Precision current Supply with voltage reference - Page 1
Forum section: electronics / projects

author: PepeChip
author_rank: Contributor
timestamp: March 20, 2017, 05:26:13 pm
content: Hi,I am trying to use the AD584 to get a high precision 10V with an output current of about 200mA. This devices can output up to 10mA.I am trying to copy the circuit which comes in the datasheet in the page number 8:http://www.analog.com/media/en/technical-documentation/data-sheets/AD584.pdfInstead of an 2N6040 I am using a 2N2905 (PNP whose Icmax is 600mA) I had around and at the output I have a 10uF capacitor.My circuit does not work as expected but honestly I also do not understand how the circuit they propose would work. Could somebody help me understand this please?Many thanksRegards

author: Kleinstein
author_rank: Super Contributor
timestamp: March 20, 2017, 06:06:07 pm
content: The idea of the circuit is to use the PNP transistor to boost the current: if the supply current of the AD584 goes over about 4 mA (with the 2N6040 darlington) the transistors turns on and supplies extra current to the output.With an 2N2905 the 470 Ohms resistor should be replaced by a smaller one (e.g. 270 Ohms) as the voltage is only about half with a normal transistor instead of a darlington.That circuit might not like a low ESR capacitive load at the output, even with the 0.1 µF cap. A small resistor at the emitter side of the transistor might help to improve stability against oscillation.It also requires a relatively stable supply with decoupling caps.

author: PepeChip
author_rank: Contributor
timestamp: March 21, 2017, 09:10:29 am
content: My idea was that we could see problems if the resistor is too small or too high but the only difference between 470 Ohm and 270 Ohm is the moment at which the PNP turns on.I tried with the resistor that you propose and face the same problem. After 10mA the output voltage stops being a perfect 10V and starts decreasing.Maybe this is something I have to count on anyway but as they present it in the datasheet I was hoping that I can get that perfect 10V.Is the behaviour I am having the expected one? or is it possible to get that perfect 10V at Iout ~ 200mA?

author: blackdog
author_rank: Frequent Contributor
timestamp: March 21, 2017, 12:12:39 pm
content: Hi PepeChip,Kleinstein already explaned wath i would say about de Darlington repalcement.But there could be another prpblem, and this is generation of the 2N2905 it need a resistor directly on de base of about 47 Ohm.But also this, the 2N2905 is not good enough for 200mA, the Hfe drops quickly above 100mA.And the dissipation is nog high enough.Just a bad part for your application!Than anything else, how "perfect" should this voltage reference be?Have you thought about the wiring problems?Can you show some pictures of your test setup?Kind regardsBlackdog

author: PepeChip
author_rank: Contributor
timestamp: March 21, 2017, 01:26:08 pm
content: Thanks to both of you for your help.With "perfect 10V" I meant the highly accurate 10V that the AD584 gives. Not 9.85V, for example.For some reason everything got solved when I removed the 100n capacitor between the emitter and base of the PNP. Just that.The power dissipated would be about 2W (Vce ~ 10V) I think (neglecting the power at base-emitter), this is way too high without heat sink but It will not work continuously at 200mA. It was simply to learn how to use those devices in case I will need it in the future. I was surprised at the application that AD shows in the datasheet and wanted to put it into practice.For the time being I am using this as an analog supply for an ADC.

author: tggzzz
author_rank: Super Contributor
timestamp: March 21, 2017, 01:32:04 pm
content: Quote from: PepeChip on March 21, 2017, 01:26:08 pmDo a sanity check to ensure any heat dissipated in the PNP transistor doesn't cause the output voltage to change too much for your purposes (whatever they are).

author: Kleinstein
author_rank: Super Contributor
timestamp: March 21, 2017, 01:38:42 pm
content: The circuit with the 100 nF capacitor in place is sensitive to supply ripple. So it really needs a low ripple supply (e.g. 15 V). With significant ripple on the supply, it could end up haven the 10 V as maximum output, but about the same ripple at the output as at the input. So on average a lower voltage.Without the capacitor the circuit gets less sensitive to supply ripple, but it can be sensitive to a capacitive load and might oscillate. Something like a 10 µF electrolytic cap at the output might be still OK as it could have enough ESR to be acceptable.

author: PepeChip
author_rank: Contributor
timestamp: March 21, 2017, 01:48:07 pm
content: Quote from: tggzzz on March 21, 2017, 01:32:04 pmi did it. And I started having problems at ~20mA. I do not think it was the heat at that pointQuote from: Kleinstein on March 21, 2017, 01:38:42 pmThe input supply is a 20V laboratory power supply. I measure the ripple of the AD584 with the oscilloscope and is < 20mV under these conditions. But I will try it out with an electrolytic 10uF to see if it has some effect.Thanks