Switching Problem...

[Colonel_Sanders]

I have a problem with my switching circuit (switching_1). I tested it in my friends dads circuit software and it works fine. When the -12V control voltage is applied to the transistor base, it pulls the transistors open, switching the relays on. If you're wondering why its a negative control voltage, its because there are N-channel fets in the amp circuit.

The -12V is supplied by a shunt zener circuit (switching_2) and is working fine, however the +12V is being supplied by a 5A regulated wall wart I had lying around and stuck in the amp. When I read voltages it seems like the wall wart ground, and amp ground are different values.

'Zener+' to 'Zener-'(amp ground) = -11.96V
'Zener+' to 'Wallwart-(WW-)' = -2V
'WW+' to 'WW-' = +12.54V
'WW+' to 'Z-' = +3V



Any explanations or ideas for possible fixes?

Do you think it would work if I ran the Zener ground to the 'Wallwart-' instead of the amp ground? Or should I just whip up a +12V Zener supply for the relays?

Cheers,
C_S

[Enzo]

Without proper power, circuits won't work. Get that right before thinking on anything else.

I don't like your math on the 12v supply, but since you are not asking much current from it, if it works, then fine. YOu didn't specify the zener. Is it 1/2W, 1W, what? To be a stable zener reference, sufficient current must flow through the thing. SO that seems to me to be about 80ma for a 1 watt.

Your 300VAC becomes more like 420VDC once rectified. I know it averages lower unfiltered like that, but the peaks are still there. And I would not be rectifying 300VAC with 50v 1N4001s if I were you. Use at least a 1N4004, or better yet, just use a 1N4007 like in the B+.

If you want 80ma - or 40ma for half watt, or whatever - through the zener, pick the resistor to provide that current. The zener is not a resistor, once it "zenes" at 12v, you can increase the current but the voltge drops stays about the same. SO unlike series resistors, you have to select both current and voltage through the resistor.

That 500k resistor will seriously limit base current to the 2N3904. Does it even turn on?


Now power grounds. You say the two supplies seem to have different grounds. ALL power supplies will have different grounds unless you connect them together. If you are using a wall wart for +12, then ground the - side of it to the circuit. Where is the - side of it connected now? Your circuit ground is the ultimate voltage reference, so the + of your negative supply is grounded, and the - of your positive supply should be grounded. Only way they can be 2v apart is of they are not connected together, ie. one is not connceted to the same ground as the other.

Actually you need not use chassis. These are just relays and lightbulbs, so all that matters is there is current through them. But the circuit still must have a common.

[Colonel_Sanders]

Sorry enzo, I haven't updated the schematic. I'm using 1n4007's and the Zener is a 1W. Ok, so at 420V I'd need a 34kΩ resistor, which means with a 24kΩ I've got current to spare.

The 1st transistor turns on in the circuit simulator. This is my first time testing it practically. I've got it 'inverting' the -12V Vc to a +12V Vc, a circuit K O'connors explained to me (he didn't mention values though, just 'a high value resistor'). Is there a better way to do this?

Ah, ground the wall wart! *lightbulb* Why didn't I think of that one... The wallwart 0V is floating by itself, connected to the relay circuit. I'll run a wire to the Power Supply main ground now.

Thanks a $mil,
C_S

[Colonel_Sanders]

Sweet bix, grounds are the same. I'd buy you a beer if I was in America and 21!

I've lost my -12V somewhere between the switching circuit and footswitch though, so onto that now...

Cheers Enzo,
C_S

[Colonel_Sanders]

I found my -12V, and I tried applying it directly to the first transistors gate. But alas, the transistor and relays don't switch on.

If I apply +12V to the first 1n4007 cathode, all relays in that circuit switch on.

I guess the problem is with the 500K resistor at the collector of the first transistor that Enzo mentioned. Should I try reducing the resistance?

Cheers,
C_S

[Enzo]

I hope you mean you applied the -12 to VC2, not directly to the base of the transistor. ("Gate" is used for FETs, no bipolars.) it is not a good idea to apply a power source directly to a base like that. The 1k resistor is there to control base current.

If you apply 12v to that first diode, it is normal all the relays pull in. I am not sure why they are cascaded like that, but it ought to work OK.

In future drawings, part numbers would be nice. For discussion, I am going to number the transistors left to right, so VC2 feeds the base of Q1. I will also call resistors numbered left to right, so the 1k is R1, the 500k is R2, 10k R3. If this confuses, then I will stop.

By the way, you apply -12 to VC2 to turn the transistor off, but what is connected to VC2 at other times? It needs to be terminated somewhere.

The transistors here are used as switches. Turn them on and they conduct. There are voltage drops and things, so they are not zero ohm connections like a switch, but let's pretend they are just switches.

If Q1 were turned off, it would be as if it were not there at all. OK. In that case, current through the 500k R2 and 10k R3 will flow through the Q2 base and hopefully turn it on.

Transistors are current devices, unlike tubes which are voltage devices. I know I know, you don't have one without the other, but those are the focus of the parts. (foci?) SO there must be sufficient current through the base to turn the transistor on. It is not sufficient just to get some voltage there. Imagine the base of Q2 was ground. You would not have 12v across 510k (R1 + R2). The current available then is (by Ohm;s Law) I = 12/510,000 which is .0000023 amp, or 0.02ma. Two hundredths of a milliamp is not much current. SO yes, I think 500k is way to large a value.

R1 and R2 must be able to turn on Q2 all by themselves here. The job of Q1 is really to keep Q2 turned off. When you turn off Q1 with -12v, it steps out of the way and lets Q2 come on. SO lets find out what we need. Put -12 at VC2 to make sure Q1 is off - or even remove Q1 for now. Q2 does not turn on the relays, right? With a piece of wire or a clip lead, momentarily short across R2 - we are bypassing the 500k. Now the +12 flows through R3, which is only 10k. Does that fire the Q2? AT 10k, R3 will allow about 1.2ma to flow. If that is enough to fire Q2, we have something to work with.

I know nothing of the relays coils that are the load for Q2, but Q2 has to have enough current gain to take the base current and turn it into enough current through the collector for the relay. SO even that lone 10k resistor could be too much.

What we need to do is connect some resistance to +12 and the base of Q2 that will allow turn-on current through Q2. A simple test would be to clip in a pot in place of the resistor, and turn it from high towards low resistance until the relay pulls in. Then remove the pot and measure it. That is the resistance you need.

[Colonel_Sanders]

Awesome post Enzo, learned a lot.

Ah, didn't know applying a voltage straight to the base was bad. And by gate, I meant base .

Got it working after a while. I ended up adding a 1M to ground from VC2, changing Q2 to 2n3904(NPN) instead of 2n3906(PNP), changed R2 to 20K and R3 to 1K. Cheers heaps!
C_S

[Enzo]

Oh geez, in paragraph 7 it should say you will NOW have 12v across the 510k. "not" was a typo.

The connection in a transistor from base to cathode is like a diode. Same from base to collector. If you connect a power supply directly to the base, what is ther to limit the base current through the transistor? Just the emitter's load in this case.