This is likely an extremely basic question for you folks. I have what I think is a simple voltage inverter that is intended to invert a 15 volt input via an NPN transistor being driven by a 555 timer with a couple of diodes and capacitors that I can then feed into a 7912 for a regulated -12 voltage. What I’m doing this for is to create a replacement +5/+12/-12 internal power supply for a vintage Apple Iic.
I’ve built the circuit and it sorta works, but my 555 is sizzling hot, so I think I’ve missed something. If you wouldn’t mind having a look, I have an image of the circuit here:
Off the top of my head the only two things I can think of is you 555 pin3 is shorting out but unlikely, the other is that you are powering it 15v, i know the datasheet says 15 or 18v but I’ve never been able to drive a 555 above 9v without complications.
If it is because you are running it at 15v but the supply is working then just get a little heatsink for the 555 or run it at a lower voltage and maybe add a transistor to do the switching from sub 12v to >12v.
I had not put a resistor between the output of the 555 and the base on the transistor. Adding a transistor sorted it right out (heat issue).
I have a different issue now… While the charge pump is running, it settles in at -4.96 volts… I’m guessing that this is because I’m driving the transistor with the 5 volt output from the 555… My intention, though, was to use that square wave to allow for -14 volts or so from the +15 volt input at the collector on the transistor.
Would I need to push the output from the 555 through a voltage doubler to increase the flow through the transistor and drive the negative voltage higher?
Thanks! Trying to add the schematic picture again:
Programability:
Off the top of my head the only two things I can think of is you 555 pin3 is shorting out but unlikely, the other is that you are powering it 15v, i know the datasheet says 15 or 18v but I’ve never been able to drive a 555 above 9v without complications.
If it is because you are running it at 15v but the supply is working then just get a little heatsink for the 555 or run it at a lower voltage and maybe add a transistor to do the switching from sub 12v to >12v.
I think there’s at least 1 mistake in your schematic. Look at my red overlay below. C4, as drawn, is shorted out. Also are you missing a connection ? Lastly shouldn’t CV be terminated w/a cap (0.01uF) to ground ?
Mee_n_Mac:
I think there’s at least 1 mistake in your schematic. Look at my red overlay below. C4, as drawn, is shorted out. Also are you missing a connection ? Lastly shouldn’t CV be terminated w/a cap (0.01uF) to ground ?
[attachment=0]inverter.png[/attachment]
Hmmm… There is something wonky in that diagram. I’m looking at the circuit on a breadboard and it’s definitely not shorted out. I’ll have to redo the schematic and repost it.
Checking the data sheet I realized that I could power the 555 with upwards of 15 volts, so I am now powering it off of the 12 volt regulated supply that I also have available, but the voltage output is still hovering at about -6 volts. Interestingly, as soon as I power off the circuit, the negative voltage “rises” to about -9 volts and then decays to zero very, very slowly.
This type of charge pump circuit is very unfamiliar ground for me. I’m usually dealing exclusively with digital logic levels and supplying adequate levels to components, not working on designing a simple power supply. What is it, exactly, that regulates the maximum inverted voltage? Clearly, it’s not the supplied voltage (meaning that I likely wasted time adding the NPN transistor to switch in a supply that wouldn’t be limited by the maximum amperage available through the 555). Is it the pulse width frequency on the square wave?
There are a number of problems with the circuit as shown:
No current limit resistor on base of Q1.
Insufficient drive voltage on Q1. (emitter limited to less than 5V)
Will not work.
When Q1 is on you charge C3 but when Q1 turns off, that charge has nowhere to go. Normally you would use a push-pull driver so that you connect that node to ground when Q1 is off.
The topology of your switched capacitor circuit is also suspect. I am pretty sure it will never work.
What you want is:
Drive a capacitor (C1) with the output of the 555. The other lead of the capacitor is connected to two diodes: one diode with its cathode grounded (D1) and the other (D2) with its anode connected to the output capacitor. (C2, other lead to ground)
When the input is high, C1 charges through D1. When the input is grounded this presents a negative voltage that turns off D1 and turns D2 on, the charge is then transferred to the output capacitor.
It is much easier to play around with this sort of thing by using a simulator like LTspice. Less chance of letting the smoke out of parts as well.
Well, the thing is that the circuit is functioning. There are clearly errors in the schematic that I posted and I had previously identified the missing resistor connected to the base.
Unfortunately, I didn’t really see anything in your response that indicates what governs the voltage output of the circuit.
And to prove that point here’s this circuit’s sim (LTSpice) output. With no load the output is closer to -Vcc (15v). With only a 500 ohm load the output drops to an average of -8v.
Drive a capacitor (C1) with the output of the 555. The other lead of the capacitor is connected to two diodes: one diode with its cathode grounded (D1) and the other (D2) with its anode connected to the output capacitor. (C2, other lead to ground)
I was aware of the limitation that the inverted voltage cannot be greater than the input. That’s why I was trying to drive an NPN transistor with a separate +12 supply and that’s also why the result surprised me (-6.17 volts at the moment). The output currently has no load on it at all.
I’ve revised the diagram to fix the transcription errors from my scrap paper to the online schematic editor. I believe that this one more accurately reflects what I’ve got running here.
Again, what is it that’s controlling the output of the voltage here? I can’t see how I’m losing nearly 6 volts with no load at all.
By the way, the primary reason that I’m tying those two grounds together is that the 5 volts driving the 555 are actually coming off of a 7805 regulator being driven from the same 12 volt supply.
dhoelzer:
Again, what is it that’s controlling the output of the voltage here? I can’t see how I’m losing nearly 6 volts with no load at all.
I'm surprised it works at all ! Look at what happens w/Q1. The 555 output is limited to it's supply, 5v (?) at best. In order for Q1 to conduct current it's base must be ~0.65v above it's emitter voltage. Given the diode (D2) in the path, C1 can only charge up so much, obviously < 5v minus 2 diode drops (1.3v).
Then when Q1 shuts off, there’s no hard ground applied to the ‘top’ of C1 … so I don’t see how you get a negative output voltage. But I’m intrigued enough to simulate it and see what happens.
ps - what kind of current draw are you expecting. Increasing C1 and C2 to > 10 uF will help … w/a proper circuit.
Mee_n_Mac:
And here it is … as expected the circuit, as depicted, doesn’t work. See if your schematic, and actual circuit, agree w/mine.
[attachment=1]555_Inverter_cktB.jpg[/attachment]
[attachment=0]555_Inverter_pltB.jpg[/attachment]
c3 and c4 in your diagram are 4700 uf electrolytics in my circuit. The diagram you have there does indeed match my circuit except for the mysterious “555” line coming off of the emitter on the transistor. I’d post a photo but it would be tough to make stuff out with jumper wires on this breadboard.
dhoelzer:
c3 and c4 in your diagram are 4700 uf electrolytics in my circuit. The diagram you have there does indeed match my circuit except for the mysterious “555” line coming off of the emitter on the transistor.
The 555 is just a label, actually leftover from the prior circuit. It should have been labelled Ctop or the like. In any case the schematic shouldn't work for the reasons previously given. The cap size won't affect those. Why not try the simple ckt UhClem suggested.
IIRC the cap on CV is there to prevent noise intrusion.
dhoelzer:
I’d post a photo but it would be tough to make stuff out with jumper wires on this breadboard.
Agreed. Unless it's a simple ckt, and laid out exceptionally well, a photo is just too hard to trace.
Well, I think it is essentially the same circuit with the addition of the transistor and second voltage source. It’s very interesting that the sim shows it non-functional. I’ll go over it tomorrow and verify that what I’ve made matches the schematic I drew up from it.
dhoelzer:
Well, I think it is essentially the same circuit with the addition of the transistor and second voltage source.
Not really. Again consider how the working ckt works. C3 charges up when the 555 output is Vcc. When the 555 output goes low it provides a short to 'ground'. That forces the cathode of C3 to go below ground. Your Q1 does the 1'st part but not the 2'nd. Q1 goes into a high impedance state when the 555 output goes low. As a result the cathode of C3 is not forced to go below ground. And that's the 'trick' of this type of ckt.
https://www.dropbox.com/s/4fyl57t6utl8ccm/inverter.png
