Elenco XP720K Pass Transistor

[Gripnook]

I'm trying to understand the pass transistor circuit used in the XP720K to raise the maximum current to 3A on the 5V power supply. The relevant schematic is attached.

The problem I have is that when Q2 is turned on by a 0.6V drop across R3+R4, the Emitter-Base voltage drop across Q1 is already 0.3V. So it only takes an extra 1.5A across R5 to turn Q1 on, and overload the 7805. This would limit the output current at 1.5A. Is there something I'm missing here?

Note that while Q2 is heatsinked, the 7805 is not and is intended to overheat and turn off when the output draws 3A.

[Paul Price]

Q2 is not turned on by a .6V drop by R3+R4. Q2 is turned on by the current flowing through the 7805 to the output.

The idea here is to allow greater than 1A to be available and maintain regulation, but also not allow this circuit to deliver 3Amps.
Something like "foldback current limiting" is accomplished as the output current approaches 3A.

[Lovely_Santa]

Just a quick little bit more specification at what Paul Price said:

The cap C8 will load up to 5V.

By drawing current, you will unload the cap. At that point the LM will produce 5V for loading the cap again..
At this point Q2 will turn on and apply voltage to the output (until 5V). The higher the output current, the faster the cap will unload and the more Q2 will turn on.
So this is to draw more that 1A for some time out of the circuit, but this isn't enough to draw 3A out of it...

[Paul Price]

Sorry, Santa,

The output cap has nothing to do with current limiting, but only needed to keep this regulator output constant under fast load changes and prevent this circuit from oscillating.

[KE5FX]

I'm trying to understand the pass transistor circuit used in the XP720K to raise the maximum current to 3A on the 5V power supply. The relevant schematic is attached.

The problem I have is that when Q2 is turned on by a 0.6V drop across R3+R4, the Emitter-Base voltage drop across Q1 is already 0.3V. So it only takes an extra 1.5A across R5 to turn Q1 on, and overload the 7805. This would limit the output current at 1.5A. Is there something I'm missing here?

Note that while Q2 is heatsinked, the 7805 is not and is intended to overheat and turn off when the output draws 3A.

Here's my take on it: initially, assuming a light load, both Q1 and Q2 are off.  All of the load current goes through R5, R4, R3, and the 7805.  As the load current exceeds 0.1A, the voltage drop across R3 and R4 approaches 0.6V, turning Q2 on.  The 7805 also begins pulling less current through R3 and R4 at this point, as Q2 handles more of the total load.   

Once the current exceeds 3A, the voltage drop across R5 will become high enough to turn Q1 on.  This will divert current from the R3-R4-R5 path, turning Q2 off.  The 7805 now has to handle all of the load current, and it will overheat and shut down.  Q2 will then remain off since nothing is drawing current through R4+R3.

Pretty neat hack, actually.

[Gripnook]

I am aware of the intent of the circuit. My problem is with the values given. As KE5FX said, initially both transistors are off, and when the 7805 draws about 0.1A through R3+R4, the V_EB drop for Q2 will reach 0.6 V, turning it on and essentially drawing all remaining current to the load.

My issue is with the current limiting circuit. It seems to me like the circuit will limit the current at around 1.5A instead of 3A, since the drop across R4 is already 0.3 V from the current through the 7805, Q1 will turn on when the drop across R5 is 0.3 V, not 0.6 V as specified on the datasheet. This is about 1.5A and not 3A.

Link to datasheet: http://www.elenco.com/admin_data/pdffiles/XP-720K_REV-I.pdf

[KE5FX]

I am aware of the intent of the circuit. My problem is with the values given. As KE5FX said, initially both transistors are off, and when the 7805 draws about 0.1A through R3+R4, the V_EB drop for Q2 will reach 0.6 V, turning it on and essentially drawing all remaining current to the load.

My issue is with the current limiting circuit. It seems to me like the circuit will limit the current at around 1.5A instead of 3A, since the drop across R4 is already 0.3 V from the current through the 7805, Q1 will turn on when the drop across R5 is 0.3 V, not 0.6 V as specified on the datasheet. This is about 1.5A and not 3A.

Link to datasheet: http://www.elenco.com/admin_data/pdffiles/XP-720K_REV-I.pdf

The circuit is "squishier" than the description in the manual suggests.  Q1 Vbe goes from about 0.5 to 0.9V as the load current goes from 1A to 3A, but Q2's Vbe only goes from about 0.66 to 0.72 over the same range.  The 7805's share of the current doesn't stay constant at all, but goes from 0.13A to 0.3A. 

With the 7805 in open air (no heat sink), 4A appears to be the circuit's real limiting threshold.  At 4A the 7805 is drawing 0.5A, which eventually raises its case temp to about 135C and causes the protection to kick in.   When this happens the 7805 doesn't shut down abruptly, it just drops its output voltage a bit.   The load current drops accordingly, the case temp goes back below 135C, the voltage goes back up, and the cycle repeats.   

[Gripnook]

So the circuit as is should be able to output 3A regulated current?
When you say Q1 Vbe goes to 0.9V, that would be an increase in current around 5 orders of magnitude from the current at 0.6V, which would mean ~100A through it (assuming 1mA at 0.6V).

[KE5FX]

So the circuit as is should be able to output 3A regulated current?

With the components I tried (I couldn't resist building it and making a few real-world measurements) it can deliver 4A all day long.  But again, this was with the 7805 in open air on the workbench surface.  An enclosure full of hot parts with marginal/no ventilation would lower its ability to stay under 135C, as would a higher input voltage.

When you say Q1 Vbe goes to 0.9V, that would be an increase in current around 5 orders of magnitude from the current at 0.6V, which would mean ~100A through it (assuming 1mA at 0.6V).

Not sure where that comes from.  Keep in mind that Vbe(on) increases with collector current.  0.6V will let 1 mA flow, but if you want to take  advantage of the MPSA70's max collector current rating of 100 mA, you need more like 1.0V to drive it into saturation.   (I used a 2N3906 with similar basic ratings but 200 mA max Ic.) 

The big drawback of this circuit from an engineering standpoint is that its exact behavior depends on its input voltage, temperature, and transistor properties in ways that aren't easy to predict and control.  It accomplishes what Elenco was shooting for -- keep the cost of goods down, don't rack up a lot of warranty returns, and don't set the customer's workbench on fire -- but it's not what most people would consider a good design for a CV/CC supply.

[Gripnook]

I was under the impression that transistor current is an exponential function of Vbe, and current increases by an order of magnitude every 60mV of Vbe. So 1mA @ 0.6V would mean huge currents at 1V.

I've been trying to get this power supply to work for a long time now, and it's never actually delivered the rated current. I'll try replacing the 7805 and giving it another go.

EDIT: Just measured the diode drop on the MPSA70 and got .75V @ 1mA, so it makes sense then to have ~0.9V @ 100mA when loaded.

[danadak]

Sim, attached.

[Gripnook]

The problem with the simulation is that LTSpice doesn't take into account thermal overload for the 7805. You'd have to measure the power dissipation in the 7805 and check when it goes above the threshold for shut down.

EDIT: That would be approximately where the output current is 0.3A, or at 2A load in the simulation. I don't think you're taking into account the specific transistors in your simulation though. Q1 has Veb=0.75V @ 1mA and Q2 has Veb=0.6V @ 1mA as measured with a multi-meter diode function.

[danadak]

I wondered the same thing, so swept the current further, got this result, attached.

Regards, Dana.

[Gripnook]

Thank you everyone for your help. Having rebuilt the circuit again, the supply delivers somewhere around 2.6A until it shuts down due to overheating. It still has some issues at high currents, such as poor regulation due to the wires and pcb traces having significant voltage drops, but it'll be more than enough for my needs.

[AG6QR]

Note that while Q2 is heatsinked, the 7805 is not and is intended to overheat and turn off when the output draws 3A.

I own the Elenco 720K, and I assembled mine from the kit.  The 7805 is most definitely heatsinked, with a fairly beefy heat sink that is pretty well ventilated on the back of the unit.  This heat sink contains Q2, as well as all three linear regulators: the 7805, LM317, and LM337.  It's detailed on pages 10 and 11 of the construction manual, http://www.elenco.com/admin_data/pdffiles/XP-720K_REV-I.pdf

Yes, the design takes advantage of the 7805's built-in thermal limiting to limit the current, but since the 7805 has a good heat sink, it should handle pretty nearly a full amp on its own, with Q2 passing the remaining current.

I assume you're not actually building the actual Elenco kit, but trying to duplicate the idea from the schematic?  If so, your problem might be that you overlooked the heatsinking of the 7805.  Leaving the 7805 without a heat sink would definitely cause the protection to kick in a bit sooner and would lower the current limit somewhat.

[Gripnook]

Note that while Q2 is heatsinked, the 7805 is not and is intended to overheat and turn off when the output draws 3A.

I own the Elenco 720K, and I assembled mine from the kit.  The 7805 is most definitely heatsinked, with a fairly beefy heat sink that is pretty well ventilated on the back of the unit.  This heat sink contains Q2, as well as all three linear regulators: the 7805, LM317, and LM337.  It's detailed on pages 10 and 11 of the construction manual, http://www.elenco.com/admin_data/pdffiles/XP-720K_REV-I.pdf

Yes, the design takes advantage of the 7805's built-in thermal limiting to limit the current, but since the 7805 has a good heat sink, it should handle pretty nearly a full amp on its own, with Q2 passing the remaining current.

I assume you're not actually building the actual Elenco kit, but trying to duplicate the idea from the schematic?  If so, your problem might be that you overlooked the heatsinking of the 7805.  Leaving the 7805 without a heat sink would definitely cause the protection to kick in a bit sooner and would lower the current limit somewhat.

The 7805 is not thermally connected to the heat sink. See page 8 of the manual. Unlike the other ICs, it is only connected through a washer and not the mica/thermal paste arrangement.

Also, drawing 1A through the 7805 would burn out Q1 which is rated at 100mA.

[KE5FX]

The 7805 is not thermally connected to the heat sink. See page 8 of the manual. Unlike the other ICs, it is only connected through a washer and not the mica/thermal paste arrangement.

The mica is only needed to insulate the tabs of the power transistor and LM317s, which aren't at ground potential like the 7805's tab is.

Also, drawing 1A through the 7805 would burn out Q1 which is rated at 100mA.

The circuit won't let the 7805A draw 1A, due to voltage drop across R3 and R4.  It doesn't limit at .2A exactly, but it's in that ballpark.

[Gripnook]

The 7805 is not thermally connected to the heat sink. See page 8 of the manual. Unlike the other ICs, it is only connected through a washer and not the mica/thermal paste arrangement.

The mica is only needed to insulate the tabs of the power transistor and LM317s, which aren't at ground potential like the 7805's tab is.

Also, drawing 1A through the 7805 would burn out Q1 which is rated at 100mA.

The circuit won't let the 7805A draw 1A, due to voltage drop across R3 and R4.  It doesn't limit at .2A exactly, but it's in that ballpark.

The 7805 is insulated both thermally and electrically from the heatsink. See the arrangement with a washer on each side for electrical insulation, and no contact between the body of the IC with the heatsink. There might be some heat transfer through the washer, but not much.

Also, for the 1A current, the R3+R4 arrangement do limit it to about 0.2A, but im referring to the additional current that will run through Q1 when it gets turned on. If the 7805 is drawing 1A as the commenter above said, then 0.8A of it will come from Q1 and thus will overheat it.