Hello , I'm new here at the forum but i"ve got some electronic skills. I've got some questions about charging a capacitor.
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If we are using a 4V6 zener diode , the voltage on the emittor would be around 3.9V . So the capacitor would charge directly to this voltage?
( My answer is untill 3.9V directly because there is no resistor)
2)
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So the current trought the resistor R2 = 4.6+0.7 / 470 = 11mA. The capacitor will load lineair . But my question is untill wich voltage will it load?
( my answer is untill around VCC - UR2 , because then the voltage on the collector is almoust > then the emittor , so there will flow no current to the collector)
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if i add a resistor in the collector branch , could I then manipulate the speed that the capacitor loads? But doesn"t there flow a constant current trough the resistor and the capacitor so it doenst matter?
1. The cap will charge rapidly to about 3.9 V since the transistor will be saturated. Then it will charge slowly towards 4.7 V.
2. initially the R2 current will be 10/470.
It will not be linear, it will decay exponentially towards zero.
3. The transistor is acting as a constant current source provided that the 1k pot is set to a low value. If it is higher, the transistor will saturate.
It would be better to use a pot in lieu of R2. Then the pot will determine the charging current.
1. The cap will charge rapidly to about 3.9 V since the transistor will be saturated. Then it will charge slowly towards 4.7 V.
2. initially the R2 current will be 10/470.
It will not be linear, it will decay exponentially towards zero.
3. The transistor is acting as a constant current source provided that the 1k pot is set to a low value. If it is higher, the transistor will saturate.
It would be better to use a pot in lieu of R2. Then the pot will determine the charging current.
ive tested it ,
for the first one ) Vcc = 5V , Vb = 3V , I replaced the zener and R with 2 resistors making a Vb = 3V. Rc = 10K , C charged to 2.6V and Vc = 5V. I think this is because if the Capacitor is fully charged his R is very high and there is no current , so Ube is getting smaller . So it reaches a value where it cant get any smaller and then Ie = 0 mA . Thats why Vc = 5V.
2)
Once again A voltage devider for the basevoltage . Vb = 3.3V ==> Ve = 4V . Re = 10K , C was loaded untill 4V , since Ic 5v - 4v/10k = 0.1 mA . So Uce is very small ; thats why Uce ~ 0V .
The last circuit i didnt test it out . I'll do that later.
Now I get it. You don't consider R1 D1 positions incorrect.
Well if Ib is small enough then it doesn't matter but then Ub is not
measured with respect to gnd but its V1-UD1.
Now I get it. You don't consider R1 D1 positions incorrect.
Well if Ib is small enough then it doesn't matter but then Ub is not
measured with respect to gnd but its V1-UD1.
1)
Yea is IC is getting smaller Ube will also be smaller , its just the graph of a diode. So for the 1st one , the capacitor charges untill 3.9V and then it will charge to the minimum Ic(µA) that is needed to still conduct the transistor . Sio like ljcox said it will be around 4.7 V
Voltage supply , the transistor will be saturated. If we lover the pot to 100 ohm U2 would be 1V.
1)
Yea is IC is getting smaller Ube will also be smaller , its just the graph of a diode. So for the 1st one , the capacitor charges untill 3.9V and then it will charge to the minimum Ic(µA) that is needed to still conduct the transistor . Sio like ljcox said it will be around 4.7 V See below
Voltage supply , the transistor will be saturated. If we lover the pot to 100 ohm U2 would be 1V. I don't understand this point. To what does it refer?
As Vbe decreases, Ic will decrease into the nA region and then the pA region.
It would be very difficult to measure the capacitor voltage as you would need a very high resistance voltmeter. Otherwise, the capacitor will charge to a voltage where all of the emitter current is going into the meter.