Good morning.
I picked up a homebuilt 12-volt 5-amp power supply at a yard sale. It seems to work well. Some of the parts are vintage, the SK9136 and the SK3083. I took the cover off and this is what I found inside. I need help with a couple of things.
1) What is the purpose of the SK3083 ? I don’t recognize that circuit.
2) Any suggestions for replacements (if I ever need to) for the SK3083 and the SK9136 ?

Thanks

Pete

 

It looks like the 3083 is there to progressively shunt the 3 Ω resistor. The resistor is there to shift some of the waste power to the resistor from regulator so that it doesn't have to dissipate at much heat. But as the load increases, the voltage drop across the resistor would become excessive. This is sense by the voltage across the four parallel resistors.

What's the DC output of the rectifier?

 

Certainly the 3083 portion of the circuit is uncommon, in fact I have not seen such before. It may also, possibly, serve as a bit of a ripple filter. The 9136 is a "wrap around" transistor to boost the current handling beyond what the 7812 regulator can supply.

 

I redrew the schematic.
R3 is the load.
When the load is small current flows through R2 (3) in the regulator, out and to the load.
When the current is larger, more than 0.6V is across R2 and some current passes E-B of Q1 causing it to turn on a little. Now Q1 Collector current helps lift the load. (transistor by pass)

Q2 is for current limiting. When the current is large and there is 0.6V across the 0.1 ohm resistor, Q2 shorts out Q1 Base to Emitter of Q1. This turns off Q1.

Error. There needs to be a Base resistor on Q2. If not Q2 base current may get too high.

 

I redrew the schematic.
R3 is the load.
When the load is small current flows through R2 (3) in the regulator, out and to the load.
When the current is larger, more than 0.6V is across R2 and some current passes E-B of Q1 causing it to turn on a little. Now Q1 Collector current helps lift the load. (transistor by pass)

Q2 is for current limiting. When the current is large and there is 0.6V across the 0.1 ohm resistor, Q2 shorts out Q1 Base to Emitter of Q1. This turns off Q1.

Error. There needs to be a Base resistor on Q2. If not Q2 base current may get too high.
View attachment 289451

I think you mean resistor on Q1..

 

I think you mean resistor on Q1..

No. Q2 needs a resistor. When you put a B-E across a shunt resistor, current flows through the resistor until you reach 0.6V then the current flows B-E without limit. It is very possible to have 6A in the resistor and 2A in the B-E junction. By adding some resistance to the Base you will save the little transistor in case of a short.

Q1 does not need a Base resistor because the 7812 can only pull a limited amount of current.

 

No. Q2 needs a resistor. When you put a B-E across a shunt resistor, current flows through the resistor until you reach 0.6V then the current flows B-E without limit. It is very possible to have 6A in the resistor and 2A in the B-E junction. By adding some resistance to the Base you will save the little transistor in case of a short.

Q1 does not need a Base resistor because the 7812 can only pull a limited amount of current.

But Q2 already has a resistor on it??

 

Q1 does not need a Base resistor because the 7812 can only pull a limited amount of current.

I think there does needs to be a resistor between the collector of Q2 and the LM7812.for proper current-limit operation and to limit the Q1 base current.

When the current is large and there is 0.6V across the 0.1 ohm resistor, Q2 shorts out Q1 Base to Emitter of Q1. This turns off Q1.

With the added resistor I suggested above, It doesn't completely turn off Q1.
It starves the base current to Q1 to limit the current to about 0.6V / 0.1Ω = 6A.

 

With no load current, both transistors are cutoff and there is no current through the 3 Ω resistor, so the input voltage to the 7812 is essentially the power supply voltage (which I don't see any indication of what that is, so I will arbitrarily assume it is 20 V).

As the load starts to pull current, it goes through the 3 Ω resistor and then through the 7812. This continues until the voltage drop across the resistor gets to about 600 mV (i.e., enough to start turning on the bypass transistor). So the load curretn at this point is about 200 mA.

At this point, the output transistor turns on and most of the additional current is now sourced through the bypass transistor until there is enough current to cause enough of a voltage drop across the current-sense resistor (about 110 mΩ) to start turning on the shunt transistor -- let's call that also about 600 mV. So the current through the bypass would be about 5.5 A. If the beta of the bypass transistor were 50 at this point, then the base current would be about 110 mA. The voltage across the 3 Ω resistor would be about 1.2 V, making the current through it about 400 mA. That, plus the 110 mA of base current in the bypass transistor, would flow through the 7812, with most of it being delivered to the load. So that would put the load current right at about 5.9 A.

The question is, what happens as the load current tries to increase further such that the shunt transistor starts turning on?

An attempt to increase the current will lower the base voltage on the shunt transistor, which will allow the excess current to bypass the 3 Ω resistor, and thus the voltage at the base of the bypass transistor does not increase. As a result, it will try to maintain the same collector current that it had before (about 5.4 A, which all goes to the load). The base current of the shunt transistor is essentially siphoned off as base current in the bypass transistor to then goes through the 7812 and onto the load, so nearly all of the additional current to the load is routed through the 7812, allowing it to use its internal thermal shutdown limiting circuitry, which starts kicking in somewhere around 1 A of continuous current.

I don't think any additional base-current limiting resistors are needed -- the limits are imposed via negative feedback.

 

The redrawn circuit seems to be different from the original circuit posted, and now, certainly, the portion with Q2 makes sense, and I have seen that before many times.
Pleas note that if the current division between the regulator and the pass transistor is set optimally, the current limit in the regulator will operate at the same point as the "wrap around" transistor is reaching it's limit. In that case the separate current limiter is not required.

 

The above Circuit is the standard "Application-Notes" Circuit found
in the Spec-Sheet of many Linear-Regulators.

If You want to increase your available Voltage or Current,
or make it quieter,
the Input, and Output, Filtering needs to be improved upon.

The whole thing can of course be replaced with
a Switching-Regulator-Chip for much less Power wasted as Heat.
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.
.
.

 

Gentlemen ..... good afternoon ..... I have redrawn the schematic, I didn't like the way it looked ...... this version makes more sense to me ..... I put labels of Q1, Q2 and U1 on it ..... I also added voltage measurements ..... transformer output is 18 VAC .... across the 6800 microfarad cap I measured 22 VDC ..... and I am going to add a voltmeter to it .... hope my redraw helps .....

 

What is the question ???

What do You want from this Power-Supply ???

Does it provide exactly what You are looking for ???
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Certainly it is important to know the requirements before starting a design.
For a power supply that can be as simple as knowing the voltage and the current required, and in addition the ripple and the regulation requirements.

 

I picked up a homebuilt 12-volt 5-amp power supply at a yard sale. It seems to work well. Some of the parts are vintage, the SK9136 and the SK3083. I took the cover off and this is what I found inside. I need help with a couple of things.
1) What is the purpose of the SK3083 ? I don’t recognize that circuit.
2) Any suggestions for replacements (if I ever need to) for the SK3083 and the SK9136 ?

Has Question #1 been answered to your satisfaction?

As for Question #2, the information I could find seems to indicate that the NTE197 is usually a suitable replacement for the SK3083 and the NTE180 for the SK9136.

I've never used any of those transistors. I don't know why they used either of those specific transistors -- may have been what he had laying around. From what little I was able to find, I would have considered using the NTE180 (or SK9136, if it's truly comparable) for both of them.

As long as the transistor can handle the current and power and has at least a modest beta (20?), the specific choice is probably not too critical.

 

Even a 2N3055 with a PNP driver could work, but it would be less efficient. But there are quite a few high power medium gain transistor types available. Frequency response is not a big challenge in this application, nor is linearity,

 

Gentlemen ..... good morning ..... and I thank all of you for the education and help with my two questions .... yes, question #1 has been answered to my satisfaction ... and beyond .... and question # 2 also ..... my intended use of this power supply is only to be a rough bench 12 volt source ..... if I need better than I've got a power supply that I removed from a computer .... I think that I'll keep it ...... thanks again ..... I think that this thread can be closed ......

Pete