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Originally posted by Qiaozhi
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Hello,
on Bandido Schematic there is a LED named VD8 (it looks like a Battery check LED). On the PCB there is a normal 1N4148 Diode. I have soldered a 5Volt LED on PCB (see attachment) but it doesen´t work. So what is wrong ??
By the way, the LED is mounted in the right direction
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ok obviously nobody use the battery test diode. No matter.
Does anyone know what is meant by the symbol in the schematic? in RX Side? Must it connected to Ground or Case?
RegardsComment
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There is no need to make this connection to ground, as 5-pin Tesoro coils already have this connection made inside the search head.Originally posted by Firebox View PostComment
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You have two diodes in your version of the battery test circuit. Are they both oriented correctly?Originally posted by Firebox View PostComment
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Thanks for the info @Qiaozhi
@porkluvr: I only see one Diode for the Battery test circuit in the Schematic. VD8 (LED) should be this one, but it doesen´t work. Which other diode do you mean?
By the way, I mean a LED Diode, not a normal diode
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The other diode is a 4.3V zener. I see it on the PCB located just above the point of the arrow you drew, but I can't really read the designator on your schematic (VD?).Originally posted by Firebox View PostThanks for the info @Qiaozhi
@porkluvr: I only see one Diode for the Battery test circuit in the Schematic. VD8 (LED) should be this one, but it doesen´t work. Which other diode do you mean?
By the way, I mean a LED Diode, not a normal diode
A question arises in my mind... judging from the footprint on your Bandito SMD PCB, I see that your battery test indicator (LED) is a surface mount device. I think it would make more sense for that LED to be a through-hole device, mounted on the front panel. eh? But that's easy enough to take care of later.
edit: OK, that 4V3 zener looks like it's designated as VD9
another edit: Firebox, exactly, what is your battery test circuit doing, or not doing, that you think is wrong?? (I think I see something wrong with the circuit, but I want to read your answer before I elaborate.)Comment
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I'll disregard the last statement of my previous post.
Q: What did you mean by "I've soldered a 5V LED..." (#17 above)
I am not familiar with your meaning!! (What is a 5V LED??) Seriously.
The way I think your battery test circuit works: it does not illuminate the LED so long as you have a good battery. The LED illuminates only after the battery has dropped to a predetermined level, set by the 4V3 zener and the resistors R70/R71.
Problem: With R70 = 4k3 and R71 = 3k3, your "battery low" indicator would illuminate at Vbat =~ 9.2V which seems awfully high to me.
I don't know why your LED won't light, because 9.2V is almost fully charged for a 9V battery... unless maybe you are using 12V?? Or unless your "5V LED" is something wierd. I think the circuit uses a normal red (or maybe yellow) LED.
Anyway, if I were you I would replace R71 (3k3* on your circuit diagram). Use maybe 7k5 or 8k2. That would make your VD8 LED trigger at a battery voltage of around ~6.3 or ~6.0V. That would still be enough juice for reliable detection but, that's just a guess. You would adjust R71 up or down if need be.
I cannot understand why on earth the original designer would have specified 3k3 for R71, but then he did add an asterisk which gives some wiggle-room.
There is a formula to determine exactly where the battery-low will trigger as determined by your circuit components but I am too rusty to pull that out of my brain right this minute.
EDIT: Oh, wait. I get it.
A "5V LED" already has a limiting resistor in one of the leads. But, you don't want or need another limiting resistor because the current limiting function is already performed by R68 (1k5). If you make the series resistance too high, the LED will be poorly illuminated (or not at all). You might try jumpering across R68 to see if the LED illuminates, but you should be aware that if there is NOT a limiting resistance present, the LED will likely be destroyed.
After you get that straightened out you still need to adjust the value of R71 imho.Comment
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Hi,
I tested as follows:
I have connected a power supply for voltage regulation with Bandido. Then I turned the power back slowly. At about 5 volts the Bandido is started "beeping" from speaker. but the soldered LED doesen´t work LED means "light-emitting diode"
I have tested the LED with the power supply. It operates on 2 volts! but not in the Bandido Battery Test circuitComment
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If you have a variable power supply you can test the battery-low circuit with a multimeter.
Place black meter lead on ground, and the red lead goes to DA8 pin-1 (battery-low output). Start the power supply at 9.5V and then lower it until the DA8 pin-1 output changes
The meter voltage should normally be +5V, and then suddenly swings to -5V as the battery voltage decays below the trip point. You should also be able to detect battery-low at the junction of R68 and the LED (that would be your blue wire).
Is that a 4.3V zener you have at VD9?Comment
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Now i have measured as shown in the picture. If I lower the voltage, this voltage is going down too. But there are no -5 volts on pins
By the way, yes it is a 4V3 Zener Diode, but it looks like an normal diode
RegardsComment
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I have found the problem. Pin1 from DA8.1 was not correctly soldered to the R68! The LED works now, but the Basic Voltage must be over 9,5 Volt to deactivate the LED and the sound. My 9 Volt Block haven´t enough voltage to do this. If the voltage drops down to 9Volt, the Bandido is beeping and the LED is on ...... so what can i do to reduce this ??? The Bandido can work until 5 Volt. I belive i must reduce one Resistor, but which one ?
RegardsComment
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Good deal. Wade through post #23 to see what you can do to lower the battery-low setpoint.
I'll try to derive the set-point formula for you later today (er tomorrow, as one day's start and another day's finish is a relative concept).
The formula is simple, based on the voltage divider formula. But, performing even that simple math is beyond me, at this time.
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I think that the Bandito beeping at 9V is another problem altogether.
I see no obvious connection between sound and the LED circuit, but I will take another look. (LATER)
edit: So, what negative voltage did you finally manage to squeeze out of the negative charge-pump circuit (VT3, VT4, VD1, VD2, and etc.?
If you can get ahold of some schottky diodes either in the SOD123, or also MELF package (the little glass cylinders) will work, then replace the 1N4148 and you could squeeze another -1V, maybe more, from your charge-pump.
Either MBR0530 or BAT65 would work (both in the SOD123 package).
Also, MBR0520, or MBR0540 should be good.Comment
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Hi,
BEVOR i soldered the Pin1 from DA8.1 to the R68, the Bandido was beeping at 5Volt, and at 5,5 Volt the Bandido was working!
Now the Bandido is beeping at 5 volt and the led is on. At 6 Volt beeping is off, but the LED is always on until 9,4 Volt. At 9,5 Volt the Bandido beeping again and LED is OFF and the Bandido is working .... under 9,4 Volt Bandido is not working and LED is on and there is a short beep from Bandido ....
Negative Voltage are -2,02 Volt
EDIT: I believe the "beeping" Battery checker works good! but the LED Checker Circuit ist set to high! it must be lower.... I will make a video, so you can see the function ....
RegardsComment
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Replace the 3k3 resistor, R71, with something around 6k8 or 7k5. That will set the trip-point at around 6.7V or 6.3V and maybe it will stop bothering you. Using a higher value for R71 might be better, but you can work on that after any other bugs have been worked out.
I hope you have a supply of SMD resistors on hand, and of course you can easily parallel resistors as necessary.
G'day. G'night (whatever).Comment
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