Can someone review this circuit design? - Electronic Circuits Projects Diagrams Free

azari · 24th November 2003, 04:54 AM

OK, I was fiddlin' around today and I made this super simple circuit for infared obstical avoidence. Well, I just want some opinions(good or bad), so I'll hand over the diagram.

Please note, all transistors are 2N3904 unless otherwise stated.

nettron1000 · 24th November 2003, 11:51 AM

Looks ok to me. Im assuming you meant to connect the second IR (bottom 1/2 556) transistor driver to +V pin 14 in the schematic.

Roff · 24th November 2003, 02:58 PM

Looks OK, except your H-bridges won't work. You have emitter follower pullups (Q1, Q3, Q5, Q7), so the emitter can only get to 0.7v below a logic "1", minus the drop in the 2k base resistor, which can be sizeable unless you have motors that only draw a few milliamps - not likely. You can solve that problem by eliminating the base resistors in the emitter followers, but a logic "1" from an LS gate is only about 4 volts on a good day, so you will NEVER get more than about 3.3v at the emitters, even with little or no load.
One solution is to run your entire circuit off 9 volts, and use CMOS CD4000 series logic. Then your emitters can get to 8.3 volts. Keep in mind that the 2N3904 is not a high current transistor, and the beta goes to hell above 100ma.
All your switches should pull to GND, with a pullup resistor to +V. Switching to +V does nothing on TTL gates (and, I believe, LM556), because the inputs float high anyway. You can switch high and use pulldown resistors, but they waste a lot of current, because the logic low input current in a TTL gate is much higher than the logic high input current.
If you change to CMOS logic, input current is essentially zero, so you can do your XOR switches either way, so long as you use a resistor to pull to the other rail. Never leave CMOS logic inputs floating. Leaving TTL inputs floating is not a good idea, but you can usually get away with it.
The only other comment I have is that the datasheet for the IR receiver (from Radio Shack) didn't show the need for a pullup resistor, but it probably won't hurt anything.

nettron1000 · 24th November 2003, 11:50 PM

I should have mentioned that it might be a good idea to place diodes across the transistors in reverse polarity to protect them from EMFs from the motor. Heres a link to a motor control tutorial which shows a bridge circuit very similar to what you have but they include protection diodes across the driver transistors.

http://web.mit.edu/~dwalker/www/Motor.htm

azari · 25th November 2003, 04:33 AM

Thanks for all the info. I didn't really think about the power consumption of the ttl chips. I was just wondering, If I used this H bridge, would it work better? I designed it with both suggestions in mind:
-Diodes for reduction of EMF interferance
-third set of transistors that will give a 9V signal to the h bridge for more power, but still run off a logical output.

Roff · 25th November 2003, 04:06 PM

The new transistors you added will go up in smoke, because you have no collector resistor. I have edited your schematic and posted it below. Note that you don't need resistors in series with the bases of the emitter followers.

azari · 26th November 2003, 03:39 AM

thanx. are the values of the collecter resistors and the one connected to the base important?

Roff · 26th November 2003, 04:32 AM

The collector resistors provide base current to the emitter followers. The base current will equal motor current divided by beta. Let's say your motor current is 100ma, and transistor beta is 50 (this is just an example). The base current will then be 2ma. If you use 500 ohm collector resistors, they will drop 1 volt. Adding this to the base-emitter voltage of ~0.8v, the emitter voltage will be 7.2 volts. You've lost 1.8 volts of your potential motor drive.
The new transistors will have to sink 18ma (9v/500 ohms in this example), so you should scale their base resistors to supply about a milliamp.
The lower transistors have to sink 100ma (in this example), so you should scale the base resistors to supply about 5ma(!). This is because beta goes to hell at low Vce, and you want to make sure they are saturated. That's 6ma from your logic gates - not impossible, but TTL gates won't source that much current unless you use a buffer or add a pullup resistor. You can see that bipolar transistors are not ideal for H-bridges. I would personally use complementary MOSFETs - P channel pullups, N channel pulldowns. I'm sure you can find examples on the Internet.

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24th November 2003, 11:51 AM	(permalink)
nettron1000	Looks ok to me. Im assuming you meant to connect the second IR (bottom 1/2 556) transistor driver to +V pin 14 in the schematic. __________________ Two roads diverged into a woods...i took the one less travelled by. R. Frost.

24th November 2003, 02:58 PM	(permalink)
Roff	Looks OK, except your H-bridges won't work. You have emitter follower pullups (Q1, Q3, Q5, Q7), so the emitter can only get to 0.7v below a logic "1", minus the drop in the 2k base resistor, which can be sizeable unless you have motors that only draw a few milliamps - not likely. You can solve that problem by eliminating the base resistors in the emitter followers, but a logic "1" from an LS gate is only about 4 volts on a good day, so you will NEVER get more than about 3.3v at the emitters, even with little or no load. One solution is to run your entire circuit off 9 volts, and use CMOS CD4000 series logic. Then your emitters can get to 8.3 volts. Keep in mind that the 2N3904 is not a high current transistor, and the beta goes to hell above 100ma. All your switches should pull to GND, with a pullup resistor to +V. Switching to +V does nothing on TTL gates (and, I believe, LM556), because the inputs float high anyway. You can switch high and use pulldown resistors, but they waste a lot of current, because the logic low input current in a TTL gate is much higher than the logic high input current. If you change to CMOS logic, input current is essentially zero, so you can do your XOR switches either way, so long as you use a resistor to pull to the other rail. Never leave CMOS logic inputs floating. Leaving TTL inputs floating is not a good idea, but you can usually get away with it. The only other comment I have is that the datasheet for the IR receiver (from Radio Shack) didn't show the need for a pullup resistor, but it probably won't hurt anything.

24th November 2003, 11:50 PM	(permalink)
nettron1000	I should have mentioned that it might be a good idea to place diodes across the transistors in reverse polarity to protect them from EMFs from the motor. Heres a link to a motor control tutorial which shows a bridge circuit very similar to what you have but they include protection diodes across the driver transistors. http://web.mit.edu/~dwalker/www/Motor.htm __________________ Two roads diverged into a woods...i took the one less travelled by. R. Frost.

26th November 2003, 03:39 AM	(permalink)
azari	thanx. are the values of the collecter resistors and the one connected to the base important? __________________ The best way to learn is by blowin\' stuff up and rebuildin\' it

26th November 2003, 04:32 AM	(permalink)
Roff	The collector resistors provide base current to the emitter followers. The base current will equal motor current divided by beta. Let's say your motor current is 100ma, and transistor beta is 50 (this is just an example). The base current will then be 2ma. If you use 500 ohm collector resistors, they will drop 1 volt. Adding this to the base-emitter voltage of ~0.8v, the emitter voltage will be 7.2 volts. You've lost 1.8 volts of your potential motor drive. The new transistors will have to sink 18ma (9v/500 ohms in this example), so you should scale their base resistors to supply about a milliamp. The lower transistors have to sink 100ma (in this example), so you should scale the base resistors to supply about 5ma(!). This is because beta goes to hell at low Vce, and you want to make sure they are saturated. That's 6ma from your logic gates - not impossible, but TTL gates won't source that much current unless you use a buffer or add a pullup resistor. You can see that bipolar transistors are not ideal for H-bridges. I would personally use complementary MOSFETs - P channel pullups, N channel pulldowns. I'm sure you can find examples on the Internet.