DC motor

I'm sorry to say Robert it still doesn't work. I don't know what I'm doing wrong. Am I supposed to have two power supplies? Here are the lastest pictures if you want to have a look.

I you guys feel like I'm wasting your time just let me know and I won't bug you anymore.

Thanks, Raul

The connection on the op-amp that you have labelled as pin 4 should really go to the inverting input, pin 2, and your pin 7 connection should really go to the non-inverting input pin 3.

The op-amp also need power supply connections, which are not shown on the schematic, Pin 4 goes to the negative power supply, and pin 7 goes to the positive supply.

The two batteries should be connected in series. The black lead of the top battery should be connected to the red lead of the bottom one. This point is "ground", and the black lead of the motor should connect to it.

Then the red lead of the top battery becomes your "+V", and connects to the 2N3904 collector and to the op-amp pin 7 and to the pot red lead.

The black lead of the bottom one becomes the "-V", and connects to the

Your basic problem is that you are confusing the + and - inputs with the power supply. Pin 7 should go to the highest voltage (the + terminal of the positive side battery pair) and pin 4 should go to the lowest voltage (the - terminal of the negative side battery pair). These are the POWER SUPPLY pins. The things marked - and + on the triangle in the schematic are really the 'inverting input' and 'non-inverting input', respectively.

Thus, your opamp needs 4 inputs, two for power, and two for the + and - inputs. The output goes to bases of the transistors.

Also, your connections for the two transistors look wrong. They should be like this (view with fixed-space font):

• | e b c 2N3904 | | motor---o o-------- ctl | | e b c 2N3906 | -

With this picture, they should both be flat-side 'down', towards negative. Look at the datasheet for the transistors on

One final thing: if you take the feedback to pin 2 from the emitters of the transistors (they are connected together) then your opamp will compesate for any wierdness in the push-pull circuit, making the transition much smoother (otherwise, there will be a big spot when the output is at ground)

One other final thing: If you use two similar circuits, and hook them up so one goes up as the other goes down, you will be able to make the motor go much faster. There is a simple op-amp circuit to do this:

VCC ---------------- | | .-. | | | R2 | | | | R1 '-' | ___ | .----|----|___|--. | | | | | R1 | | | | ___ | |\\| | Input ----|--|___|--o--|-\\ | | | >----------o-Inverted Output o------------|+/ With respect to | |/| Ground | | | | .-. | | | R2 | | | | '-' | | | ---------------- -VCC (created by AACircuit v1.28 beta 10/06/04

Thus, using an LM324 and two more transistors, you could double the voltage range across the motor.

Still wrong. The + sign inside the triangle amplifier symbol refers to pin 3, the noninverting input, and the - refers to pin 2, the inverting input. After you move the connections on those pins, connect pin 7 (V+) to the positive side of the battery and pin 4 (V-) to the negative side of the battery.

This design needs two batteries connected in series, so that the middle point between them can serve as the ground point for the motor. That way, if the upper transistor turns on, the upper battery drives the motor one way and if the bottom transistor turns on, it allows the bottom battery to drive the motor the other way. I would start with two nine volt batteries, to make sure the op amp has enough voltage to function. There are opamps that work on less, but this one needs at least 6 volts total, before it can swing its output much at all.

You have two supplies, two battery units. Connect plus from one with minus from the other to get your ground.

I don't think you are wasting anybodies time, you do a lot to explain your situation, and your pictures are very helpful.

I have reloaded the web page to get the latest version. In the first picture I see the faulty numbering of the pins is still there.

The pins marked V- and V+ in the second picture, are the power input pins, not the op-amp input pins. The inputs are marked Inverting input and non-inverting input.

In the schematics you still have only 3 connections on the op-amp. This has led you to the faulty use of the pins.

Start by adding vertical line into the op-amp triangle, from the top side, write V+ beside it. Add vertical line into the op-amp from the down side, write V- beside it. Erase these symbols from where you had earlier written them. These arrows you have drawn point to the op-amp inputs, often marked with - or +. But not with V+ and V-, they are the power connection labels.

Now you have an op-amp with 5 connections. Make sure you mark and use them correctly.

On the breadboard, in reality, you have connected the battery pack to the power rails, that is good. But you should add a wire between the minus of the upper battery pack to the plus of the lower pack. That wire and the rows of holes connected to it, is your ground.

I have not checked the rest of the circuit, you need to fix these big problems to begin with.

Here is a better schematic, that shows everything including pin numbers:

note the batteries, which are 9V. Your 3V batteries may not work, as JP pointed out. I bet the 741 can't get closer than about a volt to the top rail.

Here is a schematic for the circuit I proposed, that gets you twice the voltage across the motor

Here are some fuzzy pictures of the second circuit, which is using an LM324 opamp instead of two 741s:

Here is a circuitmaker input file of the circuit:

Good Luck.

Ok, I changed the schematic to add in the voltage. I also changed to a higher voltage as suggested incase that was a problem. I have it wired as the new schematic shows and the motor turns. Yea! Only one direction though. The 3904 resistor get very hot very quickly so I have to unplug the power. Atleast I'm making progress. Here are the new pics.

Thanks, Raul

Ok. Could you check the drawing and make sure I am connecting the transistors correctly. I'm off to the store to buy some 9v batteries.

Thanks, Raul

Last question then I'll bug off until I can get it to work. I'm confused on the ground. What is the ground on a bread board? There's no frame or structure.

Raul

Terry, sorry about that. When I got the new schematic I stripped everything off the board and started from scratch. The last photo was not valid anymore. Thanks for the info about the ground. Raul

Terry, I'm sorry about that but I had stripped the board when I got the new schematic, so the photo wasn't valid any more. Thanks for the info about the ground though.

Raul

It only goes one direction because one side of the motor is grounded. You could just get rid of the 2N3906, because it will never do anything with this hookup.

If you want it to go in both directions, you could use dual batteries, and connect the other side of the motor not to the negative rail, but to the point between the batteries. See the schematic I posted in another area of the thread.

Note that your transistors aren't going to be able to take much current. The 2N3904 is limited to something like 200mA, and 650mW.

The term "ground" is used in several different (and confusing) ways in electronics.

For portable battery operated circuits like yours, ground is just the point in the circuit that the designer decides to call "0 volts" - in this case, "ground" is the junction of the two batteries, so that we can say that one battery provides +9 volts, and the other provides -9 volts.

Your schematic shows the emitters of the two transistors (the leads with the arrows on them) connected together, and to the motor, but the wiring layout on the right shows the emitter of the 3906 connected to the collector of the 3904.

The drawing of the transistors is incomplete and looks wrong in several ways.

Remember that the emitters of the transistors are connected to each other, as well as the bases. The power lines from the batteries should be connected to the collectors. The other connection of the motor to ground, not to V-.

You have to learn to follow a system when checking connections you make.

Look at every component, and every connection to that component, and check what it is connected to, both in the schematics and in reality.

For a battery powered circuit, ground is just a node that has several things connected to it, and conceptually, is a sort of reference point, so it actually may have nothing to do with the Earth or a frame or case. Putting a ground symbol at each component that connects to that common node just saves you from having to draw so many lines on your schematic. Likewise, you could have labeled the positive 9 volt battery +9 and the negative 9 volt battery -9 and eliminated those lines by labeling each connection to that node with these labels.

For this circuit, you could just as well eliminate the symbols and connect those points (one side of the motor and the connection of the two batteries) with a line.

Your proto board has 4 long lines running along the edges. I would use the outside pair as +9 and -9 and one of the inner lines as the ground. This makes the layout look more like the schematic and makes checking the connections easier.

With a single battery: it's the negative terminal.

With two batteries connected as you've been shown, it's the point of connection.

In either case, use the appropriate stripboard section, often marked with a black line.

I was going to give you specific advice, by referring to one of your photos. But you've used the same file name for every successive update. So now the only diagram accessible appears to be the schematic. In future similar discussions, remember to upload each of them with different names, like

In most tubes, the cathode is indirectly heated - the filament or heater is inside the cathode, but electrically insulated from it.

However, there are some directly heated tubes, where the filament also acts as the cathode.

For NPN transistors, the arrow faces away from the body, and marks the emitter.

For PNP transistors, the arrow faces *towards* the body, and again marks the emitter.

The fact that they are both called the emitter, but in the NPN case is negative of the base, but in the PNP case is more positive than the base is annoying, but actually means something. In the NPN case, the emitter is "emitting" electrons into the body of the transistor (which are travelling against the arrow, oddly enough). In the PNP case, the emitter is emitting 'holes', which are gaps in the silicon where an electron should be, into the body of the transistor. This time, the 'holes' are travelling in the same direction as the arrow. Confused yet?

This is all just like tube terminology (as I'm learning). Tubes have a cathode, also called a filament, which emits electrons into the space around the filament. The grid controls the movement of these electrons towards the plate. The plate is the 'collector' of the electrons, and returns them to the supply. Thus, when they were thinking up names of transistor pins, they used the same ones; the cathode of the NPN is the 'emitter', and the anode of the NPN is the 'collector. In tubes, the 'charge carrier' can only be electrons. However, for PNP transistors, the 'charge carrier' is not electrons, it's gaps in the valence band of the atoms making up the device (which makes the atom positive). They call these 'holes', and say they are being 'emitted' by the emitter, and 'collected' by the collector, under control of the voltage at the base (which is analogous to the grid).

Way more information than you needed to switch the wiring of your 2N3904 (NPN) around, so that the emitter is negative, and the collector is positive.

Sorry Raul, I had to go out of town for a few days. It looks like you did get some good help while, I was gone. The guys here are usually very helpful. Looking at your drawing, everything looks good execpt for the way you have the two transistors hooked up. Both transistors emitters should be connected together and then go to the motor. Both transistors bases, should be conntected together and then go to the output of the op-amp, pin 6 (as you have). The 2N3906 transistors collector should go to the - side of battery V2 (as you have). The collector of the 2N3904 transistor, should go the the + side of the battery V1.

You haven't told us what the current rating of the motor is. The higher the battery voltage is, the more power that will be wasted in the transistors, which will make them get much hotter. The two transistors you picked, are only rated at 1/2 watt (at an ambient temperature of 25 deg. C). The more current the motor has to have, the more power the transistors will have to handle (with a given voltage), which means more heat. Knowing the current rating of the motor, we could tell you if these two transistors will be good enough (and if not, suggest a better choice). Brian