Sawtooth generator with PNP transistor problem

[kingh12]

Hi,
I've been lost in a big trouble with the capacitor charging. First let me summarize you how does this circuit works. When the positive square wave signal reaches pnp transistor, the transistor becomes reversed biased and it doesn't work. By time time the transistor doesnt work, the capacitor starts to charge through DC power supply until negative square wave signal comes. When the negative signal comes, transistor starts to conduct and the voltage charged on the capacitor discharges rapidly. So, we get the sawtooth signal on output. This is the purpose of this circuit.

My first question is the capacitor charges by the DC power +12V? I mean when we used the formula Vc(t)=V(1-e^-t/RC) , the symbol "V" is my DC power +12V? Or the emitter voltage Ve? Or the voltage on 15k ohm resistance?

If the answer is 12V, when I used the formula for 47nF capacitor, I found that the capacitor must be charged until 6,02V. But when I praticcally observe this situation on ossiloscope the signal on the capacitor reaches rapidly to 4V and stays on 4V until the transistor starts to conduct.This means capacitor reaches its high voltage which is 4.
How can this happen? If it will become to its high voltage doesnt it have to be 12 V?
Here is the circuit of my project on attachment

I will be very glad if you solve my problem. Thanks

 

[audioguru]

The datasheet for the Extremely Old BC161 transistor shows an absolute maximum allowed reverse Veb voltage of only 5V but you are feeding it 12V.
You are damaging it.

Use a PUT (programmable unijunction transistor) instead because it is designed to make a sawtooth waveform.

 

[throbscottle]

Try a transistor with more gain, ideally one that is designated as a switching transistor. The one you have is an audio amp, so it's not really designed to turn on and off sharply. It's possible that the one you have isn't turning fully off. Also, perhaps you need a higher signal voltage.

I'm sure other people can help improve your design, far better than I can anyway.

 

[Nigel Goodwin]

The datasheet for the [/b]Extremely Old[/b] BC161 transistor shows an absolute maximum allowed reverse Veb voltage of only 5V but you are feeding it 12V.
You are damaging it.

It's not reversed, it's biased normally and with a huge 15K resistor in the emitter.

 

[kingh12]

Thanks throbscottle, this may be the reason why I cannot get the expected values on the circuit. I will research about it.
So am I using the formula correctly with "V"=12V ? What are your thoughts about it?

 

[throbscottle]

Looking at it again, I've just realised something. As the capacitor charges, the voltage across it will turn on the transistor. This is why the voltage isn't going over 4V - and I think it's only getting that high because of the large value of bias resistor. So you need a much bigger signal in order to keep the transistor turned off.

My memory of electronics formulae is far too rusty so I'll trust that yours is correct. The charging circuit for the capacitor is your 15k resistor in series with C1 or C3 (or the two in parallel, if your switches will work that way) with 12V across the whole thing - so I suppose the V in your equation is 12V.

 

[kingh12]

Yes! I didn't look at that situation. When I found 6.02V capacitor voltage from the formula, this means it is greater than my input pulse signal. So the transistor must be conducted. That is why I get 4V on capacitor. The 6-4=2V is passes through the transistor. Am I right?
So there will be another question. Today I went to the laboratory to try that. I applied higher input signal for example 6V peak-to-peak . I got my desired capacitor voltage on 6V. But when I apply 8V peak-to-peak my capacitor charges until 8V. Actually doesn't it have to be stay in 6V? Because the input signal is greater than my capacitor measured charged voltage (6.02V) and the transistor has to be stayed on reversed bias... I am missing something like 47nF capacitor charges until the input signal voltage level anyway

 

[alec_t]

R2 biases the transistor into conduction. The time constant of R2C2 is 10 secs, so your circuit takes considerably longer than that to stabilise after power-up. The stable operating point will depend greatly on the gain of the particular transistor used.

Edit:
Try connecting 75k (or thereabouts) between transistor base and the 12V rail to bias the transistor appropriately.

 

[schmitt trigger]

I agree with AudioGuru, if you want to build a single transistor sawthooth generator, use a PUT or an UJT.
It can be free running or synchronized easily.