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Old 2nd March 2004, 11:19 AM   (permalink)
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Well, I supose I'm wrong about only one transistor, as Cyn70 may not always be saturated (depends on light intensity), so I think it may require 2 transistors? One for on-off and the motor driver?
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Old 2nd March 2004, 12:02 PM   (permalink)
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Quote:
Originally Posted by patroclus
Well, I supose I'm wrong about only one transistor, as Cyn70 may not always be saturated (depends on light intensity), so I think it may require 2 transistors? One for on-off and the motor driver?
I don't know what a cyn70 is, is it a slotted opto-switch, or a reflective opto-switch. By using the gain of the extra transistors it should help to switch the output more positively.
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Old 2nd March 2004, 12:45 PM   (permalink)
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It is a reflective optical sensor with transistor output.
I just was wondering on what transistor configuration to choose, better than these 3 kind of bad designed transistors.
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Old 2nd March 2004, 01:24 PM   (permalink)
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Look at this schematic
http://www.eurobotics.org/robotica/scav/puenteh.gif

It also uses BD136 (kind of BD140) to operate a H bridge, with 4,7K base resistor! This takes us to a even lower colector current... I supose this is not coincidence..
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Old 2nd March 2004, 03:01 PM   (permalink)
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I had a few spare minutes, so I did some tests. I used a BD132 (PNP), emitter to +6V, collector to a 6V 6W (1A) bulb, and bulb to -ve. I tried different value resistors from base to -ve, and measured the voltage drop between collector and emitter, these are the results:

1K - 2.50V
470 - 1.93V
220 - 1.35V
100 - 0.62V
47 - 0.22V
39 - 0.20V
33 - 0.19V
27 - 0.18V
22 - 0.18V
18 - 0.18V
10 - 0.18V

Certainly 1K looks a bit useless :lol: as low as 27 ohm is required for full saturation - I didn't even bother trying 2.2K!.
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Old 2nd March 2004, 04:49 PM   (permalink)
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I also tried things
But I tried with the first 2 transistors, not yet with Q3. I builded the circuit as it is, the same thing, except Q3. When Q1 is ON, it is in active region, and Q2 in full saturation. Vce in Q1 is around 4.9 V.
Well, i supose it's ok, isn't it? I supose Q1 can never be saturated, can it?..
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Old 2nd March 2004, 04:58 PM   (permalink)
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Nigel, I don't understand I don't know if I'm dumb or something. Do you understand something? What about the last circuit which link I posted last, it is the same thing! (in the H bridge.. look at those 4,7ko base resistors)
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Old 2nd March 2004, 07:00 PM   (permalink)
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Originally Posted by patroclus
I also tried things
But I tried with the first 2 transistors, not yet with Q3. I builded the circuit as it is, the same thing, except Q3. When Q1 is ON, it is in active region, and Q2 in full saturation. Vce in Q1 is around 4.9 V.
Well, i supose it's ok, isn't it? I supose Q1 can never be saturated, can it?..
Q1 doesn't need to saturate, in fact it must never saturate, if it did it will probably blow the BE junction of Q2.

As for the H-Bridge circuit, yes that seems even worse - I can only imagine they are using very low power motors?.
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Old 2nd March 2004, 08:49 PM   (permalink)
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Nigel, that's what I thought, but no. :lol:
Better laugh than cry. The author says that the desing is suitable for motors of up to 1A or even more. Theoricaly, 1.5A !!!
This is really good now, because the first circuit was for small motors of up to 300-400mA...

It must be something we are missing??
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Old 3rd March 2004, 06:01 AM   (permalink)
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Originally Posted by patroclus
It must be something we are missing??
I think all we are missing is burning our fingers on the hot transistors on the very slow moving robot :lol:
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Old 3rd March 2004, 11:31 AM   (permalink)
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Well, not. Nigel, take a look at this test I did...

I used a BD135 (same as BD140 but NPN), and conected a DC motor, that at 3V takes 320mA. I set Vcc to 3V.

Motor + to Vcc
Motor - to colector
Emitter to ground
Base to Vcc across 330 ohm

As now Vbe is 0.8V, not 5V, Ib is 0.8/330 = 2.4mA (instead of 2.27mA from the BD140 circuit). Ok?? take a look at this.

With that base current, the transistor achieved full saturation... Ic = 310mA, and Vce=0.19V
Using a larger base resistor, 1k, (just 0,8mA Ib) the motor did not start until Vcc was risen to 4.5V. then it turned ON in active region, not far from saturation (Vce=1.1V) but take a look at THIS :

Ib = 0.8mA
Ic = 310mA !!! (measured by multimeter)
This lead us to a hFE of 380!!

NOW I'm really confused
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Old 3rd March 2004, 12:13 PM   (permalink)
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Quote:
Originally Posted by patroclus
As now Vbe is 0.8V, not 5V, Ib is 0.8/330 = 2.4mA (instead of 2.27mA from the BD140 circuit). Ok?? take a look at this.
Slight miscalculation, the base current is 2.2/330 = 6.67mA, giving an Hfe of 46.

For the 1K resistor it will be 2.2V/1000 = 2.2mA, which gives an Hfe of 141 - assuming the collector current is still 310mA, but if the saturation volage has increased, it's likely the current has dropped (less voltage to tne motor).
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Old 3rd March 2004, 03:15 PM   (permalink)
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Quote:
Slight miscalculation, the base current is 2.2/330 = 6.67mA, giving an Hfe of 46.
No.. it is a common emiter, and emitter is grouned. Base is at 0.8V (I checked it with multimeter too), so 0.8/330...

Remember I use a NPN, and builded a equivalent circuit.
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Old 3rd March 2004, 04:02 PM   (permalink)
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Quote:
Originally Posted by patroclus
No.. it is a common emiter, and emitter is grouned. Base is at 0.8V (I checked it with multimeter too), so 0.8/330...
It makes no difference, the current is dependent on the value of the resistor, and the voltage across the resistor (not the Vbe of the transistor, which is pretty well constant). With your 3V supply and 0.8V Vbe, the voltage across the resistor is 2.2V.
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Old 3rd March 2004, 07:09 PM   (permalink)
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Damn, you're right...
I'm almost to give up. I'm waiting for the author of the robots to answer my question...

no one in this forum has worked with this kind of circuits??
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