Mini roulette

[SneaKSz]

Hello guys ,

I want to make this circuit :

**broken link removed**.

But i wan't to know how it works first .

So if you press the pushbutton(S1) there will be a basiccurrent so T1 will conduct. Now is my question Ie = Ic+Ib ==> Ic = Ib*hfe , so the collectorcurrent has to be constant.So there will be a constant Collectorvoltage , so the capacitor will charge.

After simulating this circuitry i saw that the capacitor is loaded exponential . i don't get this because Ic is constant. Why doenst it load lineair?

Is that because Vc reaches a point where it cant go higher so the voltage will be constant so C will load exponentially?

Thanks

 

[crutschow]

The collector current is constant only if there is nothing else to limit the current. If the available collector current is more than the current through P1 (with T1 acting as a saturated switch), then the voltage at the collector is constant and the current is determined by the voltage drop across P1. This voltage decreases as C4 charges, thus the charge current to the capacitor is exponential.

If you look at the collector voltage in your simulation, you will see what I am saying.

 

[SneaKSz]

The collector current is constant only if there is nothing else to limit the current. If the available collector current is more than the current through P1 (with T1 acting as a saturated switch), then the voltage at the collector is constant and the current is determined by the voltage drop across P1. This voltage decreases as C4 charges, thus the charge current to the capacitor is exponential.

If you look at the collector voltage in your simulation, you will see what I am saying.

Hello thanks for your reply , I've checked the collector voltage and it is constant. But i dont get why the collector current isnt constant. You say that the collector voltage is constant if there is nothing ( Resistor? Capacitor? ) to limit the current.

I understand that the current changes exponential because Vc is constant and the voltage of the capacitor rises.

I just dont get it why the rule of Ic= Ib*Hfe doenst count here , because Ib and Hfe doenst change.


If i would swap the capacitor with a resistor then the collector current would be stable isn't it ? Because here the rule counts?
Greats

 

[crutschow]

The rule Ic=Ib*Hfe only works if there is no impedance to limit the current to less than the value of the calculated Ic. Here the collector resistor does limit the current. Calculate Ic and you'll see what I mean. It's greater than the current through the collector resistor. The transistor doesn't generate current, it just acts as a valve to control the current. The transistor can't provide more current unless you raise the emitter voltage so the collector voltage can be higher.

 

[colin55]

You don't need any of the transistors in the output. The 4017 will drive the LEDs directly:
**broken link removed**

 

[SneaKSz]

The rule Ic=Ib*Hfe only works if there is no impedance to limit the current to less than the value of the calculated Ic. Here the collector resistor does limit the current. Calculate Ic and you'll see what I mean. It's greater than the current through the collector resistor.

I never heard of this :d. I though the Rule always worked. If i woul calculate Ic i would replace the side on the left of R2 with thevenin. That gives me Rth = 1M//220 = 220 Ohm and a Vth = Vcc* 220/220+1M = 0.001 ~0V So now we have R2+ Rth ~R2 so the left side of R2 is grounded. Ib = VCC - 0.7 / R2 = 5V - 0.7 /220K = 43 µA. So if i pick ,Hfe = 150 Ic would be 6,4mA. I reall don't get what your"e trying to tell me .


Collin55 thanks for the schematic !!!
Greats

 

[colin55]

Do you want to know how the circuit works, without any of the messy mathematics? I designed this circuit 30 years ago and sold hundreds of kits.

 

[crutschow]

I never heard of this :d. I though the Rule always worked. If i woul calculate Ic i would replace the side on the left of R2 with thevenin. That gives me Rth = 1M//220 = 220 Ohm and a Vth = Vcc* 220/220+1M = 0.001 ~0V So now we have R2+ Rth ~R2 so the left side of R2 is grounded. Ib = VCC - 0.7 / R2 = 5V - 0.7 /220K = 43 µA. So if i pick ,Hfe = 150 Ic would be 6,4mA. I reall don't get what your"e trying to tell me .

The rule only applies under condtions where the transistor is not saturated. I'm saying the current through the transistor can not be greater than the voltage across the collector resistor will allow. To get 6.4mA through a 2.2kΩ resistor requires 15V which is greater than your supply voltage. Don't know how to say it any clearer than that. If you still don't understand I suggest you review Ohm's Law.

 

[SneaKSz]

yea i understand if you press the button the 555 will have a certain frequentie , so the counter will change his outputs on this frequentie. if you release the button the capacitor will uncharge so the 555 will still be supplied but i'll go slower because Uc will decrease. so you're roulette will slow down

 

[SneaKSz]

The rule only applies under condtions where the transistor is not saturated. I'm saying the current through the transistor can not be greater than the voltage across the collector resistor will allow. To get 6.4mA through a 2.2kΩ resistor requires 15V which is greater than your supply voltage. Don't know how to say it any clearer than that. If you still don't understand I suggest you review Ohm's Law.

yea I understand it . so the transistor here is saturated and it's Uce will be arround 0.2V so you have almoust the voltage supply to load the capacitor. Thanks for making that bright