Hi guys, im a new electronic student, but im stucked in the basics So would appreciate some help correcting my calculations in this curcuit. D2 is 16V S1 is connected here is my calculations Ur5 = Ucc-Uzd Ur5 = 24V-16V = 8V Ir5 = Ur5/R Ir5 = 8V/1000 = 8mA Ir5 = IB 8mA = 8mA Ue = Uzd - Ube Ue = 16V - 0.6V = 15.4V Ir6 = Ue/R6 = Ir6 = 15.4V/1000 = 15.4mA IRL = Ue/RL = 15.4/150 = 102.6 mA Ie = Ir6+ IRL Ie = 15,4mA + 102,6mA = 118mA Ic = Ie-Ib Ic = 118mA - 8mA Is there some truth in these calculations? Best, Peter
You need to annotate your diagram with the definitions of the variables you use. When you say Ur5, that is ambiguous because you are not only requiring the reader to assume that it refers to the voltage across R5, but you are requiring them to guess what the polarity of that voltage drop is. Similarly, when you talk about Ir5 you are requiring the reader to guess what direction you are defining that current to be in. Engineering is not about guessing -- make your definitions clear. This would appear (again, you are forcing people to guess) to say that all of the current that flows through R5 must then flow into the base of Q4. Does this seem reasonable?
Hello, When you say that Ir5 is Ib. How does the zener diode regulate? A zener diode will need a minimal amount of current to be able to regulate. Bertus
Thanks guys. I have subtraced the .5mA zenior current from the calcs and added my values to the picture..
What is your reasoning for concluding that the zener current is 0.5 mA? What would the zener current be if the wire going to the base of the transistor were cut?
I meant 5 mA, sorry. and i've looked it up in the datasheet. If i cut the base of the transistor then the zener would pass all the current delivered to it. And with that knowledge i have no explaination for why the current wouldn't pass more than 5mA.. (most likely because it's wrong) I cant seem to find the solution at this point
Hello, Yes, the current in R5 is 8 mA. This current is split into the zener current and the base current of the transistor. The currents will be different in the case the switch is open and closed. You will have to make calculations for both cases. What is the gain of the transistor? Have a look in the datasheet. For the calculations I would use the minimum gain. Bertus
You are on the right track and making good observations. We can explore this a bit more, but your next posts provide a better opportunity.
Yes, this is correct. And you correctly tracked your units -- bravo!! Like a normal diode, a zener diode will hold a constant voltage across it (if possible) while allowing whatever current through it that is needed to hold that voltage. So if the base of the transistor is cut off, then all of the current will go down through the zener. With the transistor in the circuit, some of it will go into the base and the rest will go through the zener. The amount that will go into the base will be dictated by the circuit to the right of the transistor combined with the current gain (beta) of the transistor.
Thanks for your patience According to the data sheet the minimum gain is 40. I assume the next stage is to calculate the currents after the emitter. Is that right? Uemitter = Uzener - Ubasis/emitter diode Uemitter = 16V - .6 = 15.4V current through the resistances would be calculated as the emitter voltage devided by the resistance values. 15.4V/1000 = 15.4mA 15.4/150 = 102.6mA total current ( current-emitter) = 118mA am i on the right track? :b
The typical gain is probably in the 100 to 300 range, but using 40 will make most of your calculations more conservative, which is usually (not always) good. Watch those units. 16 V is a voltage, while 0.6 is just a number. You can't subtract a number from a voltage. It should read 16 V - 0.6 V = 15.4 V What those units! 15.4 V / 1000 is 15.4 mV. What you meant was 15.4 V / 1000 Ω = 15.4 mA 15.4 V / 150 Ω = 102.6 mA This is only IF the switch is closed -- something that you haven't indicated one way or the other. Yes, you are. So what would the base current in the transistor be, assuming a beta of 40?
Sorry, I'll make sure to add units Assuming the gain is 40. Base current is: emitter devided by 40 118 mA / 40 = 2.95 mA And as far as i know the collecter current is approximately the same as emitter. Doesn't the base current determine if the transistor is closed or open?
Hello, When there is 2.95 mA going to the base of the transistor, how much current is going to the zener diode? Also calculate the currents for the open switch. Bertus
The remaining current from the resistance (R5) will go through the zener diode. 8 mA - 2.95 mA = 5.05 mA Im not sure how to calculate the open switch.
Hello, You already have calculated the current through the 1k resistor. This is the only one connected to the regulator when the switch is open. Bertus
Bertus, Im thick headed. I dont really have a clue how to go from here. Am i supposed to change resistance values? Everything is constants to me.
Hello, Think of it like the switch and the RL do not exist, like this: This would be the situation when the switch is open. Now calculate all currents. Bertus
Ohh.. of course Current at emitter is calculated by voltage at emitter devided by R6 15.4V / 1000 Ω = 15.4 mA Base current is emitter current devided by the Hfe of 40 15.4 mA / 40 = 0.385 mA Current through zener is current in r5 subtract with current at the base 8 mA - 0.385 mA = 7.615 mA Current at collecter is current at emitter subtract current at base 15.4 mA - 0.385mA = 15.015 mA