First let me say this is a great forum; first time poster...

I am working on a project for work. I am using a 555 timer in monostable to supply power to a LM7805, 2 74HCT573's, 20 LED's, 1 MAX6816 switch debouncer and a relay. I designed in a 2N4401 (NPN - 600mA BJT) on the output of the 555 to supply sufficient current, however I am concerned that the 9V battery won't serve as a sufficient power supply (at least for very long). Does anyone know if the BJT will draw significant current causing the 9V to drain quickly or do I not need it at all. Perhaps a mosfet is better?

The flags connected to the inputs of the 74HCT573's are 5V signals being read back from the circuit this will be talking to.

If anyone can add any comments or steer me in the right direction that would great.

Thanks

Cory

 

Welcome to AAC!

A 9V batteries life will be short, if you can live with that then it is good enough.

MOSFETs are almost always better, but I still use BJTs a lot because they are so available. However, unless you get a logic level MOSFET 9V is not enough voltage. A typical MOSFET needs a minimum of 10V to turn on completely, which makes 12V ideal.

LEDs, 555s, Flashers, and Light Chasers

The 555 Projects

 

Thanks for quick reply Bill.

This will be a wireless device so it will have to run on batteries. I looked at the A23 12V battery from energizer, but read that those were horrible...

Is there anything I can do with the design to make it more efficient?

 

Relays eat power like nobody's business. The coil is a constant drain as long as it's energized.

You might think that your Q1, 2N4401 is supplying power to the input of the 7805 regulator, but all it will do is short the input to ground. There IS no source of input power to the 7805.

A typical 9v PP3 "transistor" alkaline battery is rated for 500mAh; that's for a constant 25mA load over a 20 hour period (500mAh/20=25mA). If you subject the battery to a higher current draw (like you want to do), their life will be very short.

A single typical LED and a resistor would be about a 20mA load, or 4/5 of the rated output of a typical battery.

 

I don't know what your probe is supposed to do, but you would likely be much better off to use AA batteries in a pack. Four Ni-CD or NiMH AA batteries in series would output ~4.8v. You would not need a regulator. They are typically rated for ~2200mAh or more.

Your design would be a good bit more efficient if it used a microcontroller - however, the learning curve is rather steep.

 

You can make deeper battery by using 6 AA batteries instead, and it won't take up much more space. Battery holders are cheap, and with a little work you can find one with a 9V battery clip on it.

Loose the relays, go with transistors (MOSFETs or BJT) for the switching. Relays are generally power hogs.

Look into getting some logic level MOSFETs to replace the relays with.

Chapter 10 of LEDs, 555s, Flashers, and Light Chasers covers how to use transistors, though not quite for your application.

I'm not going to ask, and you may not want to tell. Automotive applications will get your thread closed, since they are against out new TOS (Terms of Service). See the link on the bottom of this page.

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Wookie, he is using 1KΩ resistors for LED current limiting, figure under 10ma, which is good.

 

With 1k resistors, assuming the LEDs have a Vf of 2v, there will be ~3mA current per lit LED, up to ~48mA total for just the LEDs.

The 74HCT573 is an octal transparent latch/octal D-type FF with 3-state outputs. Our OP is trying to latch some kind of data and display it when some event triggers the latch.

 

This is not an automotive application. It is used to monitor depths of holes that are being drilled. LED's will go out when the depth is reached. The signals can be latched for data collection purpose. I will see what I can do to get rid of the relay's. K1 one is used to completely remove the load from the battery when the 10 minutes expires, this will stop any draw unnecessarily on the battery when not in use. I will also look into the way I have the transistor placed. Any comments on how the transistor should placed in the circuit?

 

Basically series pass. You can use power BJT types, and if the power supply voltage is big enough you can use MOSFETs. Pay attention to polarities. How experienced are you with transistors in general? I can sketch something up if you need it.

 

I would appreciate any help Bill, if you have spare time to sketch something that would be great.... Thanks

 

Example 1 is good for medium and low power applications. You do have to pay attention to wattages of the resistors.

If the voltage in is +12V then the voltage at the base is +11.3V. R2 would carry 0.0113A, so the power across the resistor is V * A = 0.128W, a ¼W would work. This power is continuous, but is still much better than a relay. As a general rule of thumb a BJT can conduct 10X the current on the base, so max current for this switch is around 0.113A.

The circuit in example 2 is better, but MOSFETs have the limitation of their bias voltage. Figure 10V on the gate as a minimum. They draw a quick surge while switching, then draw almost nothing after. Power MOSFETs can handle unbelievable amounts of current at a very low ohmage. If you need lower voltage you can use a logic level MOSFET. They are more delicate to ESD and you can't overdrive the gate even a little, but it will work.

In both cases S1 can be logic or a switch. In many cheap circuits with low power applications you can use a capacitor on the gate of the MOSFET instead of a switch. Discharge the cap, and the circuit will turn off automatically after a set duration. You can not do this with power though, because the MOSFET will linger in its linear region for a long time, generating heat.

A typical example of what would turn a circuit on/off like this on both examples is an open collector comparator.

 

Thanks for the info Bill. I have redesigned the power on/auto off circuit and the latching circuit to remove the relays. Do you see any issues? When PB1 is pressed it will supply power to both the 5V regulator and the timer. When PB1 is released it will start counting and when the timer is finished it will isolate the batteries from the entire circuit.

 

Pin 4 of a 555 (or CMOS 555) can not be left floating. You need a resistor going to Vcc on pin 4.