Application of transistor theory

[littletransistor]

Hi there. I've been studying about these theories about transistors in my university, and learning how to apply these in an example osciliator circuit.

**broken link removed**

However, I'm not sure where do the current starts flowing first - from the negative side of the battery or from the positive side instead. I can do the simple transistor calculations (like base voltage needed to saturate the transistor) but couldn't grasp the idea of how the capacitor suddenly got its polarities changed, or it's charge/discharge when it's connected.

I couldn't find the value for the minimum voltage in the NPN's base to saturate it, and then turn on the PNP - everytime I do the calculation I get a value of 1000 volts and above. Is there a way to derive current/voltage needed for the circuit? And also, how can I calculate how much is voltage stored into the capacitor when it's charging? And how about the flashing rate?

The only thing I know is calculating and the theory of saturation/cutoff. But applying these onto the real circuit is very difficult. And I couldn't get to know the flow of the current at all. Please enlighten me.

 

[RODALCO]

The current flows from the positive side of the battery. It turns via the 330k resistor the base ON of the NPN transistor.
It makes the NPN collector low and turns on the PNP transistor and hence the LED. The negative side of the 10µF will charge up. Hmmm, can't see how the NPN gets turned OFF because it is always biassed ON.

I think there are some parts missing to have this circuit working properly.

In your circuit the LED ( 1.7 + 0.6 Volts ) will have a current flow of about 168 mA through it and won't last long with the 22 R resistor. ( for a RED LED )
Increse that to a 220 R resistor to keep it under 20 mA.

 

[AllVol]

I think there are some parts missing to have this circuit working properly.

It would appear someone has attempted to alter the attached working circuit from www.circuitsonline.net

 

[Russlk]

When the PNP turns on, the 10uF cap is charged thru the base-emitter junction of the NPN. The charging current of the 10uF turns the NPN on hard but when the 10uF is charged, the current thru the 330K resistor is not enuf the keep the PNP saturated. Now, since the 10uF is charged negative at the NPN base, it starts discharging thru the 330K resistor and the base of the NPN is driven negative. Since it takes longer for the 10uF to discharge than to charge, the duty cycle (on time) of the LED is small and the average current does not exceed the 20mA rating of the LED even tho the peak current is 165mA.

 

[Hero999]

Nope, it won't work and I've tried simulating it just to make sure, it just sits there.

EDIT:
If there's a version of this circuit that work could someone please post a link to it.

And yes, I have seen this circuit before, I'm just intrigued.

 

[AllVol]

EDIT:
If there's a version of this circuit that work could someone please post a link to it.

Did you look at my post above? That circuit does work.

www.circuitsonline.net/circuits/view/12

 

[littletransistor]

Thanks for the info.

Okay - looking at the very 1st circuit I've put there on the start of the thread - I pretty much got what happened to the circuit, but I'm confused that you can charge a capacitor on the collector of the PNP, which is positive. I thought if you want to charge a capacitor, you need to connect input positive terminal to the positive terminal of the input?

And then all the charges will exit thru the positive terminal of the cap, fully saturate the NPN transistor? I read in one source that the cap is not usually charging, but changing the NPN's base into negative voltage instead.

Also, the calculations for the minimum value for the IB(min) for NPN to saturate are very complicated. Is there a special trick to calculate it? I can do only the 1st calculation, which is IB=6/(330k+22R) which is insufficient for the NPN to saturate, so the capacitor charges itself first.

 

[Hero999]

Did you look at my post above? That circuit does work.

www.circuitsonline.net/circuits/view/12

Sorry, I didn't see the attachment. I've been experiencing problems with my Internet connection recently.

 

[audioguru]

The circuit you got from "Talking Electronics" has errors:
1) The polarity of its capacitor is backwards.
2) The LED current is way too high at 180mA. The LED and the PNP transistor will blow up.
3) The base resistor for the NPN transistor has a value much too low.
4) The base of the NPN transistor doesn't have a resistor in series with the capacitor to limit the current.

The circuit should be like this one. It uses a high battery voltage of 12V so it has a diode to protect the emitter-base of the NPN transistor from a reverse voltage too high.

 

[audioguru]

In your "Talking Electronics" circuit, the 330k resistor has 6V-0.65V= 5.35V across it so its current is 16.2uA. The BC547 has a wide range of current gain but it is typically 300 so its collector current is trying to be about 4.8mA.

The 1k resistor will turn on the PNP transistor which will have a base voltage of 6V- 2V- 0.7V= 3.3V. So the 1k resistor has about 3.2V across it and will have a current of 3.2mA.
The PNP transistor also has a typical current gain of 300 so it will try to conduct 990mA which is limited to 180mA by the 22 ohm resistor.

The transistors never turn off. The circuit is wrong.

 

[Hero999]

Either way I wouldn't bother building one of these circuits, I'd use an LED with a built in flasher IC.

 

[littletransistor]

Thanks Audioguru. Meanwhile the new circuit given to me is a little bit easier, but I have to work it the another way round. The calculations are different for the circuit given - I'm sure the start of Ibase for NPN is 12v/10Mhm: and that means the current is just too little to saturate the transistor.

So the current can pass thru the PNP transistor first, and then charge up the capacitor. Then, I don't really know where should it send back the charges - must it flow thru R3, D1 and then back into the Base of the NPN transistor to saturate it?

The only thing I haven't done is getting the NPN's minimum Ibase to saturate the NPN transistor first. Going thru R1 will mean OFF for the NPN, so I'm sure the current will take another detour for this. Could it be this (assume VCE is 0 ,LED is 1.7V and forward bias for transistor is 0.7V):

Ibase(min) = (12V-0.7V-1.7V)/(470+R2)

and further calculations make it look too wrong to me. I must have missed something...

 

[audioguru]

The 10M resistor turns on the NPN transistor a little. The NPN transistor turns on the PNP transistor a little and the charging current of the capacitor adds to the base current for the NPN transistor.

When the capacitor is fully charged then the NPN transistor doesn't have enough base current from the 10M resistor so it conducts less which makes the PNP transistor conduct less which lets the capacitor help to turn off the NPN transistor.

When both transistors are off, the 10M resistor charges the capacitor to about 0.6V in reverse then the NPN transistor begins to turn on again.

 

[littletransistor]

When the capacitor is fully charged then the NPN transistor doesn't have enough base current from the 10M resistor so it conducts less which makes the PNP transistor conduct less which lets the capacitor help to turn off the NPN transistor.

I see - but is there a calculation of how the NPN's base current getting lesser? How does that capacitor charge anyway? Is it when it's fully charged, the base current will be automatically pulled down?

 

[audioguru]

The capacitor charges with current that must flow to the NPN transistor's base, increasing the tiny amount of current from the 10M resistor. When the capacitor is fully charged then the transistor's base current reduces to the amount from the 10M resistor so the transistors conduct less.

A capacitor holds a charge and cannot change its voltage instantly so it pulls down the voltage on the base of the NPN transistor which makes it turn off completely and quickly. The diode limits the negative voltage to the base of the transistor because its max allowed is only 5V.