current through series/parallel circuit

Hi,

I want to calculate current through the circuit present in attached image.

V5,V6 are the voltage drops of diodes, used directly voltage drops just for easy calculation/representation.

I need formula or the way to calculate current through this circuit.
The parallel combination troubling me a lot.

You have to use Millman's theorem. I read about that theorem in some Electric Circuits book (Schaum's Outline Series). It is a very simple formula.

https://www.allaboutcircuits.com/vol_1/chpt_10/6.html

I think Millman's theorem wont help me.
Below is my simplified circuit in which am suppose to calculate total current when the transistor is ON.
Consider diode drop as 1V, transistor Vce as 0.7.

when transistor is ON, we see 3 current loops i.e. R2,D1,R1,U1,D2,R4 and R5,R6,U1,D2,R4 and R5,R3,D2,R4.
so how to calculate total current I1,+I2+I3 ?

The circuit in post 1 is quite easy. You can simplify it like in the pic below.

The circuit in post 3 is much more difficult. There are 2 possibilities:

• If the transistor is saturated, then we can estimate Vce = 0.1V.
• If the transistor is not saturated, then Ic = Ib * hFE.

In either case you need to write out a set of linear equations describing the relationship between the various voltages and currents in the circuit, then solve the equations. I'll have a go at that later.

Calculate current through each path. Once you find Ic and Ib then voltage across R3 is equal to Vr3 = Vce + IcR6 and current through R3 is Vr3/R3.

sachinsk said:

Below is my simplified circuit in which am suppose to calculate total current when the transistor is ON...

Click to expand...

Here's the original circuit:



Let's start by assuming that the transistor is saturated. As an approximation, let's say Vce = 0.1V, Vbe = 0.7V, and the diode voltage drops are also 0.7V.

We can replace the diodes and transistor with voltage sources like this:



Then we can simplify by combining the series resistors and voltage sources on the left to get this equivalent circuit:



Looking at that circuit, we can describe the currents in terms of the voltages like this:

1. I1 = (10 - 1.4 - Vnode2) / 8K
2. I2 = (Vnode1 - 0.1 - Vnode2) / 1K
3. I3 = (Vnode1 - Vnode2) / 1.5K
4. I2 + I3 = (10 - Vnode1) / 10K
5. I1 + I2 + I3 = (Vnode2 - 0.7) / 100K

That gives you 5 equations with 5 unknowns, which should be fairly easy to solve. After calculating the values of I1, I2, I3, Vnode1 and Vnode2, you need to check that the values are reasonable, and support the initial assumption that the transistor is saturated. e.g. I1 and I2 should both be positive, and I2/I1 should be less than the normal value of hFE.

If that doesn't work out, then you have to start over, assuming the transistor is not saturated. In that case, we don't know the value of Vce, so equn 2 above changes to "I2 = (Vnode1 - Vce - Vnode2) / 1K", giving us an extra unknown variable. But this time we know that I2 = hFE * I1, so now we have 6 equations with 6 unknowns:

1. I1 = (10 - 1.4 - Vnode2) / 8K
2. I2 = (Vnode1 - Vce - Vnode2) / 1K
3. I3 = (Vnode1 - Vnode2) / 1.5K
4. I2 + I3 = (10 - Vnode1) / 10K
5. I1 + I2 + I3 = (Vnode2 - 0.7) / 100K
6. I2 = I1 * hFE

godfreyl said:

The circuit in post 1 is quite easy. You can simplify it like in the pic below.

The circuit in post 3 is much more difficult. There are 2 possibilities:

• If the transistor is saturated, then we can estimate Vce = 0.1V.
• If the transistor is not saturated, then Ic = Ib * hFE.

In either case you need to write out a set of linear equations describing the relationship between the various voltages and currents in the circuit, then solve the equations. I'll have a go at that later.

Click to expand...

this looks interesting..
If V5,V6 are not equal, what we need to do?
I mean is there any formula if 2 resistors,Diodes/voltage sources are in parallel like in this case..

- - - Updated - - -

godfreyl said:

Here's the original circuit:



Let's start by assuming that the transistor is saturated. As an approximation, let's say Vce = 0.1V, Vbe = 0.7V, and the diode voltage drops are also 0.7V.
..

That gives you 5 equations with 5 unknowns, which should be fairly easy to solve. After calculating the values of I1, I2, I3, Vnode1 and Vnode2, you need to check that the values are reasonable, and support the initial assumption that the transistor is saturated. e.g. I1 and I2 should both be positive, and I2/I1 should be less than the normal value of hFE.

If that doesn't work out, then you have to start over, assuming the transistor is not saturated. In that case, we don't know the value of Vce, so equn 2 above changes to "I2 = (Vnode1 - Vce - Vnode2) / 1K", giving us an extra unknown variable. But this time we know that I2 = hFE * I1, so now we have 6 equations with 6 unknowns:


[/LIST]

Click to expand...

I think am wrong in this case.Now in great confusion whether the transistor is ON or OFF?
How to conclude that the transistor is ON/OFF?
As per simulation,transistor is OFF. But whats reason?

Which current is causing it to turn OFF.. Ic or Ib which are in uA..
Is this uA main reason for OFF state?

sachinsk said:

t
I think am wrong in this case.Now in great confusion whether the transistor is ON or OFF?
How to conclude that the transistor is ON/OFF?
As per simulation,transistor is OFF. But whats reason?

Click to expand...

My simulation shows the transistor is on, and is saturated.



BTW: Did you try solving the equations?
If so, did you get the right answers?