peter taylor
- Joined Apr 1, 2013
- 106
Woops, I meant ' Vbe1 and Vbe2 are below about 0.6 volts', and 'I3 =[ (V0 - V3 - V4) / R1 + B.V4 / R3 ] / (1 + B + B.B) ]' should read R2, not R3.
Why do I keep doing that ?
Why do I keep doing that ?
That was my interpretation of the circuit tooAnother point of view: R1 provides the only drive current to Q1. If Q1 has enough gain, it allows current to flow through the LEDs, through Q1, and through R2 until the voltage across R2 becomes enough to turn on Q2. As Q2 begins to allow more current to ground, the supply of current from R1 becomes less available to Q1 and the circuit arrives at equilibrium.
ps, you can't get 16 volts worth of LEDs to run on 10 volts. Tear it down to one LED and the circuit should start up with a bit less than 0.018 amps of current through the LED.
If the Ice current of Q1 increases, then the voltage drop across R2 increases, which increases the Vbe of Q2, which increases the Ice of Q2, which increases the current in R1, which increases the voltage drop across R1, which lowers the base voltage on Q1, which lowers the Ice current in Q1. This forms what is known as a negative feedback loop in the circuit such that any change away from equilibrium starts a chain of events that reestablishes that equilibrium (or something very close to it).
Assume that the circuit is in equilibrium, which means that you will have Vbe of about 0.6V across R2. Size R2 to give the desired Ice you want in Q1.
Sizing R1 is a bit more complicated, but is also not nearly as critical. In this case let's say that the target current in Q1 is 20 mA, so R2=33Ω is a reasonable choice which would give about 18 mA. Assume that the transistors have a beta that is at least 100. That means that you need a base current of about 200 uA. The base voltage on Q1 is going to be about 1.2 V to 1.4 V, placing about 8.7 V across R1. Neglecting the current in Q2 for a moment, that would require R1 to be 43.5 kΩ. If the beta is actually higher, then we could use an even larger resistor. But we need to allow for some collector current in Q2, as well. So we make R1 quite a bit smaller (they chose 10 kΩ in this case). The negative feedback aspect of the circuit will simply turn Q2 on a bit more in order to sink the additional current. This has the effect of making the circuit very insensitive to variation in transistor beta.
To show how non-critical the actual choice of R1 is, if you were to use, say, 100 Ω for R1, then you would need about 87 mA of current in it. Nearly all of this would flow through Q2, which would then require a base current of about 0.9 mA. This would be supplied by the circuit increasing the drive to Q1 so that the collector current would be about 18.9 mA with 18 mA of that going through R2 (to keep the base voltage of Q2 at about 0.6 V) and the rest going to the base of Q2. With that much current in Q2, the needed Vbe would actually increase, possibly by 100 mV give or take, and this will increase the current in R2 by 10% to 20% as well. So even a radical 100x reduction in R1 will likely only increase the LED current by 25% or so.
What's the difference? I'm not sure I'm understanding the point that you are trying to make.Thats the difference between theoretical and real world.
Change of component values with age, interference pick-up.What else would upset this equilibrium ?
What's the difference? I'm not sure I'm understanding the point that you are trying to make.
There is very little math involved, to understand how it works, if you allow some reasonable assumptions, like the base current is sufficiently small, etc.wading through all of the math.
I agree.There is very little math involved
There's something wrong with your solution. How can I5 (Ic1) be equal to 75mA if I6 (Ie1 - Ib2) = 70μA ??Voltage source, V0 = 20 V
VD = 8 x voltage across LED = 8 x 2 V = 16 V
R1 = R2 = 10 kΩ
B = 400
Base-emitter Voltage = Vbe1 = Vbe2 = V3 = V4 = 0.7 V
I5 = B.I3
= 400 x 187 uA
= 75 mA
I6 = V4 / R2
= 0.7 / 10 k
= 70 uA
I'm glad V6 works out to 0.7 V, because it should be equal to V4, which is just Vbe2.
Which confirms my equations.
And these are your quiescent voltages and currents.
The equations are good if you want to know the precise working points.But no one with common sense do it this way.
R2 provides the collector current for Q1 and the base current for Q2.IR2 = (Ic1 + Ic1/β1) - Ic2/β2 = 0.07mA = 70μA
Simulators are useful if and only if you understand its limitations.I do advise you to not use a simulator
Not really (except as an academic exercise). Vbe is an unknown and is temperature dependent. Transistor beta is also unknown. So in practice you have to use approximate values and design circuits so that the values aren't critical.The equations are good if you want to know the precise working points.
Pick a reasonable number for Vbe and beta, change the temperature within a reasonable range.Vbe is an unknown and is temperature dependent. Transistor beta is also unknown.
by Jake Hertz
by Aaron Carman