Transistors and Resistance

by RhysGM
Tags: resistance, transistors
 P: 8 Hello, I have a transistor (BC547B) that is being used as a switch, connected to the base I have an 9v battery and an LDR with a maximum resistance of 1m ohms. However this is still not enough not to switch off the transistor. Can someone tell me how I can work out what resistors I need, so the transistor is on in the light and off in the dark? Links for the components I'm using; http://www.rapidonline.com/Electroni...4-1e06a4f70d0c http://www.rapidonline.com/Electroni...sistor-58-0134 Thanks for any help you can give.
 Sci Advisor P: 5,431 Show us the complete circuit diagram that doesn't work. Just saying you "connected a battery and LDR to the base" isn't enough information for us to guess what the problem is.
P: 8
 Quote by AlephZero Show us the complete circuit diagram that doesn't work. Just saying you "connected a battery and LDR to the base" isn't enough information for us to guess what the problem is.
I'm trying to replicate this, but I think I bought the wrong LDR so I'm trying to work out how to compensate.

P: 3,798

Transistors and Resistance

You should be able to adjust the VR1 to make TR1 turn off. I don't care how wrong is the LDR, if you adjust VR1 to 0, the transistor will turn off, unless, you connect the components wrong.
P: 8
 Quote by yungman You should be able to adjust the VR1 to make TR1 turn off. I don't care how wrong is the LDR, if you adjust VR1 to 0, the transistor will turn off, unless, you connect the components wrong.
Is that because resistors in parallel can be caluculated by;

R5 (47kΩ) * VR (0Ω) / R5 (47kΩ) + VR (0Ω) = 0 / 47k = 0Ω resistance?

So am I to assume if there is no resistance, the current will not even go through the transistor? Would this still be the case if there was nominal resistance say 1Ω?

If this is the case I must have wired up my circuit incorrectly.

http://www.rapidonline.com/Electroni...ometer-67-0470

This is the variable resistor I'm using, it has 3 legs and I've connected the middle leg and one of the outside legs and left the other one connected to nothing. Is this correct?
P: 5,462
 This is the variable resistor I'm using, it has 3 legs and I've connected the middle leg and one of the outside legs and left the other one connected to nothing. Is this correct?
Well that could be one way to connect a pot as a variable resistor but it does not conform to your circuit diagram, which uses the other way and connects all three terminals.

It also depends upon what you have connected the middle leg and one outer to.

Since you have made this construction error, check and double check the rest of the circuit.

Part of the learning process is to find out how often we miswire our circuits first time, which is quite often for most people.

Another possible source of construction error is to check that your transistors have the same pinout as that given in the circuit.
It is easy to identify the b, c and e leads with an ohm meter. Check the direction of the two diodes in the device.

go well
 P: 3,798 In your schematic, you have the VR1 connected with the last leg to the ground!!! It is wrong to leave one leg open. The LDR and the parallel combination of R5 and VR1 form a voltage divider. As light hit the LDR, the resistance goes lower and the voltage at the base of the TR1 goes up. When the base of TR1 goes to about +0.7V, the transistor turns on. VR1 is to use to adjust the light level when the TR1 turns on.
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P: 10,109
 Quote by yungman In your schematic, you have the VR1 connected with the last leg to the ground!!! It is wrong to leave one leg open.
It's hardly that desperate to do it that way. It looks, though, that you connected the two ends of the variable resistor into the circuit and left the slider disconnected - hence, it was always 47kR.

I'd bet the transistor would turn off if your removed the LDR completely from the circuit (it should, if the rest of the circuit is wired right).
But check, check and check again. Better men than you and I have managed to get that sort of thing wrong, initially.
P: 3,798
 Quote by sophiecentaur It's hardly that desperate to do it that way. It looks, though, that you connected the two ends of the variable resistor into the circuit and left the slider disconnected - hence, it was always 47kR. I'd bet the transistor would turn off if your removed the LDR completely from the circuit (it should, if the rest of the circuit is wired right). But check, check and check again. Better men than you and I have managed to get that sort of thing wrong, initially.
It'll work connecting EXACTLY like the schematic, it won't work if only two legs are connected. You need the ground reference to set up an adjustable voltage divider like in the schematic.
 P: 5,462 For the record there are two ways to connect a pot as a variable resistance. In Rhys' circuit the pot needs to be connected as a variable resistance. The potential divider comes from the fact that this variable resistance is in parallel with another resistor. Method 1 It can be connected as Rhys has (possibly) done with one connection to one of the fixed ends of the track and the other to the wiper. This leaves the second fixed end not connected. Method 2 It can be connected using all three terminals by conenecting one fixed end into circuit and then connecting the wiper directly to the other fixed end, thus shorting out a section of track, leaving the remaining track available as a variable resistance, depending upon the position of the wiper. The second connection to the circuit is made at the other fixed end. This is the method shown in Rhys' circuit diagram. There are advantages and disadvantages to both methods but many old time hands (yungman?) will prefer the method 2 because if the wiper lifts off (disconnects from the track) as sometimes happens methods 1 can leave section of circuit open circuit, whereas the track resistance is always in circuit in method2.
 P: 3,798 There are many ways to get the job done, I might do it totally different if I were to design it. I just follow the original schematic and trouble shoot why it does not work. As per the schematic, you open the ground, it will not work.
P: 5,462
 There are many ways to get the job done, I might do it totally different if I were to design it. I just follow the original schematic and trouble shoot why it does not work. As per the schematic, you open the ground, it will not work.
 This is the variable resistor I'm using, it has 3 legs and I've connected the middle leg and one of the outside legs and left the other one connected to nothing. Is this correct?
Yes indeed but we don't know what Rhys connected the two ends to. I remember when I first started I used to wonder why method 2 was always designed (until I found out) when method 1 was simpler to connect. I often thought I was being clever using method 1 and I was lucky enough to get away with it.
 P: 8 I'm actually using Stripboard Magic to work out the logic for me as I got a couple projects wrong on my own. I've screen printed what Stripboard Magic told me to do however I assumed it got the VR wrong so I basically moved everything on row 10 down to row 11 without the break. This did not work and after your suggestions I put a wire between VR1a and VR1w. This also doesn't work, however when the VR is set to 0 the buzzer changes pitch. I think I've tried both methods 1 and 2. I've used a sharp knife to score between the strips and checked with a magnifying glass and trippled checked everything is the same on my board as the diagram. Have I messed up using the VR? Does anyone have a simple schematic that teaches me about VRs and potentimeters etc? Stripboard Magic doesn't like using all 3 points, so something simple I can work out myself would be great. Any suggestions? Cheers. Attached Thumbnails
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