Hero999

I hope more people will post. I can't see the quote button; where is it?

It's not too critical, as long as the op-amp can tolerate 44V but unfortunately not many modern op-amps can. You could use the OP27 but you'll need to modify it because the offset trim configuration is different. http://www.analog.com/media/en/technical-documentation/data-sheets/OP27.pdf Another possibility is the NE5534 but it requires a compensation capacitor and has higher bias currents. http://www.onsemi.com/pub_link/Collateral/NE5534-D.PDF4

There have been no posts now for over a week, apart from this thread. The forum has died, since the upgrade. The new software may be more secure but it's no good if half of the old forum content is gone, along with the members. The new software, also seems to be less easy to use and lacks features or they're hidden. Where is the quote button? How about the smilies and fonts? What about the preview button to check ones reply, before submitting it?

Thanks. I suggest you wait and see how many people are using the new forum (so far it's just you and me!) before deleting the old one and switching to the new one.

Can't you just go back to the old forum software? What was wrong with it?

Since the forum upgrade, all posts made since 24 February have vanished. What went wrong?

Yes K, A G does normally mean SCR. But the socket seems to indicate it uses the whole waveform and there are only two diodes, rather than the four expected in a full wave rectifier. You really should try to draw the schematic, even if you're not sure of the components. Just draw the TRIAC/SCR as a box with the pins numbered from left to right (front view). Another thing you could do is consider stripping out the original circuit and replacing it entirely with a standard phase controller circuit.

It works, congratulations!

Are you sure it isn't a TRIAC? Have you traced the whole schematic and drawn it?

Sorry I meant SCR not TRIAC. It's difficult to know without a schematic. How is this implemented? Is it a universal motor run off rectified mains?

1) That doesn't sound right but some regulators do need a minimum load current to regulate properly. What does it say on the datasheet? 2) What resistor value are you using? That sounds like the expected short circuit current. Is the meter in series with the resistor or just connected across the meter? 3) That sounds right.

Are you sure it isn't just the TRIAC? Perhaps the motor has failed?

It can't work as a lamp dimmer because it's zero crossing. For a lamp dimmer you need a TRIAC which can be triggered in the middle of a cycle. This solid state relay can only be triggered at the start of the AC cycle when the voltage across the TRIAC is near zero. It will then remain on until the current falls to zero. It could be used to control a heater or turn a motor or lamp on and off.

It's a solid state relay with a DC input and TRIAC output. A solid state relay does not convert the voltage. It enables a low current signal to switch a higher current on and off. This type of relay has a TRIAC output. It will only turn off when the current flowing through the output falls to near zero and it has zero crossing so it will only turn on when the voltage across the TRIAC on the output is near zero. It is only suitable for switching AC. http://www.farnell.com/datasheets/1672165.pdf Here's a simplified schematic of what's inside an AC solid state relay:

It seems like your circuit is working. The current drawn from the capacitor by the switched mode should increase as the voltage drops. This is because the regulator tries to keep its output voltage irrespective of the input.

Don't confuse resolution with accuracy. They are different things. Of course it's possible to make a thermometer with a resolution of 0.001oC but it won't be necessarily be any more accurate than a cheap one with a resolution of 1oC Over what temperature range? Getting a high accuracy over a large temperature range isn't easy and it certainly won't come cheap.

It's made by lots of other manufacturers. Mouser has a lot in stock. http://www2.mouser.com/Search/Refine.aspx?Keyword=6N138&FS=True

http://www.allegromicro.com/en/Design-Center/Technical-Documents/Hall-Effect-Sensor-IC-Publications/Method-for-Converting-a-PWM-Output-to-an-Analog-Output-When-Using-Hall-Effect-Sensor-ICs.aspx http://www.cnblogs.com/shangdawei/p/3312084.html http://www.avr-asm-tutorial.net/avr_en/AVR_ADC500.html

How is the circuit physically constructed? Is it small and compact or are there lots of long leads and wires? Did you read pages 10 & 11 on the data sheet on inductor selection? http://cds.linear.com/docs/en/datasheet/3639fd.pdf

I doubt the circuit is capable of driving a relay. You can probably connect the base to the output of the circuit which drives the LED. It will already have a current limiting resistor for the LED which should also be fine for a transistor. The ZTX690 is widely available. Digikey and Rapid Electronics stock them. http://www.digikey.com/product-search/en?vendor=0&keywords=ZTX690B http://www.rapidonline.com/Electronic-Components/Ztx690b-Tran-Npn-45v-2a-Eline-81-0230

That's odd. The ON semiconductor datasheet lists a maximum IC of 800mA but Hfe is only specified at 300mA, Hfe > 60 VCE = 1V. http://www.onsemi.com/pub_link/Collateral/BC337-D.PDF 600mA with a forced beta of 30 doesn't seem unreasonable but a heatsink would be a good idea. Don't forget the minimum beta is normally specified across the temperature range (-55C to 100C) so at room temperature or warmer, it'll be much higher than the minimum. Failing that, use an higher beta transistor: http://pdf.datasheetcatalog.com/datasheet/sanyo/ds_pdf_e/2SC3807.pdf http://www.classiccmp.org/rtellason/transdata/2sc3616.pdf http://www.diodes.com/datasheets/ZTX690B.pdf

The easiest way to do that is to have two circuits: one to charge the battery and another to discharge it. A single pole RC filter is not enough to convert PWM to a steady DC voltage. You need to use two poles or more to get the voltage anywhere near smooth.

That inductor doesn't seem to be what you want. Can you post a schematic of the whole circuit.

It's more likely that each LED has a series resistor to limit the current and the LED foward voltage is 2V. If you remove an LED, you'll probably find the voltage increases to 3V. The BC337 might do. It's specified with a minimum Hfe of 40 when IC = 500mA and VCE = 1V http://pdf.datasheetcatalog.com/datasheet/philips/BC337_3.pdf

The 2V signal is probably only 2V because it has an LED across it which is driven off 3V with an appropriate series resistor. You need find out how the LEDs are wired before you can interface with the board.