Power Supply controled by 0-10V command signal?

Hi,

I am having an extremely difficult time finding a suitable ac to dc power supply.

I am looking for one that can be controlled by a 0-10V command signal to give a variable current output at ~5V.


I would also like for the power supply to be linear and open frame, if possible. Any suggestions on unit close to these requirements is greatly appreciated.


Thanks!

Actually, it sounds like he wants a voltage controlled current source, but the ~5V requirement is a bit confusing. Unless, this is a voltage-controlled-power source where the output voltage is constrained to around 5V -- very unusual, indeed -- and seemingly impossible -- I mean, how would the load know to change in order to comply with the power regulation requirement -- unless it, also, is altered by the 0-10 control voltage?!?

Ah-ha! I thought of a way: the LOAD is a voltage controlled resistor [VCR]! That way, the voltage across the VCR is always ~5 and the resistor causes the current to vary in terms of the 0-10 control voltage.

Ah sorry if the question was confusing - I am very new to this field.

I am trying to save some money and design a heating element for use in biological research. The temperature is controlled by a PID controller with input from a thermoprobe and output of a linear 0-10V signal. I want the 0-10V signal to control a linear power supply from 0-2A. The 5V was because the heating pad connected to the power supply can have a max of 10W.


I like the idea of having a fixed output 2A power supply and regulating the output with a VCR from the 0-10 control voltage.

I would like to use the 0-10V signal to regulate the output (5V @ 2A) from this power supply "HAA512AG (http://www.digikey.com/product-detail/en/HAA512-AG/179-2304-ND/1634147)":http://www.digikey.com/product-detail/en/HAA512-AG/179-2304-ND/1634147

Does anyone have suggestions on what to do from here?

Then, how about the attached. The op-map circuit is the most stable and accurate, but you might be able to get away with the single transistor version. The single tran version will put a max load of around 8ma on the PID output, so if that is too much, then use the op-amp circuit.

Now, this isn't a direct [linear] transfer of voltage to current, because the resistance of the heating element will vary with temperature, but this would be the most efficient way to do this [at least using analog components - rather than PWM] and the PID should be able to make the corrections.

Let us know if you actually need a linear transfer of voltage to current.

Forgot the attachments...

Tough to regulate 5V and have a dynamic range that includes 5V, because some "headroom" is needed due to circuit losses. But, the single transistor (darlington) circuit will come close, but you would sill need around 5.7V minimum (the label in the circuit is wrong). SOmetimes the output voltage on a power supply can be "tweaked" and perhaps you can run it up to 5.7V or beyond. See there is a datasheet for the supply and if it specifies that -- but make sure that it's still guaranteed to supply 2Amax at the higher voltage.

Actually if you are using a PID with a thermal couple to keep a constant temperature you would not need the op amp. Besides if you only have 5V, the circuit with the op am will not work. What would work better in this case so you can get around the Vce drop if that's a problem is a logic level power MOSFET. The gate-source could be driven directly from the 0-10V PID, and you don't care what that is because it regulates the temperature. The heating element resistance is all over the place anyway as it heats up, minimum when cool and max when hot so a constant current is not what you want, you want the MOSFET R_dson to be quite small with respect to the heating element, say less than 100mOhms.

If this were really efficient it wouldn't be a linear either, it would be switched PWM, but that would be another design.

So just replace the NPN with an N-channel logic level MOSFET, and the gate-source drive comes from the PID, which share the same common return, like shown with the NPN darlington, but without the base resistors.

Of course -- MOSFET...duh! So used to thinking of them in terms of PWM and SMPS

Well either one would work, it just may be a logic device MOSFET may work better perhaps. In very high power apps , bi-polars are still choice because R-dson is too high even today and parallel MOSFET devices have their own headaches.

Thanks for the advice!

I have attached the circuit as of my understanding so far.


What type of N-channel MOSFET would I use?

Also, why are the resistors 420 and 820 ohms.

Sorry if my questions are elementary, but I really appreciate the help.

2n7000 MOSFET is not the correct device for this app and will not work.

Can you give me advice on which one to choose?

I imagine I would want a MOSFET operating in the linear region/ohmic mode, so the threshold voltage is 10 volts?

Here is a page URL from IR, makers of many MOSFET power devices:

http://www.irf.com/whats-new/nr090428.html

Without knowing the specs on your heating element, I can't tell you how to do your MOSFET thermal design steady state. If you designed for 5W, which is over kill I would be sure, you are probably ok. Any of the 40V devices listed on this page will work and you may have to experiment to determine how much cooling you need for the MOSFET.

As for the gate drive you don't need any gate resistors.

You can look around for alternatives to the IR devices. This is just a guide.

There are plenty of MOSFETs with low enough Rdson. It's only 2Amps and the frequencies are quite low, so I'm not sure what the concern is.

No the idea is to use a logic MOSFET so that at 4.5V the MOSFET is fully inhanced not like 8 - 9 V. Yes the fet will operate in the ohmic region and perhaps in the saturated constant current region. This is controlled by your closed loop PID.

As Earl pointed out, you don't need the gate resistors. BUt, for the bi-polar application, here's how I come up with those values:

When the input voltage is at 10V, we want 5V at the heating element. For that to happen, the voltage at the base of the transistor needs to be around 5 + 1.4 = 6.4 [1.4Vbe].

I figured the Hfe of the darlington to be around 5000, so at 2 Amps, the base current [Ib] is going to be around 2Amps/5000 = 0.4ma. Set the current through the voltage divider (i.e. the resistors) to 0.4ma * 20 = 8ma (multiply it by 20 to swamp out the effect of the base current).

So, the lower resistor (the one between the base and ground) has 6.4 volts across it, so it's value is: R = 6.4V/8ma = 800. Nearest standard value is 820.

Which will but the divider current at: 6.4/820 = 7.8ma

Then, to get the upper resistor, subtract 6.4 from 10 = 3.6V, which is the voltage across the upper resistor. If we ignore the influence of the base current: R = 3.6V/7.8ma = 462

Now, lets try that with the expected base current and see how different it is: R = 3.6V/(7.8ma + 0.4ma) = 439, so I went with 420. But, perhaps 470 would be a better choice. It all comes down to what the base current will actually be, and that will depend on the Beta of the transistor.

I was using 5% standard values but you could also use 1% resistors and come closer to the calculated values. Also, if you up the ratio between the divider current and the base current (i.e. instead of 20 use say 50 or 100, etc), the base current will have even less influence on the divider voltage, BUT that will put more of a load on the PID output.

The advantage of using a MOSFET is there is no need to divide the PID output voltage -- thus no need for the dividing resistors. You could also achieve the same thing (i.e. eliminate the divider resistors -- replace them with a single current limiting resistor) by applying around 5.9 to 6.4 volts to the collector of the transistor (depends on what the Vce-sat is).

This just in: I took a look at a transistor selection guide (http://www.mouser.com/catalog/supplier/library/pdf/FairchildBipolarPower.pdf) and discovered that an Hfe of 1000 to 2000 is more realistic.

I would pick a MOSFET that allows an Id of 2Amps at around 8 to 9 volts on the gate (take into consideration effects of heat), that way you get the best dynamic range.

I am having trouble understanding what the optimal values for Vdss, Id, and Rds should be.

I would run the setup so that at all times 5v at 2A is connected to the drain, and a variable resistor (heating pad) that can withstand a maximum of 5V, 10W is connected to the sink. The 0-10V is connected to the gate.

Any help on figuring out these parameters would be greatly appreciated

Am I right in thinking that I want a Vds of 5V and Id 2A? This way I can maximize my movement in the Ohmic range with the 0-10V command signal?

Vdss needs to be at least 1.5 to 2 times the maximum voltage that will occur across the Drain and Source. Sounds like the max voltage across the drain/source is going to be 5V, so the Vdss needs to be at least 15, but I would call it 20 (bare in mind that I said "at least" -- it can be more with typical values of 30V, 40V etc.).

Id max is 2Amp but consider that the max power dissipation the FET will need to endure is something less than 5W (it's hard to predict because the heating element resistance is not constant -- in fact, the current could go much higher than 2A while the element is heating up, but the voltage across the FET will likely be fairly low during this time, so most of the power will be delivered to the element). So, for safety margin, use a FET that can handle 10W or even 20W.

Rds needs to be low enough to allow 2 Amps to flow without dropping appreciable voltage when the FET is in saturation. Lets say we want the full on resistance of the FET to drop 0.1V at 2 AMP, that's 0.1V/2 = 0.05ohm. 0.1V at 2 amp means a power loss of 200mw in the FET (0.1V x 2A).

If the current in the heating element were to rise to, say 5 amps during the brief time it is heating up, the voltage drop across the FET will be 5amps x 0.05 ohms = 0.25 volts and the power lost in the FET will be: 5amps x 0.25 volts = 1.25 watts -- but only briefly. Once the element heats up, the PID will cause the FET to throttle back the current (unless, of course, the heating element is inadequate for the job).

Get a FET with an even lower Rds and even less power will be lost in it and more delivered to the element.

I already supplied the person with a MOSFET part number from IR that would work fine, just drop it in and test it.

I think he wanted to know how to figure it out for himself, hence the follow-up question.

You know, teach a man to fish...

But, Earl is right, any of those MOSFETS should do the trick, with plenty of margin!

And, you might also want to try one of the ones optimized for 10Vgs:

* http://www.irf.com/whats-new/nr081216.html#specs

True if the MOSFET has enough gm (assumes in saturation region constant current region) even a 10V Vgs would work. Just thought a logic MOSFET would give more options. No details if someone doesn't understand what I'm getting at. I have been doing analog design for some 20 years. I learned a lot by just trying and failing and then trying to figure it out, more so than if it actually worked. Yes I got the shingles, several of them, so what.

If someone wants a quick answer that's what they get, not a 4 year degree in Analog design. Try the IR part and then if you want more then feel free to explore what won't work and why.

Then you should listen to Earl over me, because I'm more of a digital/MCU guy with only secondary experience with analog

Thanks Steve and Earl!

I just ordered a few MOSFETs to test which ones work the best and my design is all complete.