Using a simple light circuit to power a motor, when using 5 transistors, the motor spins fast increasing in speed and amps as each transistor is added.

However, when I added a 6th transistor, the speed (and amps?) decreased, and continued to decrease as I added more transistors.

I was using 6V, with a 6V motor and bc548 npn transistors - the transistors where connected from emitter to base, and the collectors all connected in series with the motor.

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    \$\begingroup\$ Yo dawg, I heard you liked darlington tansistors, so I put a darlington in your darlington, in your darlington, in your darlington, so you can have current gain in your current gain in your current gain in your current gain. \$\endgroup\$ – Connor Wolf Oct 23 '11 at 6:19
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    \$\begingroup\$ what you need is one halfway decent MOSFET \$\endgroup\$ – JustJeff Oct 23 '11 at 14:34
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    \$\begingroup\$ Darlingtons gone wrong.. \$\endgroup\$ – m.Alin Oct 23 '11 at 16:17

Here is a fixed version of your picture to make it easier to talk about. Your original was unnecessarily large.

The problem is you took a good idea and went too far with it. You apparently heard that putting two like transistors together with the emitter of the first driving the base of the second makes a overall transistor with the gain of the two combined. That is true, but there is also a drawback. The minimum voltage drop accross the combined transistor is higher than a single transistor. One transistor can get down to about 200mV if driven on well, and each additional transistor in a darlington configuration adds about 600-700mV.

With a single transistor in your setup, the circuit was limited by its gain. All the base current has to come thru the top resistor in your schematic (next time add component designators to make it easier to talk about). Apparently that doesn't allow for enough current to turn on the transistor fully and therefore run the motor at its maximum speed. When you added the second transistor, the additional gain allowed more current to flow thru the motor. Eventually, the gain was sufficient to turn on the combined darlington transistor as much as possible, so adding more transistors only increased the voltage drop, which took voltage away from the motor. It also increased the voltage required from the voltage divider between the top resistor and the LDR to turn on the motor. Both these effects together started making the motor current go back down after some number of transistors were added.

It would be useful to know what part values you used. Again, draw the schematic properly next time. This includes component designators and part values.

Assuming this is a CdS light-dependent resistor, it will decrease resistance as more light shines on it. In your cicuit, that means the motor will go on as it gets dark. It that what you intended? If so, here is a circuit that should work:

This will support up to about 1A motor current in full dark when the LDR is at 300kΩ. The available motor current will decrease as the light gets brighter and should shut off altogether when the LDR gets down to about 60kΩ. Adjust R2 for different light threshold levels.

This circuit works similarly to what you intended, except that the collectors of the extra gain transistors are tied to the power supply, not the collector of the power transistor. This avoids adding extra on-state voltage per stage. Three stages is plenty of gain. Conservatively, the power transistor can be counted on for a gain of 15, and the other transistors for a gain of 50 each. 50 * 50 * 15 = 38k. To allow for 1A to flow thru the motor therefore requires only 1A / 38k = 27µA thru the base of Q3. That's small enough that the relatively high resistances of R2 and R3 can provide it.

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    \$\begingroup\$ +1 for not just explaining how good concepts go bad, but trying to help the OP out of the situation too. \$\endgroup\$ – JustJeff Oct 23 '11 at 14:37
  • \$\begingroup\$ @Olin what did you use to draw that schematic? \$\endgroup\$ – gallamine Oct 26 '11 at 13:04
  • \$\begingroup\$ @gallamine: Eagle. \$\endgroup\$ – Olin Lathrop Oct 26 '11 at 13:07

What you have constructed is sort of an n-th order darlington configuration. The normal idea there is to use a pair of transistors constructed on the same die (sharing their collector) to obtain more gain than is possible with one alone. But with six, you've probably exceeded the point of diminishing returns and found a point of negative returns!

You might find it useful to review the wikipedia article on the normal paired darlington http://en.wikipedia.org/wiki/Darlington_transistor and contemplate what will happen with the mentioned disadvantages when you expand the idea to six transistors.

Your effective base-to-emmiter voltage of the 6th-order configuration may be reaching nearly half of your supply voltage.

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A BC548 is not suited to power-transistor use. They are good transistos in the their proper applications but this is probably not a suitable application. A single transistor of a different type will be able to do the whole job well. We can comment usefully on what's needed when we know the maxiumum motor current when operated on 9 Volts. Olin's circuit will work but may be overkill depending on motor current.

BC548 datasheet here

It can control a maximum current of 100 mA
At 100 mA it may waste over 0.5V across the transistor.
At 5V supply that means > 10^ of the energy is wasted in one transistor.

There are ways of causing N x BC548s to control up to N x 100 mA BUT you may be better off with a different (better) transistor.

Using a proper" transistor you cam use 1 x transistor to do your whole job.

In your diagram - Each transistor in series C to E will waste 0.2V to 0.6V.
This comes off your PV output. Vb is up to 0.7V per transistor

More once we know more.

enter image description here

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  • \$\begingroup\$ Schematic is not correct. Please see the OPs new uploaded drawing. \$\endgroup\$ – Connor Wolf Oct 23 '11 at 6:20
  • \$\begingroup\$ @Russell - I may be wrong, but I don't think the OP is using a PV panel, I think by "light circuit" he just meant the LDR. \$\endgroup\$ – Oli Glaser Oct 23 '11 at 15:43

In order to turn a transistor on, you must apply about 0.7V across its base and emitter. For your original circuit in the drawing to work, you need 5 x 0.7V = 3.5V at the base of the uppermost transistor. For 6 transistors, you need 6 x 0.7V = 4.2V at its base.

And so on, as you connect more and more transistors, your circuit will start turning on in more bright environment, therefore your motor will rotate slower compared to a fewer transistor scheme.

Also, don't forget that there is a transition region between the on and off states of a BJT. In other words, if you connect 7 transistor you will need 4.9V at the base, but that doesn't mean that motor will not run if you apply 4.5V to the base; it will still run, but will be running slower. The transition region will become larger as you connect more transistors, which is an undesirable situation when using a transistor as a switch.

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