Relatively simple Kirchoff problem with transistors!

[pyrohaz]

Hi all,

Its been an absolute age since I last did Kirchoff so I'm a little rusty. I'm trying to find the equations for voltage at two points in a simple circuit. Any help is appreciated!

As a transistor is involved, I'm assuming I should use the small signal model where the transistor can be replaced for a behavioural current source (where the current is equal to Beta*ib). In my image, VBase would be driven by a resistor connected between the transistor base and a constant power supply. For now, I'll be assuming that on voltage for the base of the transistor is 0.6V meaning Ib = (Vp-0.6)/Rb.

Knowing this, I can therefore replace the transistor for a current source where the current is equal to Beta*(Vp-0.6)/Rb. The problem for me now is finding the voltage equations at point Va and Vb. Could anybody enlighten me on how to do so? For Beta, I'll be using a quoted (from the datasheet for the BC337-40) value of 250.

Thanks!

[IanB]

It's difficult to know how to answer when you present an abstract problem without a stated objective or application requirement.

However, it is clear that VB = VA/2.

It is also clear that VA = V⁺ - R1 * I_R1

Furthermore it is clear that I_R1 = I_R2 + I_C(Q1)

Can you explain where you are stuck?

[pyrohaz]

It's difficult to know how to answer when you present an abstract problem without a stated objective or application requirement.

However, it is clear that VB = VA/2.

It is also clear that VA = V⁺ - R1 * I_R1

Furthermore it is clear that I_R1 = I_R2 + I_C(Q1)

Can you explain where you are stuck?

Hi Ian,
Thank you for your fast reply! If the transistor is operating in its active region, will there not be some current sharing happening between the potential divider and the transistor, meaning that Vb = Va/2 won't hold true?

EDIT: Sorry, I didn't actually explain what I'm trying to figure out. I'm trying to find the TOn and TOff time dependency of a 555 timer by varying the amount of current sank from the control input. Va and Vb represent the upper and lower trip points (respectively).

[IanB]

If the transistor is operating in its active region, will there not be some current sharing happening between the potential divider and the transistor, meaning that Vb = Va/2 won't hold true?

Vb is always equal to Va/2 because there is a potential divider with R2 = R3. Potential dividers divide potential differences regardless of what the potential difference happens to be.

If you doubt this, try solving directly for Vb using Ohm's law.

[pyrohaz]

If the transistor is operating in its active region, will there not be some current sharing happening between the potential divider and the transistor, meaning that Vb = Va/2 won't hold true?

Vb is always equal to Va/2 because there is a potential divider with R2 = R3. Potential dividers divide potential differences regardless of what the potential difference happens to be.

If you doubt this, try solving directly for Vb using Ohm's law.

Doh! Of course! Well I feel stupid now...

Vb = Va/2
Va = (Vp - R1*Ic[Q1])/(1 + R1/(R2+R3))

After calculating on paper then simulating, this is the equation! Thank you very much for clearing this one up for me.