Yes. Good. Now do the next steps as outlined in my previous post.Would that be 24 V - 5.6 V = 18.4 V
18.4 V / 1000 ohm = 18.4 mA
?
Yes. Good. Now do the next steps as outlined in my previous post.Would that be 24 V - 5.6 V = 18.4 V
18.4 V / 1000 ohm = 18.4 mA
?
Seldom any need to bump threads until they are a few days old. Definitely no reason to bump a thread when the more recent post isn't even three hours old.bump
Thanks a lot. It's very well explainedVery good!!!
I think you've got it down.
If the base current has a value of relevance then you need to take it into account. What constitutes a value of relevance depends on the circuit and the specs, but a general rule of thumb is that if the base current is less than 1% of the currents "adjacent" to it then it can be ignored while if it is more than 10% of those currents then it can't. Between those limits is a judgment call, so if it doubt, don't ignore it.
Once you have worked the infinite-beta case and then gone back and figured out the base currents assuming that they don't affect overall circuit operation (which is the point you are at now), you can usually get a pretty accurate accounting of the actual effects through just one more round of iteration. For instance, let's say that the base current in Q3 turned out to be 300 uA, which would be more than 10% of the 2.82 mA flowing through R4+R3, you would look at the circuit and conclude that the voltage at the junction would still be clamped at 6.2 V and so you would still have 2.82 mA in R3, but you would have 3.12 mA in R4 giving you a Uout of 16.50 V, or one volt higher. You would then follow the analysis around just like you did before and get an updated value for the Q3 base current. If the change in that base current is small relative to the currents in the adjacent resistors, then you are probably safe calling it done. If they aren't, then you go another round and keep doing that until they are. You seldom have to make a second round and if you designed the circuit well to begin with you seldom even have to make the first update round and can stop at the point you got to.
Hi,Thanks a lot. It's very well explained
I'll attach another curcuit, and do the calculations when i find time.
For now, I have other homework to do, with deadline tomorrow.
Here it is.
View attachment 101653
Input 24Vac, 50 Hz
Output 12VDC, 48W
Ripple voltage is Max. 5% of Input
The components are of my own choise.
You're okay up to this point, but then you go off the rails.In order to make this simple. Let me use the same components as before.
Q1 gain is 50
Q2 gain is 20
Q3 gain is 110
Zener diode 5.6 V
I'll start by finding current in the output.
48W / 12 V = 4 A
Q3 base voltage is calculated by adding the zener diode voltage requirement to the diode in transistor
5.6 V + 0.6 V = 6.2 V
R4 does not establish the Q3 base current. That is determined by the current in R2 (less the current needed to feed the base of Q2) and the current gain of Q3. By setting the current in R4 to such a small value, you ensure that the base current in Q3 will be a significant fraction of the current through R3 and R4 and, hence, that they will have significantly different currents in them (because a significant fraction of the current in R3 is diverted to the base of Q3 and never goes through R4).Now i pick a relative big resistor at R4 to make sure the current in Q3 base is rather small.
6.2 V / 500 µA = 12400 Ω
Adjusted to a real resistor
~ 6.2V / 12 kΩ = 516 µA
Yes. The actual Q1 emitter current will be higher due to the need to feed R3, but you can (at least initially) assume that this current will be negligible compared to the 4 A.Makes tatally sense.
To move on from here, would it be better to assume the Q1 emitter is 4 A taken from the output and use the Q1 and Q2 gains to calculate Q1 base?
4A / 20 / 50 = 0,004 mA
and after that using calcs from the Input to find R2?
Okay, so the ripple at C1, not the ripple at the output.Trying to calculate the voltage generated over the capacitor and finding the ripple at C1 is part of the task.
by Jake Hertz
by Duane Benson
by Jake Hertz