Explanation needed for this circuit...

[boylesg]

**broken link removed**

I have implemented this circuit in multisim 11 and have been exploring it but I am not certain whether or not this software accurately simulates the behaviour of the real circuit.

In multisim of you use a virtual oscilloscope to probe the circuit at the 555 out put pin then you get the expected square wave with a low of 0V and a high of about 9V.

However if you probe just after the 2.2kR resistor you get a reduced amplitude square wave with the low raised to about 5V or so.

I do not understand this. If you simply connect a resistor to a source and probe this simple circuit after the resistor then you find that the horizontal line lowers from 12V to roughly 5V or what ever.

So I would expect that in the above circuit that the resistor would lower the voltage peaks from the 555 to less than 9V. But, according to multisim, it actually raises the low voltage.

Why does this occur?

Or is that multisim is not quite accurately simulating the behaviour of the circuit or I am not using its virtual oscilloscope correctly?

I am intending to implement a very simple voltage divider interface to a pc sound card line in socket so that I can roughly examine the behaviour of the real cicrcuit and compare it to multisim.

- - - Updated - - -

Also, the designer of this circuit have specifically chosen to invert the signal from the 555 via the BC327 PNP.

Does anyone know why it is advantageous to do this? Why not a NPN transitor?

 

[boylesg]

Here is a similar circuit I am working with using a MOSFET rather than a Darlington. This circuit is wired up on a bread board and actually drives a tv flyback transformer successfully. But I can't reconcile this with what multisim 11 is telling me about the operation of the circuit.

Here is the circuit:



And here is the output of the circuit:

Cannel A is on the 555 output pin before the 470R resistor and channel B is on the 555 output pin after the 470R resistor.

So why is the pulse been lost according to multisim?

 

[BradtheRad]

About the 555 layout...

Can you confirm whether Q2 and Q3 are oriented properly?

The arrow in a transistor icon points in the direction of a pn junction. In other words, the direction of current flow.

 

[Darktrax]

The 2K2 in the base circuit of Q1 is unlikely to have a nice voltage square wave at the Q1 base. A transistor is a current input device, and the base current is set by the 2K2 and the 555 driving it.

Note that Q1 is a PNP, which which is turned on by the negative-going part of the 555 output. The main Darlington Q2 is then turned on by the output of Q1 pulling its base positive.

I think the orientations seem correct.

I am confused about "Q3". There seems to be Q1 and Q2 - or maybe did I miss something?

 

[boylesg]

About the 555 layout...

Can you confirm whether Q2 and Q3 are oriented properly?

The arrow in a transistor icon points in the direction of a pn junction. In other words, the direction of current flow.

Thanks For that. You are right of course, the transistors in the multisim version are around the wrong way. I swapped them and now the signal before and after the resistor are identical. But I was still expecting the peak voltage to be lower. The idea of the 2.2k resistor was to reduce the peak voltage to about the 5V max of the BD1## base voltages.

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There are two version of the circuit.

One using a mosfet and that is a re-design of a similar cicruit off the net that I could not get working. This one os wired up on a bread board and working.

The one that uses the Darlington was in a kit that I bought from Jaycar. This is soldered up on a PCB but it is not working. Probably an errant solder blob or something. Will re-solder it on a protoype board and then see if I can get it working.

But in both cases I am trying to understand the interaction between the 555 signal and the resistors on output pin 3 as observed in multisim 11.

To be clear.....

Mosfet based circuit


Mosfet based output before and after the 2.2k resistor showing identical output.
Shouldn't the resistor reduce the peak voltage of the 555 signal?


Darlington based circuit


Darlington output.
Why is the low voltage raised rather than the peak voltage reduced? If you use a large resistor eventually the signal disappears altogether and you get just a steady voltage - horizontal line.
Please note that I have moved channel 2 below the zero line so that I can see both signals seperately. However without the displacement, the peaks of the channel b signal correspond with the peaks of the channel a signal. So, as previously stated, the signal lows have been raised to about 7V or something like that.

 

[Darktrax]

Since I do not have Multisim, some of my perceptions about your circuit may be at a disadvantage - but here goes..

If you are to simulate switching the current into the primary of an ignition coil, the first step is to leave out the 555 and make a simplified version, where the FET is switched with an ideal (forced) voltage waveform. Check out the FET data. Are you absolutely sure you should be connecting the gate to Vcc via a transistor to turn it on?

Next is to try and get the right parameters into the transformer model. You can, if you like, put its (measured) DC resistance in as a separate component for simulation purposes. Next is to know the inductance of the coil. This too, you can find out by experiment in a number of ways, if you do not have an instrument to measure it. The inductance of the secondary, and the mutual inductance are all basic parameters needed for this.

Then, you need to understand that generating high voltage in this way is what is known as a flyback circuit. The duty cycle of the 555 waveform has to be non-symmetric, in that when the main FET goes ON, it first has to charge the FET gate and the (Miller) effect from the output. A power FET like this one is like millions of FETs gates in parallel. Just because you offer it a pulse through a resistor, does not mean you see the same pulse at the gate.

The current in the primary of the coil has to increase to store energy in the magnetic field, and it will take the time to rise, depending on its inductance, to approach a limit set by its resistance, and the Vcc. Then, when you try to cut off the FET, there is a large reverse voltage generated across the coil, which will be taken care of by the built-in energy recovery diode. It is the *rate of change* of this current, and how the diode clamps the ringing overshoot that will decide what voltage happens at the secondary.

The next stage is to add in the transistor drivers. Simply connecting them straight across the whole supply as shown risks there being some point when both are biased ON at the same time, so short-circuit across the supply. Commonly, the bases can be separated by a bias diode. Timing is everything here! many circuits use separate, non-overlapping base drive signals.
Another thing is that one should not be wasting precious base drive current driving a LED that will not be very useful unless the circuit has already stopped.

Finally, one does not get much information by ever placing an oscilloscope probe directly on the base of a transistor, because it is clamped to about 0.7V Vbe by the transistor junction. Either a differential set of probes placed across the base resistor, or some other way of checking the current waveform is much better - if you need it at all. The voltage at the transistor emitters is much better way of knowing when it starts conducting.

For me, the free simulator LT-Spice IV from Linear Technology lets you get the current waveform easily.

 

[boylesg]

Since I do not have Multisim, some of my perceptions about your circuit may be at a disadvantage - but here goes..

If you are to simulate switching the current into the primary of an ignition coil, the first step is to leave out the 555 and make a simplified version, where the FET is switched with an ideal (forced) voltage waveform. Check out the FET data. Are you absolutely sure you should be connecting the gate to Vcc via a transistor to turn it on?

Next is to try and get the right parameters into the transformer model. You can, if you like, put its (measured) DC resistance in as a separate component for simulation purposes. Next is to know the inductance of the coil. This too, you can find out by experiment in a number of ways, if you do not have an instrument to measure it. The inductance of the secondary, and the mutual inductance are all basic parameters needed for this.

Then, you need to understand that generating high voltage in this way is what is known as a flyback circuit. The duty cycle of the 555 waveform has to be non-symmetric, in that when the main FET goes ON, it first has to charge the FET gate and the (Miller) effect from the output. A power FET like this one is like millions of FETs gates in parallel. Just because you offer it a pulse through a resistor, does not mean you see the same pulse at the gate.

The current in the primary of the coil has to increase to store energy in the magnetic field, and it will take the time to rise, depending on its inductance, to approach a limit set by its resistance, and the Vcc. Then, when you try to cut off the FET, there is a large reverse voltage generated across the coil, which will be taken care of by the built-in energy recovery diode. It is the *rate of change* of this current, and how the diode clamps the ringing overshoot that will decide what voltage happens at the secondary.

The next stage is to add in the transistor drivers. Simply connecting them straight across the whole supply as shown risks there being some point when both are biased ON at the same time, so short-circuit across the supply. Commonly, the bases can be separated by a bias diode. Timing is everything here! many circuits use separate, non-overlapping base drive signals.
Another thing is that one should not be wasting precious base drive current driving a LED that will not be very useful unless the circuit has already stopped.

Finally, one does not get much information by ever placing an oscilloscope probe directly on the base of a transistor, because it is clamped to about 0.7V Vbe by the transistor junction. Either a differential set of probes placed across the base resistor, or some other way of checking the current waveform is much better - if you need it at all. The voltage at the transistor emitters is much better way of knowing when it starts conducting.

For me, the free simulator LT-Spice IV from Linear Technology lets you get the current waveform easily.

Is this better? With a 100R current limiting resistor across the totem pole? Also added a voltage divider on the 555 output to bring the peak voltage down 5V on the base of the transistors. Works much better than just a single resistor......at least in multisim.

What would a diode accomplish exactly that is not already done by the transistors?



I was connecting the fet gate to the 12V rail because I was reading they need at least 10V on their gate for them to fully switch on. Otherwise they are operating in the ohmic region which puts stress on them. They all seem to have a 30V maximum gate voltage but the 555 is powered by 9V and olnly puts out about 8V as measured by my multimeter.

The FET in multisim is not the FET I am actually using.
The FET I am actually using is a 1500V max (or something like that) salvaged from a tv set - not available in multisim. I was using new irf630 and some other small 200V salvaged FETs but they kept burning out.

Thanks for the info about transistor base voltage and clamping. So the 555 sends out a pulse, turns on the BD139 and then base emitter voltage drop gets mixed up with the 555 signal. So perhaps the raised low voltage I am seeing in multisim is actually the base emitter voltage drop of the conducting BD139 etc. Is that the general idea of what multisim is showing me?

The LED is on the 555 output for 2 reasons. The first is that it indicates the circuit is on and also I can connect to a larger timing capacitor such the the LED visibly flashes and indicates that the timing circuit is working. I could probably use a bigger resistor so it draws less current.