Simple Time Delay Circuit Using Op Amp

lC1a is provided with re- sistive and capacitive feedback to form an integrator with initial conditions. lC1b is in an "open loop" mode so that its output is either high or low depending on its inputs, and changes state when the output of lC1a goes more negative than the voltage set at ZD2.
When . the output of IC1 b goes positive the transistor Q1 biases hard on switching the SCR on. Diodes D1 -D4 are to make the SCR conduct on both halves of the mains wave form. The delay period is set by the components ZD1, ZD2;·C, RVl, and  R. lf ZD1 is chosen to be OV5 and ZD2 at 5V, then the maximum delay period is given by T= 10CR
RV1 = ZD2/ZD1 * R<10.r

 The meter is a voltrneter with as fsd equal to the value of ZD2. The switch then operates when the meter reaches fsd. The meter can therefore be calibrated to show remaining delay with OV equal to T and fsd equal to zero. SW2 changes round the inputs of the op—amp so that the output either swings from high to low, or, low to high. SW3 is to reset the time delay which it does by discharging the capacitor. ZD3 should be chosen  to be a value slightly higher than ZD2, this is to stop the capacitor charging beyond a set limit and therefore overloading the meter. SW1 is the run—hold switch. When the switch is at +12 volts the integrator charges the capacitor. When the switch is set to OV the charging of the capacitor is stopped until the switch is set back to 12 volts. . Q1 is a buffer to avoid loading on the IC and to trigger the SCR. The supply voltage should be 12-0-12 and does not need to be well smoothed as the zener diodes set the timing function,


Simple Time Delay Circuit Using Op Amp

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12 Volt 30 Amp PSU circuit and explanation

Using a single 7812 IC voltage regulator and multiple outboard pass transistors, this power supply can deliver output load currents of up to 30 amps. The design is shown below:

Circuit diagram

Notes:
The input transformer is likely to be the most expensive part of the entire project. As an alternative, a couple of 12 Volt car batteries could be used. The input voltage to the regulator must be at least several volts higher than the output voltage (12V) so that the regulator can maintain its output. If a transformer is used, then the rectifier diodes must be capable of passing a very high peak forward current, typically 100amps or more. The 7812 IC will only pass 1 amp or less of the output current, the remainder being supplied by the outboard pass transistors. As the circuit is designed to handle loads of up to 30 amps, then six TIP2955 are wired in parallel to meet this demand. The dissipation in each power transistor is one sixth of the total load, but adequate heat sinking is still required. Maximum load current will generate maximum dissipation, so a very large heat sink is required. In considering a heat sink, it may be a good idea to look for either a fan or water cooled heat sink. In the event that the power transistors should fail, then the regulator would have to supply full load current and would fail with catastrophic results. A 1 amp fuse in the regulators output prevents a safeguard. The 400mohm load is for test purposes only and should not be included in the final circuit. A simulated performance is shown below:

Calculations:
This circuit is a fine example of Kirchoffs current and voltage laws. To summarise, the sum of the currents entering a junction, must equal the current leaving the junction, and the voltages around a loop must equal zero. For example, in the diagram above, the input voltage is 24 volts. 4 volts is dropped across R7 and 20 volts across the regulator input, 24 -4 -20 =0. At the output :- the total load current is 30 amps, the regulator supplies 0.866 A and the 6 transistors 4.855 Amp each , 30 = 6 * 4.855 + 0.866. Each power transistor contributes around 4.86 A to the load. The base current is about 138 mA per transistor. A DC current gain of 35 at a collector current of 6 amp is required. This is well within the limits of the TIP2955. Resistors R1 to R6 are included for stability and prevent current swamping as the manufacturing tolerances of dc current gain will be different for each transistor. Resistor R7 is 100 ohms and develops 4 Volts with maximun load. Power dissipation is hence (4^2)/200 or about 160 mW. I recommend using a 0.5 Watt resistor for R7. The input current to the regulator is fed via the emitter resistor and base emitter junctions of the power transistors. Once again using Kirchoffs current laws, the 871 mA regulator input current is derived from the base chain and the 40.3 mA flowing through the 100 Ohm resistor. 871.18 = 40.3 + 830. 88. The current from the regulator itself cannot be greater than the input current. As can be seen the regulator only draws about 5 mA and should run cold.

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Build a Thermo Fan To Keep Your Amp Cool Circuit Diagram

How to Build a Thermo-Fan To Keep Your Amp Cool Circuit Diagram



 Thermo-Fan To Keep Your Amp Cool Circuit Diagram

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