20-30V input 12v high current output regulator, not working

[Andrew Mc]

Im more of a digital electronic person, this anolgue stuff is a bit confusing.

im trying to figure out how this works.

i have taken this circuit from the LM317 data sheet and just deleted/changed some of the details, i tried to make this work but i dont think the transistors are switching on, as the 78L12ACZ was starting to get warm.


in the end i want to have: Vin = 20-30V, Vout = 12V, to supply a maximum of 1A of Current

The 78L12ACZ has a maximum output current of 100mA.
The BC556 (PNP) has a max current of 100mA (Ic) and a gain of 800
The TIP120 has a max current of 5A and a gain of 1000.


before using 20-30v, i tried this on a 18V battery supply, the voltage regulator started to get warm.

what values i chose and how a calculated the values.

Ra resistor was 70ohms ( wanted the 78L12ACZ to have a maxium current flow of around 10mA, the voltage drop of across the resistor would be 0.7v)
Rb resistor was 13.8Mohms ( base of BC556 restricted to 1.25uV with a voltage drop of 17.3V to give a gain of 800, 1mA)
The 500ohm, i wasnt sure about, so i left it as it is.

Im not 100% sure on the full working of this circuit, my understanding is:

the Ra resistor will delevop a voltage drop of 0.7V when the 78L12ACZ starts to draw 10mA, this will give the base of the
BC556 the 0.7v drop and so Rb will have the rest of the voltage (17.3V), to restrict the current to 1.25uA the 13.8M resistor is used (17.3/1.25uA = 13.8M), As the base current comes out of a PNP transistor. The BC556 will amplify the 1.25uA to 1mA (gain of 800) and so the TIP 120 will amplify the 1mA to 1A (gain of 1000).
typing this out i have neglected the 500ohm resistor, if 1mA is flowing then, 1mA * 500ohm = 0.5V, not enought for the TIP120, is this my only mistake?

can some one point out any mistakes i have made and help me to understand how to get circuit working correcly.

 

[Cirkit]

Do you have to specifically use a linear regulator in your design? Using a switching buck-converter would be more efficient. There are relatively cheap modules you can just buy and use:

https://www.amazon.co.uk/Gaoxing-Tech-LM2596HVS-Adjustable-Converter/dp/B06W9K7GK7

 

[Audioguru]

A BC556 has a current gain of from 110 to 800. Only some might have a gain of 800. A BC556C has a current gain of from 420 to 800.

The regulator warms with only (18V - 12V) x 10mA= 60mW in your circuit but it is allowed to be fairly hot at 600mW. Why not allow it to conduct 30mA? Change R1 to be 0.7V/30mA= 23.3 ohms (use 22 ohms).

The BC556 with a current of 10mA heats with only (18V - 13V) x 10mA= 50mW.
It will be only slightly warm conducting 30mA. Then its base resistor R2 should be 5.6k ohms to limit its base current to 3mA when the output is shorted.

 

[duke37]

Rb at 13.8MΩ seems to be wildly excessive. Use say 10k to protect the BC556 base from too much current.

I have not looked up where the capacitors should be fitted for stability. A diode across the circuit is often used to protect against reverse polarity.

 

[Andrew Mc]

A BC556 has a current gain of from 110 to 800. Only some might have a gain of 800. A BC556C has a current gain of from 420 to 800.

The regulator warms with only (18V - 12V) x 10mA= 60mW in your circuit but it is allowed to be fairly hot at 600mW. Why not allow it to conduct 30mA? Change R1 to be 0.7V/30mA= 23.3 ohms (use 22 ohms).

The BC556 with a current of 10mA heats with only (18V - 13V) x 10mA= 50mW.
It will be only slightly warm conducting 30mA. Then its base resistor R2 should be 5.6k ohms to limit its base current to 3mA when the output is shorted.

What was wrong with just 10mA for the regulator? i picked this value as the transistor were going to provide the extra current and i wanted the regulator to stay as cool as possible. ( but yes i could have the regulator drawing 30mA).

I understand about the voltage drop across Ra needing to be 0.7, but i dont really understand what the voltage drop would be for Rb, i figured it to be 18v - 0.7 (voltage drop across Ra) is this the case?

I assume that the 5.6K you put was the best match resistor.

18V -0.7V (Ra drop) / 3mA = 5.766K ~ 5.6K best match.

im a bit confused about what you have put about the BC556, why would i want it to conduct 30mA as i only need around 1 - 2 mA for the TIP120, as its gain is around 1000? then i would change the 500ohm to a 2K,
18V - 4V (safe drop for the base of TIP120) / 2mA = 2K.

I know i am asking a lot of questions but im trying to learn.

 

[Andrew Mc]

Do you have to specifically use a linear regulator in your design? Using a switching buck-converter would be more efficient. There are relatively cheap modules you can just buy and use:

https://www.amazon.co.uk/Gaoxing-Tech-LM2596HVS-Adjustable-Converter/dp/B06W9K7GK7

i have checked out the web page, the converter is an alternative that i can consider for other projects (would try to build it myself), i used the above circuit as i have the parts already and its more about learning how to use them.

 

[duke37]

You do not need any voltage drop across Rb. The resistor is there to limit the current in the case of a fault condition so it should be quite low.
The gain of the overall circuit is colossal and you could run into instability problems.

The 500Ω will control the current through the BC556, only a very slight extra curent will be needed to generate a large output current. 5A out equals 5mA TIP120 base current.

 

[Andrew Mc]

You do not need any voltage drop across Rb. The resistor is there to limit the current in the case of a fault condition so it should be quite low.
The gain of the overall circuit is colossal and you could run into instability problems.

The 500Ω will control the current through the BC556, only a very slight extra curent will be needed to generate a large output current. 5A out equals 5mA TIP120 base current.

I tested the BC556, it looks like it doesnt work properly, not some thing you want to happen when you are unsure.

any way, i have measured point in the circuit and this is what i get:

used a 39ohm resistor for Ra, voltage drop of 0.64V (calculated current 16.4mA)

used a 6.45K (var res) for Rb, voltage drop of 14.8mV (calculated current 2.3uA)

used a 2K for Rc (was 500ohms), voltage drop of 1.25V (calculated current 0.625mA)

the load i used has always been 200 ohm (12V / 200 = 60mA)

so if 16.4mA is flowing through Ra and 0.625mA is flowing through the 2K (total is just over 17mA), then the TIP120 must be supplying the rest (43mA).

instability problems you say regarding the gain!

if i work backwards from the TIP120 and make sure that the base only gets 1-2mA max, and so the BC556 would then need the Rb resistor increased to limit its base, do you think this would help?

 

[Audioguru]

You do not want Rb to produce a substantial voltage drop. It simply limits the base current in the BC556 when the output is shorted to ground. Ra should be used to limit the current in the BC556 instead.

I recommended a current of 30mA in the regulator because your 10mA is too close to the 6mA max current used by the regulator when it has no output current.

Your simple circuit does not need a PNP transistor driving an NPN darlington. A PNP darlington can be used instead.

 

[duke37]

There is no need for the 500Ω resistor, there are resistors inside the Darlington to do this job.

The circuit which I found used a TIP73 npn transistor with a gain of 20 to 180.
So, assume a gain of 50 and a load of 5A, then the TIP73 base current will be 100mA and the base current of Q1 will be 1mA. Rb is irrelevant unless too much current is demanded.

 

[Audioguru]

When a transistor has a minimum gain of 20 then why assume its gain is 50? Don't you want every circuit made with that transistor to work or only a few of the circuits made with it to work?? I design everything with "worst case" parts (minimum gain calculations) so that every circuit works perfectly, not just a few of them.

 

[duke37]

I used a gain of 50 which is a lot less than the maximum of 180 and we are still in the realm of how the circuit works. I can divide 5000 by 50 in my head. For a final design, the loss of current through the resistors must be taken into account as well as the specification of the components.

I have not looked up the maximum permitted current and dissipation of the BC556, it may be a problem if the output transistor has a gain of only 20 is used. I always design conservatorily.This will not be a problem when,a Darlinton is used.

 

[Andrew Mc]

You do not want Rb to produce a substantial voltage drop. It simply limits the base current in the BC556 when the output is shorted to ground. Ra should be used to limit the current in the BC556 instead.

I recommended a current of 30mA in the regulator because your 10mA is too close to the 6mA max current used by the regulator when it has no output current.

Your simple circuit does not need a PNP transistor driving an NPN darlington. A PNP darlington can be used instead.

I have used these parts in my circuit, as I already have them.
The NPN tip120 was going to be used for my output from my op amp (the circuit thats going to be connected to this 12v) but the load has to be connected from V+ to the collector, I didn't want to waste it.

 

[Andrew Mc]

Thank you for all your help with this, i think i understand most of it now, i will try this out with a much higher output current to see what happens.