Charging a capacitor

[SneaKSz]

Hello , I'm new here at the forum but i"ve got some electronic skills. I've got some questions about charging a capacitor.

1)
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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)


3)
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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?


Thanks for the support

 

[ljcox]

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.

And you should transpose R1 and D1

 

[vlad777]

3) You should add that var resistor in the emitter branch providing PNP and transposed R1 and D1 .

 

[SneaKSz]

3) You should add that var resistor in the emitter branch providing PNP and transposed R1 and D1 .

Thanks for the mark

 

[SneaKSz]

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.

And you should transpose R1 and D1

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.


Does there explanationsmake any sence?

 

[vlad777]

1)
(Ub is getting smaller because UC2 is getting bigger.) Edit: sorry this line is not correct.

Ube is always 0.7V .
When UC2 rises to UD1-0.7 base current can not flow anymore so the charging stops.

Ub=UD1=Ube+UC2

UC2=UD1-Ube=3V-0.7V=2.3V (or maybe 2.6V)

 

[vlad777]

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.

2) UR2=V1-UD1-0.7 (0.7 is Vbe)

 

[SneaKSz]

1)
(Ub is getting smaller because UC2 is getting bigger.) Edit: sorry this line is not correct.

Ube is always 0.7V .
When UC2 rises to UD1-0.7 base current can not flow anymore so the charging stops.

Ub=UD1=Ube+UC2

UC2=UD1-Ube=3V-0.7V=2.3V (or maybe 2.6V)

I just think its weird that Ube = 0.4V , the C is fully charged and there will flow no current through it thats why URc = 0V and Vc = Vcc.

Thanks

 

[SneaKSz]

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.

2) UR2=V1-UD1-0.7 (0.7 is Vbe)

Yes this is correct so the C will load lineair?

 

[vlad777]

I think it will charge linear until UC2 >= V1-UR1 .
After that V1 can't provide enough voltage for constant current.
Or something like that.

 

[SneaKSz]

I think it will charge linear until UC2 >= V1-UR1 .
After that V1 can't provide enough voltage for constant current.
Or something like that.

Yes I think that also

 

[vlad777]

I just think its weird that Ube = 0.4V , the C is fully charged and there will flow no current through it thats why URc = 0V and Vc = Vcc.

Thanks

Are you using simulator or measuring?
I also noticed V1 has 1Vac ; is that intentional, do you expect something from it?

 

[SneaKSz]

Are you using simulator or measuring?
I also noticed V1 has 1Vac ; is that intentional, do you expect something from it?

Ive tested it on my test board.

greets

 

[ljcox]

1)
(Ub is getting smaller because UC2 is getting bigger.) Edit: sorry this line is not correct.

Ube is always 0.7V . This is not true.
When UC2 rises to UD1-0.7 base current can not flow anymore so the charging stops.

Ub=UD1=Ube+UC2

UC2=UD1-Ube=3V-0.7V=2.3V (or maybe 2.6V)

The simulation & resultant graphs show how Ic varies with Vbe.

Graph 2 is with an expanded scale

Graph 3 is a further expansion.

The theory of bipolar transistors shows that the collector current is an exponential function of the base - emitter voltage.

 

[SneaKSz]

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.

 

[ljcox]

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.