 100. Quite reasonable. Your analysis looks fine to me, although I haven't double checked anything.

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