Constant current power supply

Reply to Thread

Hello and welcome to Electronics Point

It would be helpful to know the operating voltage of the circuit. If you don't know, can you tell us the secondary voltage for the transformer.

The output (anodising) current is determined roughly by the emitter voltage and the emitter resistance using Ohm's Law, I = V / R. (This neglects the base current, which can be as much as 10% of the emitter current, possibly more!) The emitter voltage is equal to the base voltage minus the base-emitter junction forward voltage V_BE, which for a power transistor, is roughly 0.8~1.0 volts, depending on base and collector currents. By setting the base voltage, you set the emitter voltage to a voltage that's roughly 0.8~1.0V lower.

The usual way of making this simple current source variable is to use a fixed base voltage and vary the emitter resistor (R4). For example, putting 1.9V on the base gives you about 1V on the emitter, and therefore 1V across the emitter resistance. Neglecting base current, the anodising current is I = V / R where V = 1, so I = 1 / R.

But with such low resistance values, and high currents, you can't just use a standard carbon potentiometer for the emitter resistor, because of the need for fairly accurate, low resistance, and the power dissipation, which (for the emitter resistance) can be calculated as P = V × I where V is the voltage across the emitter resistor, in volts, and I is the anodising current, in amps.

Have a look at these options for the emitter resistor:

www.digikey.com/product-search/en/potentiometers-variable-resistors/adjustable-power-resistor/262971?stock=1
Adjustable power resistors. Not easy to change the setting. But you could use several of these, with a rotary switch to select between them. Make sure you use a good quality rotary switch with low contact resistance!

http://www.digikey.com/product-search/en?pv1=1345&pv1=1025&pv1=250&pv1=382&FV=fff40004,fff80123,400fa,4017e,40401,40404,40433,4045a,40460,4048e,404b0,40523,40541,40546,4060f&stock=1&quantity=1
Low-resistance rheostats (variable resistors). This would give you a rotary control for your anodising current.

The circuit you have there is pretty simple and only gives rough control of the anodising current. The transistor's base-emitter voltage varies over a significant range, and with such a low emitter voltage determining the current, this variation affects the current significantly.

You also need to ensure that the base drive circuit can supply significant current. Large transistors that can pass several amps have low current gain figures, and this means you need to supply quite a lot of base current to get the desired collector current to flow. R3 = 22k is far too high; it will limit the base current to less than 1 mA, and therefore will limit the collector current to around 10 mA (0.01 amps)!

You also need to consider power dissipation in the transistor, which could be dozens of watts. You will definitely need a heatsink.

There are some much better options available for this circuit. Have a look at http://www.google.com/search?q=op-amp current sink&tbm=isch or let us know the operating voltage, and the output current range you want, and I will design something for you.

 

That's a very rough "current source". Only slightly better than a big resistor.

What is the maximum anodising current you want?

 

OK, here's my suggested circuit.


The AC voltage from the mains transformer is bridge rectified by BR1 and smoothed by C1, which should be at least 10,000 µF. The thick lines represent connections that should be made with thick wire. For these wires, follow the topology shown in the diagram to ensure stable operation of the circuit.

U1 is a TL431CLP voltage reference. These are more accurate and stable than zeners. VR1 taps off a variable voltage, and op-amp U2 controls the bias on QD and QP to produce an equal voltage at RF, i.e. an equal voltage across RE1 and RE2. Therefore, the voltage at VR1's wiper determines the emitter current in QP. QD and QP are connected as a Darlington transistor, which has a current gain of at least 1000, so its base current is negligible. Therefore, by regulating the emitter current, the circuit regulates the collector current.

It is possible to make a good constant current circuit without using an op-amp but the op-amp gives predictable operation and requires no offset adjustment. The other op-amp in the package is not used.

RB limits the current into QD and QP so the op-amp does not try to provide the full 5A when the collector circuit cannot pass that current. RF protects the op-amp inputs against high differential voltage during transients.

QD and QP will both require heatsinking as noted on the schematic. RE1 and RE2 may also require heatsinking. The amount of heatsinking needed on QP depends largely on the characteristics of the load (the anodising current path) and I don't know how this will behave. Worst case power dissipation in QP will occur at maximum anodising current and minimum voltage drop across the anodising circuit.

The small components should be assembled on stripboard. Use point-to-point wiring for the high-current connections.

All components are available from Digi-Key (http://www.digikey.com).

 

Use a 7805 or an LM317 with a variable resistance between output and adjustment pins. Take output from the adjustment pin. This will provide a useful and simple constant current source. Surely you will need a TO-3 to stand 0-5A, and a big heatsink.

 

An elegant and simple design, very nice Kris. Why are pins 5,6 of the unused op-amp connected?

 

Thanks chopnhack Really it's pretty conventional. I grounded the pins of the unused op-amp just to stop them picking up noise. For CMOS ICs it is important to connect unused inputs to something, but not really for op-amps. It does avoid one track cut on the stripboard though (between pin 4 and pin 5). (Yes, I do think about those things when doing a circuit design.)

Well, it's the simplest way that I can think of, given the design requirements that I used, which include using standard components, not requiring any setup adjustment, having a linear current adjustment from 0A to 5A, being fairly consistent from one run to the next, and being comfortably over-engineered so it's unlikely to fail.

Miguel has suggested a simple alternative to the regulator section in post #10 and it meets most of those criteria. The problem with that design is R1. For the LM317 current regulator, the R1 resistance for a given output current can be calculated as 1.24 / I_OUT. This gives the following values for typical output currents:

I_OUT = 0.1A --> R1 = 12.4Ω
I_OUT = 0.5A --> R1 = 2.48Ω
I_OUT = 1.0A --> R1 = 1.24Ω
I_OUT = 1.5A --> R1 = 0.83Ω
I_OUT = 2.0A --> R1 = 0.62Ω
I_OUT = 2.5A --> R1 = 0.5Ω
I_OUT = 3.0A --> R1 = 0.41Ω
I_OUT = 4.0A --> R1 = 0.31Ω
I_OUT = 5.0A --> R1 = 0.248Ω

So there are several practical problems with this approach.

If you need low anodising currents, you need a fairly high value variable resistor, and almost all of the useful adjustment will be at one end of the travel. You could use a logarithmic variable resistor, if you can find one with such a low resistance (I doubt it), but then the high current end would be the fully anticlockwise end.

Variable resistors with such low resistance values are pretty specialised components. Digi-Key are probably the biggest component supplier; look at their range and pricing: http://www.digikey.com/product-sear...=1028&FV=fff40004,fff80123&stock=1&quantity=1

Power dissipation in R1 will be over 10W at some settings.

You could use a switch, or several switches, to select various fixed resistors, as mentioned by me and Steve in posts #2 and #3.

The short answer is that that design is my best attempt at what I would use if I wanted to do what you want to do.

 

I think that a continous current adjustment won't be required (just my humble opinion). A step-selector will do the job at the exact value of the several required current. Fixed resistors can be selected to stand high power dissipation. This way, an ammeter is not needed.

 

Yes you can use 2N3054 and 2N3773. They are both a bit low on gain but that's OK. Yes any TL431 will work; the CLP suffix I specified means commercial temperature range and TO-92 suffix, so I think that's what you'll end up with anyway.

No you don't need to use stripboard.

Is that layout viewed from the top or bottom side? If it's a bottom view, the pin numbers on the op-amp are mirrored.

The spacing for the op-amp looks wrong. Pins are spaced 0.1" apart, and the two rows are spaced 0.3" apart. Yours looks like the rows are spaced 0.2" apart.

Ideally CD should be very close to the op-amp and the track to the op-amp and the track from the rest of the circuit should connect on opposite sides of the CD pad. Keep the loop enclosing the op-amp and CD as small as possible.

You may need to heatsink BR1.

I would avoid putting mains voltages on your PCB. It means the PCB is dangerous to work on. The fuse, switch, and incoming power connector can be easily wired using insulated wire with heatshrink to prevent shock hazards.

I would keep all the high-current stuff in one area of the PCB. It means the high-current tracks can be nice and short. I don't like the long track from the negative of C1 to the rest of the circuit.

Personally I would run some reinforcing along all the high-current tracks, unless you used 2 oz copper perhaps. You can use braid, or just thick copper wire - at least 0.5 mm diameter; I use about 0.8 mm.

I don't see any other obvious problems.

 

Lamma, what software are you using for your board layout/schematic?