Discrete transistor voltage regulator

I would like some help figuring out this linear regulator.


I'd like to break it down into the three main parts as I see it.

I understand that there's a pass transistor Q3.
D1 and C1 form to create a reference voltage.

I understand that Q4 forms some type of current limiter but I am unsure how it works.

pot R4 & resistor R5 are used to set a reference voltage and I think Q1 is some type of error amplifier.

Going back to the beginning, I'd like to first tackle the variable voltage limiting.

Unreg Voltage is regulated by the pass transistor along with R4 & R5 and the reference voltage.

Can anyone help me piece these pieces together to understand how this circuit works?

Sort of. Q2 & Q3 form a Darlington pass transistor. C1 has nothing to do with creating a reference voltage; it's only for filtering. As the voltage across R5(proportional to output voltage) rises, Q1 conducts more current away from the base of Q2, causing Q2/Q3 to conduct less, lowering the output voltage. That's your feedback loop.

Current limiting works like this: as long as the voltage across R2 (proportional to output current) is below Vbe of Q4, Q4 is off. As output current increases, Q4 starts to turn on, drawing base current away from Q2 (and, thus, also Q3) limiting the output current.

Q1 is essentially an error amplifier

the be of Q1 and D1 form a (poor) voltage reference of 2 diode drops, or around 1.2V
because the reference is diode drops, variations in the forward voltages due to temperature
and current will effect the output voltage.

the output voltage sample is determined by R3, R4 and R5

barry described the operation

Note what wwfeldman stated about the poor regulation from Q1 and D1 being used as a reference.
Also note that the current limit occurs when the drop across R2 starts to make Q4 conduct. It could be more than 1A which exceeds the maximum rating of the TIP29. It isn't a particularly well designed regulator and it's performance will be significantly worse than a 7805 based design.

Brian.

Additionaly.... There is no fast and direct method to measure the die’s temperature.

Even if the series pass transistor is within its SOA boundaries, poor heatsinking, thermal interfacing or simply high ambient temperatures, may cause to exceed the thermal limits.

Having said this, I would encourage you to build the circuit, and compare it against a lowly 7805. It is a real eye opener.

betwixt said:

Note what wwfeldman stated about the poor regulation from Q1 and D1 being used as a reference.
Also note that the current limit occurs when the drop across R2 starts to make Q4 conduct. It could be more than 1A which exceeds the maximum rating of the TIP29. It isn't a particularly well designed regulator and it's performance will be significantly worse than a 7805 based design.

Brian.

Click to expand...

I am more trying to understand the workings of the circuit. I have seen the internal schematic of the 7805 but that's a bit too complicated for me to follow at this moment.

barry said:

Sort of. Q2 & Q3 form a Darlington pass transistor. C1 has nothing to do with creating a reference voltage; it's only for filtering. As the voltage across R5(proportional to output voltage) rises, Q1 conducts more current away from the base of Q2, causing Q2/Q3 to conduct less, lowering the output voltage. That's your feedback loop.

Current limiting works like this: as long as the voltage across R2 (proportional to output current) is below Vbe of Q4, Q4 is off. As output current increases, Q4 starts to turn on, drawing base current away from Q2 (and, thus, also Q3) limiting the output current.

Click to expand...

Thank you, I am re-drawing this out and working my way through to understand the circuit. Once it's sunk in I'll see if I have a full understanding. It is a bit difficult to follow sometimes because the schematics vs the actual hardware are bit different.

Not only that but I do not have access to standard parts like these. I am using parts salvaged from a 1984 audio amplifier so I have parts like;
2SA1302
2SC3281
C1815
A1015
B669A
D669A
C3298B
D2061
B1187
78M05

I'd like to build a linear regulator that can go 0-5 volts.
The transformer has one output which is labelled as 6 volts but reads as 7.89 volts unloaded.

Once I understand how the regulation works, I should be able to build a linear regulator that can go from maybe 0.7 volts to 5 volts using a PNP pass transistor.
I would like to be able to measure down to the milliamp range and 0-5 volts.
AN7805

Again, I encourage you to build a discrete transistor regulator, because it is a *very valuable* learning experience.

Start with the circuit you have. Once that it is working, poke around the different nodes with a DMM and read the different DC values.
Change the load, change the input voltage and repeat with the DMM.

Put your circuit in a freezer, then reconnect it immediately and see how the output voltage changes as the circuit warms up.

Start improving the circuit. For instance, replace the "reference" diode with a zener. You will require to change the feedback resistor values to achieve the same output voltage.

Additional learning bonus: simulate the circuit with any of the available circuit simulators.

Regulation works like this: If output voltage is high enough it turns on Q1. Turning on Q1 pulls down on the base of Q2 which turns it off to reduce current to the load (thus voltage). If output voltage droops the opposite happens. Thus the circuit stabalizes at an output voltage that holds Q1 right at its turn on threshold which is the forward drop of D1 + The Vbe threshold of Q1.

belektrik said:

Once I understand how the regulation works, I should be able to build a linear regulator that can go from maybe 0.7 volts to 5 volts using a PNP pass transistor.
I would like to be able to measure down to the milliamp range and 0-5 volts.
AN7805

Click to expand...

That circuit uses the base-emitter voltage of Q1 and the forward diode drop of D1 as a voltage reference, so the minimum output voltage would be about 1.3V.
If you eliminated D1 and connected the Q1's emitter to ground, then the minimum output voltage would be about 0.65V (with R3 removed and the pot set all the way to the top).
But the temperature stability of either configuration is poor, with the output voltage varying about 0.3% per degree C change.

What do you mean "measure"?

asdf44 said:

Regulation works like this: If output voltage is high enough it turns on Q1. Turning on Q1 pulls down on the base of Q2 which turns it off to reduce current to the load (thus voltage). If output voltage droops the opposite happens. Thus the circuit stabalizes at an output voltage that holds Q1 right at its turn on threshold which is the forward drop of D1 + The Vbe threshold of Q1.

Click to expand...

This is a bit confusing to me and here's why. These all seem to be NPN transistors.

The unregulated voltages goes into the circuit and runs into the collector for every transistor; Q3, Q2, and Q1.

This might seem basic but I just don't see a way for any transistor's base to get enough current to allow current to flow from collector to emitter.

Is it that the current comes though R1 slightly open Q2 and allow current to flow past Q4 down through R3 and Pot R4 to the base of Q1 that would allow collector emitter current from Q1?

crutschow said:

That circuit uses the base-emitter voltage of Q1 and the forward diode drop of D1 as a voltage reference, so the minimum output voltage would be about 1.3V.
If you eliminated D1 and connected the Q1's emitter to ground, then the minimum output voltage would be about 0.65V (with R3 removed and the pot set all the way to the top).
But the temperature stability of either configuration is poor, with the output voltage varying about 0.3% per degree C change.

What do you mean "measure"?

Click to expand...

I am doing small electrolysis experiments and need to test the amount of current needed to conduct through a solution.

I don't need a lot of volts but having accurate current measurements would be beneficial. I have another power supply that is switched mode power supply but it doesn't show milliamps so when using it all I see is 0 for current with 2 volts; I would like be able to accurately record current draw.

I'll try to explain in my words:

Q2 and Q3 are wired as a "darlington pair", you could replace them with a single 'darlington' transistor with equal results. The special property of this configuration is high current gain, this is because the effective 'base pin' is that of Q2 and all the current it controls through it's collector/emitter is passed to the base of Q3. The effect is that you get what looks like a single transistor with a gain equal to the gain of Q2 multiplied by the gain of Q3.

Because of the high gain, little base current is needed so R1 can provide more than is needed. Q1 diverts some of the current to ground through its own collector/emitter and D1. You might think of R1 being the top of a potential divider and Q1/D1 being the bottom of it. The more Q1 conducts, the lower the voltage at its collector will be and therefore Q2/Q3 conduct less and you get less output voltage.

Because Q1 has a fixed emitter voltage and it is wired in common emitter configuration, increasing its base voltage will decrease its collector voltage, in other words it works to invert the voltage direction.

The base of Q1 gets a proportion of the output voltage from R3, R4 and R5. Note that more voltage means more base current which in turn produces less output voltage, in other words it tries to stabilize the output. If the output voltage goes higher, Q1 conducts more and makes Q2/Q3 conduct less. The opposite happens if Q1 conducts less. This negative feedback is what tries to hold the output steady.

As current is drawn from the supply, a voltage is dropped across R2. It will be 0.47V per Amp of load current (V=I/R). The same voltage appears between the base and emitter of Q4. As that voltage increases, Q4 will start to conduct and by doing so will divert some of the bias current away from Q2/Q3 which drops the output voltage. As Q4 only starts to conduct when the voltage between base and emitter exceeds about 0.5v it will have no effect at low currents. If the output is overloaded, Q4 will conduct harder and pull the output voltage lower, hence limiting the available current. This is why I was concerned about Q3 being overloaded, although it looks like a heavy duty transistor it is only rated at 1A maximum current so it is quite possible it will pass more than that even if Q4 is trying it's hardest to keep it low.

Brian.

belektrik said:

This is a bit confusing to me and here's why. These all seem to be NPN transistors.

The unregulated voltages goes into the circuit and runs into the collector for every transistor; Q3, Q2, and Q1.

This might seem basic but I just don't see a way for any transistor's base to get enough current to allow current to flow from collector to emitter.

Is it that the current comes though R1 slightly open Q2 and allow current to flow past Q4 down through R3 and Pot R4 to the base of Q1 that would allow collector emitter current from Q1?

Click to expand...

Yes that's it. R1 turns on Q2 which turns on Q3 which which charges C2 and the load until Vout is high enough that Q1 starts turning on and prevents Vout from rising any higher (by turning off Q2).

Likewise Q4 can shut things down when current is high enough to make R2 turn it on.

As suggested download LTspice and simulate this so you can see how all the pieces work.

I had to modify a the circuit based on the equipment that I have.

I replaced Q3 with a PNP Mosfet which is always open.

The base of Q3 has a N type transistor with a "reference voltage"; made up of a diode and 330uF capacitor.

Q3 output has a voltage divider consisting of 2 1K resistors; The first one is a 1K ohm potentiometer and another 1K ohm.

Here's an image; its from the program everycircuit.


When I adjust the potentiometer the output voltage goes from around 600mV up to around 6 V, which would be good if the circuit actually performed that well.

I think that I understand the working of this part of the circuit.

I'd like to ask a few more questions about this part of the circuit.

How could I create a more stable reference voltage? I'd like to try and avoid zener diodes because my lowest value is 2 volts, I'd like to be able to get as close to zero voltage as possible.

What else could I do to stabilize this circuit as it stands?

Would adding bias resistors to any of the transistors help?

What about adding a second NPN transistor like Q2 in the original schematic shared in the original post?

Again; as suggested previously, download LTSpice and simulate the circuit!

Simulations allow you a few things:

-Change components at your heart's delight. Nothing will blow up, no magic smoke will be released.
-You are not bounded by the few available components in your drawer.
-You can simultaneously see all the node voltages and mesh currents and their interactions.

I'd suggest a 1.24V TL431 variant as a replacement for Q1 (with D1 removed). A TL431 can be thought of like a transistor with a very accurate threshold voltage.

If you want to go lower than 1.2V you'll need a different feedback topology - an opamp error amplifier.

The transformer has one output which is labelled as 6 volts but reads as 7.89 volts unloaded.

Click to expand...

Is that AC or DC?

Making a simple discrete regulator whose output can go below its reference voltage is problematic, unless you use an op amp.

I'll see if I can find this part in my area.

I do not need to go below the reference voltage, I actually won't be needing a ton of voltage so 2-5V should be fine, I would like to have accurate current measurements.

Just use a LM317 regulator with one 120 ohm resistor and one 500 ohm pot (plus a couple decoupling caps).
That will allow adjustment of the output from 1.25V (the stable internal reference voltage) to about 2.5V below the input voltage.