Charge consumed by a circuit

1. Nov 19, 2009

mathew086

I have circuit ( see figure below). I need to find the total charge consumed by this circuit.

Supply voltage = 9 V
R1 = 1kOhm
R2 = 47 kOhm
R3 = 100 kOhm
T = NPN Transistor (BC 337)
C = 22 uF
IR (E) = IR emitter (IRL 80 A); FOrward current = 60mA;
IR (R) = IR receiver (LPT 80 A);
Relay = 12V relay.

WHat does the relay do?

Also i would like to know how to choose appropirate values for R1, R2 R3 and C.

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2. Nov 19, 2009

Bob S

When the IR light beam is broken, the transistor base goes high, the emitter follower turns on and the relay turns on. In order to calculate resistors, need to know what the relay resistance is and at what voltage relay closes. Why do you want a 12 volt relay on a 9 volt battery?

The forward current on the IR emitting diode will discharge a standard 9 V battery very fast. Is this a 9 volt power supply or a 9V battery? The inductive voltage overshoot from the relay coil could damage transistor. Maybe a large capacitor is adequate. How short an IR beam interruption must the circuit react to?

Bob S

Last edited: Nov 19, 2009
3. Nov 19, 2009

mathew086

The power supply is from a 9vDC battery.I am not quite sure whther a 12 V rely or 9V relay is needed. I was jus in desiging this circuit.
IR beam should emit at a frequency of say 6 times in an hours. ( at each 10 minutes)

4. Nov 19, 2009

Bob S

I think you are much better off using 3, 4.5 or 6 volts, and a correspondingly lower voltage relay, depending on the sensitivity of the IR receiver. How far apart are the IR transmitter and receiver? A high voltage on the IR transmitter just wastes power. You might have a better sensitivity using another transistor geometry, either a common emitter npn or pnp, with a collector output to the relay.
Bob S

5. Nov 19, 2009

mathew086

The emitter and receiver is 5 cm apart. Only a 9VDC supply is available. Is is possible to adjust the parameters to user 9 VDC?
Or if a 9VDC is used. how much power is consumed by the transmitter? ( wasted energy)

6. Nov 19, 2009

Bob S

Hi
The 9 volt power supply will have to be a regulated supply and not a battery, because the turn-on voltage of the relay will be about 8 volts, so you will have only about 1 volt to spare. This also means that you will have to use a common-emitter rather than a common-collector geometry. Also, you will need to omit the series diode out of the 9 volt supply.
For the IR transmitter, use a ~1000-ohm series resistor to give ~ 8 mA diode current. If this is insufficient current, reduce the resistor to ~200 ohms minimum. For the IR receiver string, use a 10,000-ohm series resistor to the IR receiver anode, and the cathode is grounded. The npn transistor (BC-337) has the emitter grounded, and the relay connected between 9 volts and the npn collector. I am assuming that the relay coil resistance is about 1200 ohms. For a ~10 millisecond time constant. Put a 10 uF capacitor in parallel. The IR receiver anode is tied to a diode (e.g., 1N914), which is tied to the npn base. The npn base should also be tied to ground through a 10,000-ohm pull-down resistor. When the IR receiver is ON, the npn base will be LOW, and the npn collector high. When the IR receiver is OFF, the transistor will be ON (collector low). See thumbnail.

0.75 mA current in the IR receiver should be enough to turn the NPN OFF. The npn gain is >100, so the 1000-ohm pull-up should be able to sink 50 mA.

I hope this helps.

Bob S

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7. Nov 19, 2009

Bob S

Here is a circuit simulation of the above IR receiver. Input current from IR diode: 800 microamp pulse (red), right scale. 9 volt pulse (black) on transistor collector output, left scale. The relay turns ON when collector drops below about 1 volt, and OFF when the collector rises above about 5 volts.
Bob S