Lambda Probe Readout For Carburetor Tuning

A lambda probe (or oxygen sensor) can be found on the exhaust system of most cars running on unleaded fuel. Having reached its normal operating temperature (of about 600 degrees Celsius!) the lambda probe supplies an output voltage proportional to the amount of residual oxygen measured in the exhaust gas.

This information is indicative of, among others, the air/fuel ratio supplied by the carburetor(s) and hence the combustion efficiency. In modern car (and motorcycle) engines, this information is used to (electronically) adjust engine parameters like ignition timing and fuel injection. The indicator described here is intended for permanent installation on a motorcycle of which the air/fuel ratio needed to be watched, with the obvious aim engine power tuning after fitting a different set of carburetors. Apart from this obvious technical use the unit’s bright LEDs will no doubt attract the attention of curious motorcyclists.

At the local junkyard a single-wire lambda probe may be salvaged from a wrecked car. Once a suitable nut has been found, the probe can screwed into the exhaust pipe of the motorcycle, at about 30 cm from the cylinders.  Since we’re talking of welding and drilling in an expensive (chrome-plated) exhaust pipe, you may find that actually fitting the probe is best left to specialists!  The starting point for the design of a suitable electronic indicator is that in the noble art of carburetor tuning an air/fuel ratio of 14.7 to 1 is generally considered ‘perfect’, the range covering 16.2 to 1 (‘lean’) to 11.7 to 1 (‘rich’). The perfect ratio typically corresponds to a probe output voltage of 0.45 V. 

 Referring to the circuit diagram, that is the input level at which 5 of the 10 LEDs will light, including the green one, D5. If one of the red LEDs lights, the mixture is definitely too rich. Note that in general it is better to have a mixture that is a little to rich than one that’s on the lean side, hence a yellow LED lights between the green LED and the first red one. Also note that the engine needs to be at its normal operating temperature before a meaningful indication is obtained.

10 000x With One Transistor

For a collector follower with emitter resistor, you’ll often find that the gain per stage is no more than 10 to 50 times. The gain increases when the emitter resistor is omitted. Unfortunately, the distortion also increases. With a ubiquitous transistor such as the BC547B, the gain of the transistor is roughly equal to 40 times the collector current (Ic), provided the collector current is less than a few milliamps. This value is in theory equal to the expression q/KT, where q is the charge of the electron, K is Boltzmann’s constant and T is the temperature in Kelvin.

For simplicity, and assuming room temperature, we round this value to 40. For a single stage amplifier circuit with grounded emitter it holds that the gain Uout /Uin (for AC voltage) is in theory equal to SRc. As we observed before, the slope S is about 40Ic. From this follows that the gain is approximately equal to 40I cRc. What does this mean? In the first instance this leads to a very practical rule of thumb: that gain of a grounded emitter circuit amounts to 40·I c·Rc, which is equal to 40 times the voltage across the collector resistor.

If Ub is, for example, equal to 12 V and the collector is set to 5V, then we know, irrespective of the values of the resistors that the gain will be about 40R(12–5) = 280. Notable is the fact that in this way the gain can be very high in theory, by selecting a high power supply voltage. Such a voltage could be obtained from an isolating transformer from the mains. An isolating transformer can be made by connecting the secondaries of two transformers together, which results in a galvanically isolated mains voltage.

Circuit diagram:

10,000x With One Transistor Circuit diagram

That means, that with a mains voltage of 240 Veff there will be about 340 V DC after rectification and filtering. If in the amplifier circuit the power supply voltage is now 340 V and the collector voltage is 2 V, then the gain is in theory equal to 40 x (340–2). This is more than 13,500 times! However, there are a few drawbacks in practice. This is related to the output characteristic of the transistor. In practice, it turns out that the transistor does actually have an output resistor between collector and emitter.

This output resistance exists as a transistor parameter and is called ‘hoe’. In normal designs this parameter is of no consequence because it has no noticeable effect if the collector resistor is not large. When powering the amplifier from 340 V and setting the collector current to 1 mA, the collector resistor will have a value of 338 k. Whether the ‘hoe’-parameter has any influence depends in the type of transistor. We also note that with such high gains, the base-collector capacitance in particular will start to play a role.

As a consequence the input frequency may not be too high. For a higher bandwidth we will have to use a transistor with small Cbc, such as a BF494 or perhaps even an SHF transistor such as a BFR91A. We will have to adjust the value of the base resistor to the new hfe. The author has carried out measurements with a BC547B at a power supply voltage of 30 V. A value of 2 V was chosen for the collector voltage. Measurements confirm the rule of thumb. The gain was more than 1,000 times and the effects of ‘hoe’ and the base-collector capacitance were not noticeable because of the now much smaller collector resistor.

 

 

Author: Gert Baars

12 V AC Dimmer

The circuit described here is derived from a conventional design for a simple lamp dimmer, as you can see if you imagine a diac connected between points A and B. The difference between this circuit and a normal diac circuit is that a diac circuit won’t work at 12 V. This is the fault of the diac. Most diacs have a trigger voltage in the range of 30 to 40V, so they can’t work at 12 V, which means the dimmer also can’t work.

Circuit diagram :

12-V AC Dimmer Circuit Diagram

The portion of the circuit between points A and B acts like a diac with a trigger voltage of approximately 5.5 V. The network formed by R1, P1 and C1 generates a phase shift relative to the supply voltage. The ‘diac equivalent’ circuit outputs a phase-shifted trigger pulse to the triac on each positive and negative half-cycle of the sinusoidal AC voltage.

This works as follows. First consider the positive half of the sine wave. C1 charges when the voltage starts to rise, with a time constant determined by C1, R1 and P1. T1 does not start conducting right away. It waits until the voltage across D2 reaches 4.7 V and the Zener diode starts to conduct. Then current starts to flow, driving T1 and T3 into conduction. This produces a pulse at point B. The same principle applies to the negative half of the sine wave, in this case with D1, T2 and T4 as the key players.

The trigger angle can be adjusted with P1 over a range of approximately 15 degrees to 90 degrees. C2 provides a certain amount of noise decoupling. Depending on the load, the triac may need a heat sink. You can use practically any desired transistors; the types indicated here are only examples. If the circuit does not dim far enough, you can change the value of P1 to 25 kΩ. This allows the trigger angle to be increased to 135 degrees.

Note: this circuit works fine with normal transformers, but not with ‘electronic ’ transformers.

http://www.ecircuitslab.com/2012/02/12-v-ac-dimmer.html

Power On Indicator

Some types of electronic equipment do  not provide any indication that they are  actually on when they are switched on.  This situation can occur when the back-light of a display is switched off. In addition, the otherwise mandatory mains  power  indicator  is  not  required  with  equipment  that  consumes  less  than  10 watts. As a result, you can easily forget  to switch off such equipment. If you want  to know whether equipment is still drawing power from the mains, or if you want  to have an indication that the equipment  is switched on without having to modify the equipment, this circuit provides a solution. 

One way to detect AC power current and  generate a reasonably constant voltage  independent of the load is to connect a  string of diodes wired in reverse parallel in series with one of the AC supply  leads. Here we selected diodes rated  at 6 A that can handle a non-repetitive  peak current of 200 A. The peak current  rating is important in connection with  switch-on  currents.  An  advantage  of  the selected diodes is that their voltage  drop increases at high currents (to 1.2 V  at 6 A). This means that you can roughly  estimate the power consumption from  the brightness of the LED (at very low  power levels). The voltage across the diodes serves as  the supply voltage for the LED driver. To  increase the sensitivity of the circuit, a  cascade circuit (voltage doubler) consisting of C1, D7, D8 and C2 is used to double  the voltage from D1–D6. Another benefit  of this arrangement is that both halve- waves of the AC current are used. We use  Schottky diodes in the cascade circuit to  minimise the voltage losses. 

Circuit diagram :

Power On Indicator Circuit Diagram

The LED driver is designed to operate the LED  in blinking mode. This increases the amount  of current that can flow though the LED when  it is on, so the brightness is adequate even  with small loads. We chose a duty cycle of pproximately 5 seconds off and 0.5 second  on. If we assume a current of 2 mA for good  brightness with a low-current LED and we can  tolerate a 1-V drop in the supply voltage, the  smoothing capacitor (C2) must have a value of  1000 µF. We use an astable multivibrator built around two transistors to implement a  high-efficiency LED flasher. It is dimensioned to minimise the drive current of  the transistors. The average current consumption is approximately 0.5 mA with a  supply voltage of 3 V (2.7 mA when the  LED is on; 0.2 mA when it is off). C4 and  R4 determine the on time of the LED (0.5  to 0.6 s, depending on the supply volt-age). The LED off time is determined by  C3 and R3 and is slightly less than 5 seconds. The theoretical value is R × C × ln2,  but the actual value differs slightly due to  the low supply voltage and the selected  component values.

 

Diodes D1-D6 do not have to be special  high-voltage diodes; the reverse volt-age is only a couple of volts here due  the reverse-parallel arrangement. This  voltage drop is negligible compared to  the value of the mains voltage. The only  thing you have to pay attention to is the  maximum load. Diodes with a higher  current rating must be used above 1 kW.  In addition, the diodes may require cool-ing at such high power levels.  Measurements on D1–D6 indicate that  the voltage drop across each diode is  approximately 0.4 V at a current of 1 mA.  Our aim was to have the circuit give a  reasonable indication at current levels  of 1 mA and higher, and we succeeded  nicely. However, it is essential to use a  good low-current LED.

 

Caution: the entire circuit is at AC power potential. Never work on the circuit with the mains cable plugged in. The  best enclosure for the circuit is a small,  translucent box with the same colour as  the LED. Use reliable strain reliefs for the  mains cables entering and leaving the  box (connected to a junction box, for  example). The LED insulation does not  meet the requirements of any defined insulation class, so it must be fitted such that it  cannot be touched, which means it cannot  protrude from the enclosure.