Ignoring the fact that your two transistors has already ballooned to two transistors and four resistors...
Put some values on those resistors. Assume some beta values for those transistors and plot the current vs voltage. convince me that there's not an unstable region that oscillates in combination with other components in the power supply circuit as the switcher tries to compensate.
The on/off part is pretty simple. What happens when the power glitches? How do you guarantee meeting the reset requirements of the low voltage lockout circuit. Don't forget the temperature dependence of the threshold.
it hasn't ballooned, that was the circuit to begin with
You can if you want. I'm off to work in a minute. The 10mA or so the bleed conducts is orders of magnitude less than the psu can provide, so there is no risk of them fighting each other. If it switches 2 or 3 times during startup when the psu is under heavy load it's an irrelevance.
you can define that if you want
by picking your base resistors for right tr
10mA at 6v is 60mW. That's somewhere around 20C rise plus Tcase variation for a jellybean tr.
Interesting, but of no use for my application as the output range is too large. Some installer will think they can just crank it up a bit and send 7 - 12VDC through the TTL logic letting out the magic smoke.
I need that in a 5VDC output style (adjustable over a small range), that recovers from partial shutdowns (supply side doesn't fall much below output side), which could be a problem with some designs.
Thanks!
John :-#)#
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John's Jukes Ltd. 2343 Main St., Vancouver, BC, Canada V5T 3C9
(604)872-5757 (Pinballs, Jukes, Video Games)
www.flippers.com
"Old pinballers never die, they just flip out."
On a sunny day (Wed, 10 May 2017 06:59:56 -0700) it happened John Robertson wrote in :
You could replace the trimpot by a fixed resistor for 5V output. The chip it uses is a XL402E1:
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the voltage is set by 1 resistor (in this case a trimmer). See datasheet for the value. Output current sense is done with a shunt and a LM358 dual opamp. There are also some transistors on board. Cannot beat the price.
Turn lamp current : constant current value * ( 0.1 ) , turn lamp current and constant current value linkage. For example, constant current value
3A, turn lamp current is set to 0.1 times constant current(0.1 * 3A =
0.3A). When adjust the constant current to 2A , then turn lamp current is 0.1 times constant current (0.1 * 2A = 0.2A). Turn lamp current is fixed to 0.1 time in this version( actually turn lamp current value is probably not very accurate ), it is for identify whether the battery is fully charged.
I see this on a number of similar but not identical units too.
What do you see if you plot output voltage versus input voltage going up, 0->12V, and down, 12V->0 ?? ...Jim Thompson
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| James E.Thompson | mens |
| Analog Innovations | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| STV, Queen Creek, AZ 85142 Skype: skypeanalog | |
| Voice:(480)460-2350 Fax: Available upon request | Brass Rat |
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Thinking outside the box... producing elegant solutions.
"It is not in doing what you like, but in liking what you do that
is the secret of happiness." -James Barrie
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| James E.Thompson | mens |
| Analog Innovations | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| STV, Queen Creek, AZ 85142 Skype: skypeanalog | |
| Voice:(480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |
Thinking outside the box... producing elegant solutions.
"It is not in doing what you like, but in liking what you do that
is the secret of happiness." -James Barrie
I am way too rusty for transistor work, but why does one actually need C1? Consider that R2 is pulled towards 0VDC as the Vout side has collapsed won't that circuit happily discharge the Cin as soon as Vin is reduced enough that Vout fails? Vout always is loaded after all...it just needs Q1 biased off when the voltage rises above 0.7VDC at Vout, right?
On the powering up side, Q1 nd R1 only work for the fraction of a second it takes for R2 to bias the transistor off...
I just don't have time right now to breadboard this as we are moving the shop - adding about 3,000sq ft to the current foot print...
Thanks, I think this will work exactly as I want but must test first...
John :-#)#
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John's Jukes Ltd. 2343 Main St., Vancouver, BC, Canada V5T 3C9
(604)872-5757 (Pinballs, Jukes, Video Games)
www.flippers.com
"Old pinballers never die, they just flip out."
I really should learn to use Spice, but too many other projects. It looks nice though. Thanks for taking the time to generate that.
John
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John's Jukes Ltd. 2343 Main St., Vancouver, BC, Canada V5T 3C9
(604)872-5757 (Pinballs, Jukes, Video Games)
www.flippers.com
"Old pinballers never die, they just flip out."
No part of Q1 is grounded so the absolute voltage on Vout is not relevant, only in regards to the voltage on C1. When Vout powers down Q1 will not be turned on unless C1 provides current into the base. Otherwise the base and emitter are at the same voltage through different resistances. The idea is to turn on Q1 anytime Vout drops significantly. If Vout doesn't get pulled down by the load this won't work.
I still think a diode in parallel with R2 is better. It brings the cap up with the supply although if it goes right back down again Vout has to drop about 1.4 volts to turn on Q1 (or about 1 volt if you use a Schottky diode).
I was
I'm not sure what you are looking at. Q1 won't be turned on at all by powering up the circuit. A quick enough down and up again before C1 discharges will leave Q1 on until, as you say, Vout is brought up enough. C1 only needs enough charge to bring Vin below the under voltage lockout threshold minimum. It really doesn't need to discharge Cin totally.
If the power is off for some milliseconds C1 will have discharged to a point that Q1 is essentially off. If power is restored before that it can draw current (limited by R2) until Vout starts to rise again. So it will be good to size C1 so it does the job without an excessive hold up time.
Simulation would be easier to test the various cases I think. Maybe I'll try this later if I have time.
Cin is 10 - 20K (20,000ufd), depending on the game - it is right off a pair of diodes making a bridge rectifer.
John :-#)#
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John's Jukes Ltd. 2343 Main St., Vancouver, BC, Canada V5T 3C9
(604)872-5757 (Pinballs, Jukes, Video Games)
www.flippers.com
"Old pinballers never die, they just flip out."
snip Given all the work necessary to get the parts and construct the adapter to make the 5V supply work and package the whole thing... Why not just eliminate the middle man and run the thing off AC?
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Is the total installed cost any greater? Presumably, this one knows how to restart. And the good news is that if one fails or you can no longer get it, you can easily replace it with another standard supply that just works.
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| James E.Thompson | mens |
| Analog Innovations | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| STV, Queen Creek, AZ 85142 Skype: skypeanalog | |
| Voice:(480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |
Thinking outside the box... producing elegant solutions.
"It is not in doing what you like, but in liking what you do that
is the secret of happiness." -James Barrie
With the huge size of the input capacitor I figured you didn't want to use too large a hold cap in this circuit, so I drew it up as a Sziklai pair to get more gain allowing a smaller cap. It appears to work, but the draw down time is pretty long with a 220 ohm resistor (20 mA). Even at 200 mA it takes a while. One advantage with this circuit is the current is limited by the voltage retained on the C1 cap rather than drawing a current proportional to the supply input voltage.
It cuts in an out like I expected. I can't tell if the current draw will be a problem or not. That depends on the details of your circuit and the timing. Also, I believe your circuit will draw down the input cap voltage a lot before the output goes out of regulation. That helps with the timing.
So I added a 4.7 volt zener diode in series with a 1 ohm resistor to provide a pulldown similar to the load. Now the input voltage drops quickly to a point where the PSU would stop providing output current (approximately 6 volts input) and the output voltage drops off with a 50 ms fall time. 150 ms later the input voltage has dropped below 4 volts.
With these values the C1 hold cap seems to hold up ok. It drops with the voltage on Vin since the transistors are essentially an emitter follower and if the collector current voltage drops it draws more current from the cap, but that's ok since that still brings the input down below 4 volts.
If you want to try simulating this circuit, here is my schematic. LTSpice is a bit awkward to learn, but being able to simulate this stuff can be very useful.
Version 4 SHEET 1 1060 680 WIRE -48 48 -144 48 WIRE 32 48 -48 48 WIRE 192 48 96 48 WIRE 224 48 192 48 WIRE 400 48 224 48 WIRE 624 64 576 64 WIRE 640 64 624 64 WIRE -144 96 -144 48 WIRE 192 96 192 48 WIRE 640 96 640 64 WIRE -144 208 -144 176 WIRE 192 208 192 160 WIRE 192 208 -144 208 WIRE 640 208 640 176 WIRE -144 240 -144 208 WIRE 400 240 400 48 WIRE 272 288 240 288 WIRE 336 288 272 288 WIRE -32 336 -112 336 WIRE 80 336 32 336 WIRE 128 336 80 336 WIRE 176 336 128 336 WIRE 80 368 80 336 WIRE 368 384 240 384 WIRE 400 384 400 336 WIRE 400 384 368 384 WIRE 400 416 400 384 WIRE 80 464 80 432 WIRE -112 528 -112 336 WIRE 400 528 400 496 WIRE 400 528 -112 528 WIRE 576 528 576 64 WIRE 576 528 400 528 FLAG 80 464 0 FLAG -144 240 0 FLAG 640 208 0 FLAG 224 48 IN FLAG 624 64 OUT FLAG 128 336 CAP FLAG 272 288 JUNC FLAG 368 384 LOAD FLAG -48 48 PSU SYMBOL voltage -144 80 R0 WINDOW 3 -202 -92 Left 2 WINDOW 123 0 0 Left 2 WINDOW 39 -202 -61 Left 2 SYMATTR InstName PSUINPUT SYMATTR Value PULSE(0V 15.3V 0 100ms 100ms 100ms 500ms) SYMATTR SpiceLine Rser=0.01 SYMBOL voltage 640 80 R0 WINDOW 3 -253 -72 Left 2 WINDOW 123 0 0 Left 2 WINDOW 39 0 0 Left 2 SYMATTR InstName PSUOUTPUT SYMATTR Value PULSE(0V 5V 20ms 50ms 50ms 200ms 500ms) SYMBOL npn 176 288 R0 SYMATTR InstName Q1 SYMATTR Value 2N3904 SYMBOL pnp 336 336 M180 WINDOW 0 57 32 Left 2 WINDOW 3 57 68 Left 2 SYMATTR InstName Q2 SYMATTR Value 2SAR552P SYMBOL cap 176 96 R0 SYMATTR InstName C1
SYMBOL cap 64 368 R0 SYMATTR InstName C2
SYMBOL schottky -32 352 R270 WINDOW 0 32 32 VTop 2 WINDOW 3 0 32 VBottom 2 SYMATTR InstName D1 SYMATTR Value BAT54 SYMATTR Description Diode SYMATTR Type diode SYMBOL res 384 400 R0 SYMATTR InstName R1 SYMATTR Value 4.7 SYMBOL schottky 32 64 R270 WINDOW 0 32 32 VTop 2 WINDOW 3 0 32 VBottom 2 SYMATTR InstName D2 SYMATTR Value BAT54 SYMATTR Description Diode SYMATTR Type diode TEXT -86 562 Left 2 !.tran 1000ms
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