Common emitter to drive analog tach adapted

Thread Starter

Voltar1

Joined Mar 15, 2020
9
Hello all, new member here looking for advice and help.
I have a mag pickup into a common emitter amplifier with that output going into a switching transistor. The load being a coil that on collapsing is intended to replicate an ignition coil to drive the tach. Amplifier works great giving a nice output when tested by itself. Tach adapter works as it should when it is used by itself and fed a 3 volt ppk sine wave or square wave. BUT tie them together, the tach initially reads then fizzles to zero.
Any ideas out there? Really want this to work
Cheers
Walter Glover
Alberta, Canada
 

MrChips

Joined Oct 2, 2009
20,884
I would tweak the circuit a bit first.

The input resistors of 1K and 4.7K seem a bit on the low side by about a factor of 5.
I wouldn't know for sure until I have an idea of the impedance of the pickup.

What is the purpose of the diode?

I would put a reversed biased diode across the coil to protect the output transistor.
 

Thread Starter

Voltar1

Joined Mar 15, 2020
9
I would tweak the circuit a bit first.

The input resistors of 1K and 4.7K seem a bit on the low side by about a factor of 5.
I wouldn't know for sure until I have an idea of the impedance of the pickup.

What is the purpose of the diode?

I would put a reversed biased diode across the coil to protect the output transistor.
Chips, thank you for your response. The input resistors as in go to 5k and 25k?
The diode might have been to protect the engine tach output circuitry. Using a free wheeling diode would defeat the spike the tach needs to trigger adequately.
Update: had an epiphany this afternoon. Eliminated the amp and input diode to remove a 0.6ish voltage drop the mag pickup struggled to overcome. Tach works as it should currently!
Would still like to know how to marry two circuits together.
cheers
 

MrChips

Joined Oct 2, 2009
20,884
What is the max rpm of the tach?
Is the tack sensitive to changing duty cycle?

I would be inclined to scrap the amplifier circuit and use a 555-timer circuit instead.

1584396678190.png
 

MrChips

Joined Oct 2, 2009
20,884
In the dotted box is an optical coupler or optical switch.
(Originally I thought it was an optical isolator, but it is not since the emitter is powered from the same source as the receiver.)
 

Thread Starter

Voltar1

Joined Mar 15, 2020
9
Thanks for moving this thread I think?
The question was the electronics part and has little to do with automotive.
Anyhoo been thinking that diode would rectify the sine wave input and therefore reduce the input by half
 

BobaMosfet

Joined Jul 1, 2009
1,063
@Voltar1
You need to remember that your inductor coil has to be rated for the frequency you intend to drive it at. An inductor, like a capacitor is a means by which the flow of current is limited depending on frequency. If you're driving it at too high a frequency, it's going to 'fizzle' because the coil cannot charge and discharge fast enough to have a 'range of output' it will graduate to a neutral range.
 

Thread Starter

Voltar1

Joined Mar 15, 2020
9
Thank you Boba I think 200hz is a very low frequency for this use. Once I eliminated the amp stage and the input diode the tach is working with the mag pickup as I need it to.
 

BobaMosfet

Joined Jul 1, 2009
1,063
Thank you Boba I think 200hz is a very low frequency for this use. Once I eliminated the amp stage and the input diode the tach is working with the mag pickup as I need it to.
As long as you're happy with the results. I prefer to do the calculations before implementation IMHO.
 

BobaMosfet

Joined Jul 1, 2009
1,063
Would love to see the calculations to do it properly. No idea where to begin
It's a bit long-haired but will get you there:

Inductance Calculations
Author: Frederick W. Grover
ISBN-10: 048647440-2

An easier read that will help you understand inductors better and how to do some basic calcs:

Title: Understanding Basic Electronics, 1st Ed.
Publisher: The American Radio Relay League
ISBN: 0-87259-398-3

And then of course, the educational section on this site.
 

BobaMosfet

Joined Jul 1, 2009
1,063
@Voltar1
Understand what your circuit is doing:

1584481208945.png

You have created a voltage divider between your inductor and your 2N5551. While you can view this 2N5551 as a 'common emitter' configuration, it's more valuable to understand that it is also an 'open collector' output. The 2N5551 is not there to provide power. It is there to ground the inductor. Every time, the 2N5551 sees a sufficient positive voltage on its base, relatively, it conducts current from the inductor to ground. In every other case, the inductor acts as both a pull-up, and a conductor of current via the 12VDC power-supply.

You can look at the 2N5551 like a variable resistor in this case, controlled by the rest of the circuit. As such, you can then determine your range as an RL circuit. Simply calculate the minimum loss through 2N5551 pinched off, and then at max. The 2N5551 is being operated in its linear region. Derate for thermal. Figure out your resonance, your inductance, and then you can determine how/when current will be limited based on frequency.
 

BobaMosfet

Joined Jul 1, 2009
1,063
Here's some additional- I have no idea what your inductor is rated at, so I have a 1uH inductor, but this will give you an idea of wave-form what's happening to current & voltage in the circuit:

1584483605035.png

Hopefully this will help.
 
Last edited:

Thread Starter

Voltar1

Joined Mar 15, 2020
9
In that section of the circuit the transistor is simply a switch. Replacing the points as it were in the ‘ignition’. The inductor is simply acting as the ‘coil’ in the ignition and provides the spike as the field collapses.
 
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