# Help needed in finding current

Discussion in 'General Electronics Chat' started by mguptamel, Dec 14, 2012.

1. ### mguptamel Thread Starter Member

Dec 14, 2012
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I have following circuit (diagram). I am having trouble finding current in Leg A. What I am getting confused because of diode which has voltage drop of 0.7V.

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2. ### Dodgydave AAC Fanatic!

Jun 22, 2012
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You need to provide the supply voltages across R1 and R2...

3. ### mguptamel Thread Starter Member

Dec 14, 2012
37
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10V, for example, is supply voltage.

4. ### JMac3108 Active Member

Aug 16, 2010
349
67
mguptamel,

Do you know basic circuit analysis? If so, just replace the diode with a 0.7V voltage source with the polarity arranged so its a voltage drop. Then perform normal circuit analysis and you'll get the answer. Give it a try

5. ### mguptamel Thread Starter Member

Dec 14, 2012
37
0
JMac3108

Do I have to apply KVL loop? Trying hard still unable to get an answer. Help will be really appreciated as I have been struggling a lot. Answer could be really simple; so far I unable to connect dots.

Nov 3, 2012
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7. ### #12 Expert

Nov 30, 2010
18,076
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I=(v-.7)/300 + v/300.

8. ### mguptamel Thread Starter Member

Dec 14, 2012
37
0
From simulation, I am getting 5.9mA in right branch and 12.7mA in left branch

I=(v-.7)/300 + v/300 - Doesn't seem to match with simulation result.

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9. ### JMac3108 Active Member

Aug 16, 2010
349
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Yes, you'll apply KVL around the circuit. Draw it out and you'll see two loops. Just treat the diode a 0.7V voltage source. Make sure the polarity of the 0.7V source makes it a voltage drop.

Try redrawing the circuit and give a shot at writing the KVL equations.

10. ### JMac3108 Active Member

Aug 16, 2010
349
67
Your simulation is for 5V and you stated 10V earlier.

Here is my solution at 10V. See the diagram attached.

(1) Write KVL around the loop for current I1.

10 - 200I1 - 200I2 - 100I1 = 0
300I1 + 200I2 = 10

(2) Write KVL around the loop for current I2.

10 - 200I1 - 200I2 - 0.7 - 100I2 = 0
200I1 + 300I2 = 9.3

(3) Now we have two equations in two unknowns which have to be solved simultaneously. You can do this many ways ... I was lazy and punched them into an on-line calculator. The results are,

I1 = 22.8mA
I2 = 15.8mA

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11. ### JMac3108 Active Member

Aug 16, 2010
349
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See attached simulation which matches above. Do you see now what I meant by treating the diode as a voltage source with the polarity arranged to make it a voltage drop?

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12. ### #12 Expert

Nov 30, 2010
18,076
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You're right. I didn't account for the interaction at points A and B.

13. ### mguptamel Thread Starter Member

Dec 14, 2012
37
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Thanks a lot JMac3108 for helping me out.

Is KVL only method of doing this calculation? What I mean is in parallel circuits, if diode or any voltage drop source is added, is KVL the only method?

14. ### #12 Expert

Nov 30, 2010
18,076
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I cheated. I built the circuit and measured it.

That drawing sucked! Anyway, the voltage after the 200 ohm resistor was 2.29 volts and the voltage between the diode and the 100 ohm resistor was 1.593 volts. After that, E=IR for 15.93 ma on the right and 22.9 ma on the left.

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15. ### JMac3108 Active Member

Aug 16, 2010
349
67
#12, its nice when the calculation, the simulation, and the REAL circuit all agree!

MGUPTAMEL,
You can use any circuit analysis method you want as long as you include the diode in the circuit as a voltage source and get its polarity right. In circuit analysis there is always more than one way to solve a problem. But usually one or the other method is easier. People also have their favorites that they feel comfortable with. As an exercise, you could try to solve this problem using nodal analysis. Write the current equatins using KCL and try to solve.

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16. ### mguptamel Thread Starter Member

Dec 14, 2012
37
0
Thanks all for helping me out earlier.

In attached screenshot, I have 2 example circuit. Circuit on left is with Vout where as circuit on right is Vout attached to a diode.

In left circuit, Vout (top) is 8V. As soon as I connect Vout to a diode (in right circuit), Vout has changed. The question is Vout changes depending on how I am completing the circuit. Textbooks refer to Vout a lot. If Vout is changeable, why Vout is referenced in textbooks?

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17. ### #12 Expert

Nov 30, 2010
18,076
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Something has to be used to describe the output voltage of a circuit, so it's called, "Vout".

The problem here is that the load on a circuit always affects the circuit because it becomes part of the circuit when you attach it. Under optimum conditions, the load is insignificant compared to the circuit it is attached to. Example: high input impedance opamps. In this case, you have attached a diode which characteristically passes all voltage greater than about .7 volts. Above .7504 volts (as per your post) the diode is essentially a short circuit. If you really, really, tried to get it to support 4 volts, it would melt.

18. ### mguptamel Thread Starter Member

Dec 14, 2012
37
0
The problem here is that the load on a circuit always affects the circuit
Total understandable. This is confusing me for quite a while because as soon as a load is added for low resistance, for example, voltage will drop. Now if output of 1 circuit is an input of 2nd circuit, 2nd to 3rd ... so on and so forth, wouldn't the last circuit be affected as calculations will be mostly depend on Vout unchanged starting from 1.

As you mentioned that diode is a short >= 0.7 volts. The example I presented was to get an idea. It's not real life circuit though .

One more question...my apologies for so many questions...as these questions are haunting me for quite a while.
In textbooks transistor circuits talk in term of voltages but not current. Why would that be? From voltage, for example, Vout in above example, one can never know what is output current unless ammeter is connected to it? Please correct me if I am wrong. If circuit relies on some sort of input current, Vout won't help (again correct me). An example would be transistor where beta = 100 x alpha. I am trying to understand AC circuit where Vout and Vin (AC signal) is used.

19. ### #12 Expert

Nov 30, 2010
18,076
9,678
1) yes, the last stage will be affected by errors in all the stages before it.
What you do to fix this is to design the circuit as if the load was already attached. As I have said before, most circuits are designed backwards...from the output to the input. You start with what the load is going to be and design the circuit to drive that load. In the example above, let us assume that the load is a 1k resistor and you expect to see 8 volts on it when you are done. You started out thinking that 12V through (3) 100 ohm resistors will produce 4 volts and 8 volts for outputs.

What you have to do is draw the circuit again, but with the 1k resistor attached to the alleged 8 volt output. When you do the math (Ohm's Law) you find that the alleged 8 volts is really 7.5 volts. You assume that the output is right, so it must be 8 volts. Eight volts applied to a 1k resistor causes .008 amps to flow. Meanwhile, 8 volts applied to (2) 100 ohm resistors will cause .04 amps to flow. Add the load currents to get .048 amps. You need the top resistor to allow .048 amps when 4 volts are pushing electricity through it. From Ohm's Law, R=E/I so R = 4/.048

The top resistor must be 83.333 ohms to get the output you desire.

2) Textbooks talk about voltage, but not current, because voltage and current are interdependent. It's just a simplification to speak of volts instead of amps. It could be done the other way just as easily, but doing both at the same time would actually confuse you more. You can calculate the current by knowing the voltage across a resistor, and you are expected to do that.

Now I have come to the point where I am confused as to what more you need. I will quit now and you can ask another question if you want to.

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