Struggling to Understand...

Hi,

I have seen this foldback circuit and would like to use it however I am struggling to understand how it works?


The rather brief writeup with it says:

In normal operation, Q1 is saturated. When load attempts to draw more than this saturation value, base current of Q1 cannot maintain saturation so voltage across unmarked resistor drops and current through Q1 drops correspondingly. When load is shorted, Q2 goes off and short-circuit current folds back to safe lower value. Choose value of unmarked resistor to ensure saturation of Q1 at load current

Click to expand...

Could someone please explain how Q1 switches on in the firstplace. I 'thought' I understood that a PNP transistor would require -0.6 volt to switch on but I cannot see how it would get this until Q2 switches on but it can't until Q1 is on :???::???::???::???:

I also do not understand why Isc is 20mA. How is that derived?

What is the saturation voltage expected to be able to calculate the unknown resistance value? I am guessing you'd use the 20mA for the current?????

Perhaps is someone could provide a better, more concise explanation of the operation of the circuit it would be greatly appreciated.

Thanks for your help.

I never saw a foldback circuit without current sensing. There are other well known and working circuits to choose if you need this function.
I am wondering that if this circuit will work at all.

It will work but it's not what I would call a good circuit, as it relies on the hFE of Q1 to set the current limit. Here's how it works:

With the output shorted, 0.2mA flows through R2 into the base of Q1. If we assume hFE = 100, then Q1's collector current = 20mA.

If the output voltage is increased, then Q2 switches on and Q2's collector current flows into Q1's base, so Q1's collector current will increase.

Higher output voltage causes higher current through Q2 into Q1's base, which causes higher output current from Q1's collector. The limit occurs when Q1 or Q2 saturates.

Could someone please explain how Q1 switches on in the firstplace. I 'thought' I understood that a PNP transistor would require -0.6 volt to switch on but I cannot see how it would get this until Q2 switches on but it can't until Q1 is on

Click to expand...

For a PNP transistor, if Base is negative, w.r.t Emitter.. it always conducts.. (A simple P-N forward biased junction).. With the resistive divider, ur enabling the base to go negative w.r.t emitter.. So, Q1 is always ON under normal conditions..

Consider a case where you short E,B without a resistive divider, then your Transistor goes OFF.

Thanks guys for your input.

At first glance I must admit I didn't think it'd work however multisim kind of indicated otherwise, however that wasn't brilliant so I could see that it might have worked in some fashion, I just didn't know how.

I can see what you mean about it not being a good circuit. Interesting circuit now that I sort of understand how it works. :-D

Basically, all I need is a variable current foldback circuit which is good for around 12v 50A, in case anyone knows anything. I was looking to understand this in order to modify it it to suit. I think i'll look at something a little easier to understand, or perhaps I say better built, for my purposes. :???:

Thanks once again for your assistance guys. Much appreciated.