-5V converter outputs -2.5V instead - Page 1
Forum section: electronics / beginners

author: ppTRN
author_rank: Regular Contributor
timestamp: August 31, 2023, 01:20:43 pm
content: Hi everyone,I needed a -5V source to power some op amps and I brutally copied the one unsed in this devicehttps://www.hobby-hour.com/electronics/s/tesoro-metal-detector.phpThe 5kHz are correctly generated by a PIC microcontroller. The problem is that the output voltage is not correct being half of what needed. All components seems fine and no one is overheating.It is my understanding that the final stage is an emitter follower, and the BJT's Veb should be subtracted to the voltage on the base. So on the base of T7 I should have at least -5.6VI checked the circuit multiple times and seems fine.Load resistence is 1kThanks to anyone who can help me!

author: magic
author_rank: Super Contributor
timestamp: August 31, 2023, 01:32:45 pm
content: This makes no sense.T3 charges the capacitor through the grounded diode, the capacitor can't get more than maybe 4.4V across it. It could be a little more if the diode were Schottky type.Then left side of the capacitor is grounded by T4, its right side is at -4.4V and the output is pulled through another diode down to no more than 3.8V.So this part works as expected, and you won't get better than -3.8V, or maybe -4.something with Schottky.The final transistor may frankly be pointless, unless you find the 5kHz "leaking" into the signal path. It certainly won't make the output more negative.If -5V is strictly required, try a voltage doubler like Cockcroft–Walton.

author: Ian.M
author_rank: Super Contributor
timestamp: August 31, 2023, 01:51:22 pm
content: You loose a bit in T3,T4 as they cant drive rail-to-rail, so at best there will be something like 4.8V pk-pk of drive to the charge pump. Its two diodes (if silicon) will drop another 0.6V to 0.7V each, so approx 3.3V to 3.5V out, *if* the flying capacitor C4 is big enough for the load current at that switching frequency.T7 is configured as a capacitance multiplier, which drops 9% + Vbe, so 2.5V out is within expectations for this circuit, and *if* it works the '-5V' rail is misnamed . . .The *BIG* question is how much current do you need at -5V?If its under 10mA an ICL7660 Switched Capacitor DC-DC converter or equivalent may well do the job. Beware of cheap Chinese knockoffs of the '7660' as I've seen some that failed at under 5V in vs max. 10V operating for the genuine ICL7660.Edit: corrected '9660' to '7660'

author: ppTRN
author_rank: Regular Contributor
timestamp: August 31, 2023, 01:58:11 pm
content: Thank you both for the replies, I now understand why this circuit cannot work as intended. Maybe there is a typo on the schematic and the voltage driving the push-pull transistors is 9V (available in that circuit form a battery) and not 5V. I will try with 9V and I guess that it'll be just fine.

author: Ian.M
author_rank: Super Contributor
timestamp: August 31, 2023, 02:17:19 pm
content: It wont be fine as it would need rail-to-rail base drive. With only 5V drive from your PIC and 9V supply, T3 never turns off so at best it and T4 will be hot and bothered (assuming Ic is hFElimited) and at worst it will let the magic smoke out!I'd go with magic's suggestion of a voltage doubling charge pump. Duplicate the C4,C5 D6,D7 circuit, but make its 'ground' the top (negative end) of the existing C5. That will give you close to -5V, unregulated. Swap the diodes out for Schottky ones to get another volt or so headroom and swap R68 for a 5.6V Zener, cathode to ground to regulate the output at approx -5V.

author: ppTRN
author_rank: Regular Contributor
timestamp: August 31, 2023, 02:25:24 pm
content: what if I drive T3 and T4 using another transistor that will shift the square wave to 9V? This way T3 should turn off.

author: Ian.M
author_rank: Super Contributor
timestamp: August 31, 2023, 02:29:45 pm
content: That will get proper turnoff for T3 but may not work well, unless the new resistor is *much* smaller than the R5, R6 series base resistors - try 1K.Of course that's an over 4mA increase in average current consumption, so I *hope* you aren't powering it from a PP3!

author: Benta
author_rank: Super Contributor
timestamp: August 31, 2023, 02:30:02 pm
content: You prolly mean 7660. Sinple, cheap and easy.If more current than 10 mA is needed, the LM2660/61 are available but cost quite a bit more.

author: Zero999
author_rank: Super Contributor
timestamp: August 31, 2023, 06:45:06 pm
content: If only 1mA or so is required and it can be a little less than 5V, just connect a voltage doubler to the output of the microcontroller.

author: RFDx
author_rank: Regular Contributor
timestamp: August 31, 2023, 07:36:00 pm
content: Quote from: ppTRN on August 31, 2023, 01:20:43 pmTwo more diodes and capacitors are required. After the C-multiplier stage you will get at least -5.5V @ 1k load.

author: Zero999
author_rank: Super Contributor
timestamp: August 31, 2023, 07:42:13 pm
content: Quote from: RFDx on August 31, 2023, 07:36:00 pmOnly two diodes and capacitors are required, as per my schematic.The problem with that circuit is both transistors will turn on, when the output from the microcontroller is changing from 0V to 5V, causing large supply current surges. If the MCU faults and has its output set to high impedance, both transistors will turn on continuously and it'll smoke.

author: Benta
author_rank: Super Contributor
timestamp: August 31, 2023, 09:11:56 pm
content: Quote from: Zero999 on August 31, 2023, 07:42:13 pmThis will also happen during reset and boot, where most MCU I/Os will be either input or three-state until configured.Happy smoking.

author: Zero999
author_rank: Super Contributor
timestamp: August 31, 2023, 09:38:38 pm
content: Is a negative rail actually required? What's the op-amp doing? If it's just amplifying an AC signal and the output is under 4V peak to peak, or so, just use a single supply, AC coupled op-amp configuration.

author: RFDx
author_rank: Regular Contributor
timestamp: September 01, 2023, 03:21:30 am
content: Quote from: Zero999 on August 31, 2023, 07:42:13 pmThe requirement was -5V, wich you don't get with just two diodes/capacitors.

author: ppTRN
author_rank: Regular Contributor
timestamp: September 01, 2023, 07:11:28 am
content: Thank you all for the suggestions.Quote from: Zero999 on August 31, 2023, 09:38:38 pmI may be able to design my circuit using only single supply op-amp, but having a dual rail will surely make my work easy.Quote from: Benta on August 31, 2023, 09:11:56 pmYes, I actually noted an increase in heat when leaving unconnected the 5kHz signal, but no smoke yet.Quote from: RFDx on August 31, 2023, 07:36:00 pmI like this idea of having a voltage doubler embedded in the original circuit. I will try this configuration and I will let you know.

author: magic
author_rank: Super Contributor
timestamp: September 01, 2023, 07:18:41 am
content: Simultaneous conduction is prevented by choice of the four resistors near bases. The central ones should be higher in value than the top and bottom.ICL7660 is the most straightforward solution if "almost -5V with no more than a few mA load" is good enough.

author: Zero999
author_rank: Super Contributor
timestamp: September 01, 2023, 09:40:44 am
content: Quote from: RFDx on September 01, 2023, 03:21:30 am-4V might be near enough and is closer to meeting the -5V requiremnt than your circuit which gives -6.76V out.None of these simple circuits provide an accurate, regulated output voltage, but that's probably unimportant.

author: ppTRN
author_rank: Regular Contributor
timestamp: September 01, 2023, 10:41:55 am
content: Quote from: Zero999 on September 01, 2023, 09:40:44 amExactly, I will just make sure that the signals will keep their amplitude inside the output dinamic range

author: ppTRN
author_rank: Regular Contributor
timestamp: September 01, 2023, 12:23:12 pm
content: Quote from: RFDx on August 31, 2023, 07:36:00 pmThis works perfectly! No transistors overheating durin operation, they do overheat when I force the PIC in reset state. Again, this will never actually happend, but I'll try twiggle a bit the base series resistors and try to fix it.

author: Ian.M
author_rank: Super Contributor
timestamp: September 01, 2023, 12:44:09 pm
content: Tweaking the base resistors probably wont fix the overheating in Reset problem without introducing other issues. If you increase the series resistors and reduce the shunt (B-E) resistors to keep the transistors off when the input is floating mid-rail, depending on your desired output current, you are likely to have difficulty getting enough base drive when the input is driven to either rail.One option would be to feed the squarewave from the PIC through a logic buffer, with a pullup or pulldown resistor on its input to set the desired output state when the I/O pin tristates.Another option if the PIC peripheral is capable of multiple synchronous squarewave outputs and you've got a spare I/O pin the peripheral can use, would be to generate separate high side and low side drive signals from the PIC, so both transistors turn off when the I/Os tristate. This may also let you add deadtime, eliminating shoot-through, and thus improving efficiency and reducing unwanted supply transients.

author: ppTRN
author_rank: Regular Contributor
timestamp: September 01, 2023, 01:48:38 pm
content: Quote from: Ian.M on September 01, 2023, 12:44:09 pmthe uC is a dsPIC30F2010 that should be able to generate simoultaneously two identical PWM signalswith opposite polarity. If I generate such PWM signals with 50% D.C. and feed them separately to the transistor it might work.If it won't work I'll try a buffer with a pulled input. Wether I pull the input up or down should not make any differece at this point

author: Ian.M
author_rank: Super Contributor
timestamp: September 01, 2023, 05:47:17 pm
content: Its worth investigating how to add deadtime so the high side turns off before the low side turns on and visa versa. This is a commonly available feature in advanced PWM peripherals, so I'd be surprised if your dsPIC doesn't have that capability.

author: Zero999
author_rank: Super Contributor
timestamp: September 01, 2023, 06:40:59 pm
content: AC coupling the transistors will limit the length of time when both transistors turn on. If the MCU's output is held low for long enough for the capacitors to charge, it will prevent it. AC coupling also speeds up the turn-off time because the base voltage is taken above/below the emitter voltage.