I'm powering a Model 15 Teletype current loop. The Teletype has a selector magnet resistance of 55 ohms and needs 60mA, so it only needs 3.3V. But because of the huge inductance of the selector electromagnet (4H), a huge voltage is needed to pull in the electromagnet fast enough to follow the incoming data stream.
I'm using the traditional solution, a 120VDC supply through a 2K
10W resistor. It's really dissipating 10W, continuously. A 10W resistor gets hot enough (250C!) that you can't mount one on a PC board. I'm designing an interface board, and would like a better solution.
Basically, I need a constant-current regulator circuit that takes in 120VDC and limits the output current to 60mA. Preferably one that doesn't dissipate too much heat. Suggestions?
John Nagle
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T
Tim Wescott
Switcher. Possibly with an oversized coil to give your control circuit enough time to _really_ keep things at 60mA. You'll need the voltage overhead to supply that relay inductance, but you'll only be supplying that power when you need to.
Of course, if you have to ask it'll take you some time to make it work, but that ain't _my_ cross to bear (heh heh heh).
http://www.wescottdesign.com
J
Joerg
Perhaps a SEPIC switcher that has enough oomph to ramp it's voltage from near zilch to tens of volts within a reasonable speed. AFAIR teletype never ran more than 100bd so this could work. You'd have to regulate for current and keep a cap on the voltage (in case the loop connection goes open). 20:1 is a stretch but possible so you should be able to get to
60V or so.
Another option would be to look at what's available in the stepper motor driver world. They operate in similar fashion, huge voltage overhead to force current into the winding inductances.
Regards, Joerg
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S
Spehro Pefhany
selector
What's the shortest time that the coil is "off"?
S
Spehro Pefhany
selector
Just a thought here.. what if you had an inductor that had similar characteristics to the selector magnet. And a 55 ohm resistor to ground and a 6.6V supply. Now connect one side of the inductor to 6.3V and use the data to switch (using something like a 400V SPDT optomos relay with some TVSs) the other side from the 55 ohms to ground (0) to the selector magnet (1).
W
whit3rd
has a selector
t
2K
How about a class G amplifier: use a sense resistor and op amp to drive two output circuits, one with a modest (12V) power supply, the other with your 120VDC. Use an isolator diode on the 12V output, so it doesn't blow up when 120V shows up on the Teletype...
On turn-on, the sense resistor shows less than 60 mA current, so the op amp output rises; at 3V, the low-voltage output circuit turns on, but (because of the inductance of the load) current is still too low. The op amp output rises to 4V, which turns on the 120V section of the output drive, and the output current rises to 60 mA. A few microseconds later, the current hits 60.1 mA and the op amp output drops, until it's once again at 3V and only the 12V drive circuit is drawing current.
Peak power in the 120V section is the same, but the duty cycle is much lower, and the power efficiency of the 12V (carrying) amplifier is better.
J
James Arthur
A hysteretic switcher could be pretty simple. Here's one for inspiration:
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Cheers, James Arthur
M
MooseFET
has a selector
t
2K
r
The inductance of the coil can be the working inductor of the regulator. A simple comparator can be your controller. You sense the current in the coil and turn off a MOSFET when it passes some upper limit. A diode lets the current continue. When the current drops below some limit, you switch the MOSFET back on.
When the plunger pulls in, the current in the coil will attempt to decrease suddenly. Using the comparator as suggested, makes sure that the MOSFET is on and forcing the current back up while this happens.
R
Rich Grise
An NPN transistor with a regulated voltage at its base (like two or three diode drops) and an emitter resistor to regulate the current?
Hope This Helps! Rich
J
Joerg
That would dissipate the difference, like the resistor does. Except now you need a large heat sink ;-)
Regards, Joerg
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F
Fred Abse
On Tue, 03 Feb 2009 09:56:48 -0800, John Nagle wrote:
The crux of the issue is current rise time, which is equal to L/R, hence the higher series resistance, the faster the rise of current. If you use a resistor (in your case 2K), you're going to need 120 volts to push enough current through it. That's the reason for the "traditional" teletype power supply, which goes back to the early 20th century.
An electronic current source mimics a high resistance, while allowing a lower compliance (ie. open-circuit) voltage. The following LTSpice simulations show firstly a "traditional" teletype circuit, and secondly a suggestion for a simple current source using only a 24 volt supply. Dissipation is less than 2 watts total.
The pulse source represents a TTY signal of 1:1 mark/space (RYRYRYRY)at
50 baud (AFAIK, the Model 15 did 45, 50, or sometimes 56). Compare the rise times of current in L1 in the two examples, whilst not quite as fast in the suggested circuit, it should prove adequate.
Traditional circuit:
Version 4 SHEET 1 880 680 WIRE 64 176 -64 176 WIRE 272 176 144 176 WIRE -64 224 -64 176 WIRE 272 224 272 176 WIRE -64 368 -64 304 WIRE 96 368 -64 368 WIRE 272 368 272 304 WIRE 272 368 96 368 WIRE 96 400 96 368 FLAG 96 400 0 SYMBOL voltage -64 208 R0 WINDOW 3 -429 54 Left 0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR Value PULSE(0 120 0 100n 100n 40m 80m) SYMATTR InstName V1 SYMBOL res 48 192 R270 WINDOW 0 32 56 VTop 0 WINDOW 3 0 56 VBottom 0 SYMATTR InstName R1 SYMATTR Value 2k SYMBOL ind 256 208 R0 SYMATTR InstName L1 SYMATTR Value 4 SYMATTR SpiceLine Rser=55 TEXT -494 506 Left 0 !.tran 100m TEXT 320 264 Left 0 ;R (series) = 55 ohm
LTSpice is free, if you didn't already know that, from
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"Electricity is of two kinds, positive and negative. The difference is, I presume, that one comes a little more expensive, but is more durable; the other is a cheaper thing, but the moths get into it." (Stephen Leacock)
R
Ross Herbert
: I'm powering a Model 15 Teletype current loop. The Teletype has a selector :magnet resistance of 55 ohms and needs 60mA, so it only needs 3.3V. But :because of the huge inductance of the selector electromagnet (4H), a :huge voltage is needed to pull in the electromagnet fast enough to :follow the incoming data stream. : : I'm using the traditional solution, a 120VDC supply through a 2K :10W resistor. It's really dissipating 10W, continuously. A 10W resistor :gets hot enough (250C!) that you can't mount one on a PC board. I'm :designing an interface board, and would like a better solution. : : Basically, I need a constant-current regulator circuit that :takes in 120VDC and limits the output current to 60mA. Preferably :one that doesn't dissipate too much heat. Suggestions? : : John Nagle
Back in the days of relay based telephone exchanges it was often required to lower the power consumed by a relay yet have it operate with a snap action. This was achieved using a second relay to maintain a capacitor charged to exchange battery voltage (-48V) in the normal condition. When the main relay was required to activate the secondary relay would, via its contacts, arrange for the charged capacitor to be connected in series aiding with the main relay coil which had
-48V behind it. Effectively, a double voltage pulse was applied to the main relay coil which snapped in fast but thereafter remained operated via normal
-48V at reduced current.
A modernised version of this arrangement has been devised by Maxim and it may be applicable to your application.
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You will probably need to tailor it a bit but it appears as if it would be a neat solution.
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