LDR circuit ( Application of transistor as switch)

[vick5821]

Hey guys, I plan to make this simple circuit into a teaching aid for my secondary school teacher for them to let student understand better how LDR works and as an applicationof transistor as switch. So , I plan to make a DIY PCB for that. How am I able to design my circuit so as to easy measurement of the base current ? And what type of connector should I use so that the two point series in the circuit connected to an ammeter ?

Attachment :

 

[john monks]

I'm afraid you will have to cut the circuit leading to the base and insert a microamp meter in the circuit. Looking at the circuit you may have to adjust the 100K resistor up or down the get the transistor to turn on at the right point.

 

[vick5821]

I'm afraid you will have to cut the circuit leading to the base and insert a microamp meter in the circuit. Looking at the circuit you may have to adjust the 100K resistor up or down the get the transistor to turn on at the right point.

Yayaya, cut at the leading to the base.then insert two header to be connected to ammeter right ?

 

[john monks]

Yes sir.

 

[vick5821]

what type of header?

 

[john monks]

It mask no difference. I woud probably use banana jacks so that I could easily place a current meter in series with the bas and short out the banana jacks when no meter is in place.

 

[KrisBlueNZ]

It might be simpler to put the current shunt on the board and use a multimeter measuring voltage to display the current.

For example if you choose a current shunt resistance of 1K, you put a 1K resistor in series with the base, with a 2-pin connector connected to the two ends of it. The circuit will work just the same whether or not the base current is being measured. Every microamp of base current will cause a millivolt of voltage drop across the resistor, which can be measured with a multimeter set to a voltage range.

Another possibility would be to use a jack socket with a normally closed contact, so when nothing is plugged into it, the socket is shorted out and the circuit is complete. When you want to measure the current, you plug in a jack plug that is connected to a microammeter, and the action of plugging the plug into the socket breaks the short circuit and the base current flows through the milliammeter.

I prefer the first suggestion though.

 

[vick5821]

you mean I put a resistor connected to the base ? and then I calculate the current based on theorectical value ?

 

[CDRIVE]

Maybe they're not that common any longer but I'm sure they're still available. I'm speaking of Probe Jacks, also called Pin Jacks. Your DMM probes plug directly into them. Also, have you considered placing a 1K in series with the base and measuring the voltage drop instead of current? This way the circuit will work with or without the meter connected. A 1K resistor will give you a direct uA to mV conversion. Example 52uA will read 52mV. Just connect the jacks to opposite ends of the resistor.

 

[vick5821]

Maybe they're not that common any longer but I'm sure they're still available. I'm speaking of Probe Jacks, also called Pin Jacks. Your DMM probes plug directly into them. Also, have you considered placing a 1K in series with the base and measuring the voltage drop instead of current? This way the circuit will work with or without the meter connected. A 1K resistor will give you a direct uA to mV conversion. Example 52uA will read 52mV. Just connect the jacks to opposite ends of the resistor.

Ya, now I get the idea. then I think I do not need the connector for the ammeter anymore. But I think I should make it easy to measure the voltage for the 1K resistor right ? Any easy implement method ? I plan to make this in quantity

 

[CDRIVE]

It might be simpler to put the current shunt on the board and use a multimeter measuring voltage to display the current.

Looks like we doubled.

 

[vick5821]

Eagle file :


Havent finalised yet. Still need to think what to add to make this product better

 

[CDRIVE]

Ya, now I get the idea. then I think I do not need the connector for the ammeter anymore. But I think I should make it easy to measure the voltage for the 1K resistor right ? Any easy implement method ? I plan to make this in quantity

I don't know what your resources are in Malaysia but Digikey will carry the Pin Jacks that I mentioned.

 

[vick5821]

I don't know what your resources are in Malaysia but Digikey will carry the Pin Jacks that I mentioned.

Can I have a look perhaps ? Maybe I can purchase it ? Thank you

 

[CocaCola]

Why not just get a panel mount analog ammeter (about $4 on Ebay) and make it part of the circuit? You could get a digital one for nearly the same cost, but will need an external power supply to run it...

 

[vick5821]

Why not just get a panel mount analog ammeter (about $4 on Ebay) and make it part of the circuit? You could get a digital one for nearly the same cost, but will need an external power supply to run it...

Can you provide me with link ? but I dont think measuring the current is a better way since the current will be very very small indeed

 

[CocaCola]

Ebay search for "ammeter", selected lowest price first and you should get pages of them for about $4-$5 delivered... Pick one of the appropriate rating...

 

[vick5821]

http://www.ebay.com/itm/DC-1mA-Ampm...ltDomain_0&hash=item1e6fce09b6#ht_2676wt_1037 ?

Is it this ? Dont you think measuring base current is somehow not a good idea ? Like what they said, add a shunt resistor connected to the base and measure the potential difference and claculate for the current ?

Thank you

 

[vick5821]

Why not just get a panel mount analog ammeter (about $4 on Ebay) and make it part of the circuit? You could get a digital one for nearly the same cost, but will need an external power supply to run it...

Some of the digital one do not need extra power supply

 

[CocaCola]

Some of the digital one do not need extra power supply

You might want to double check your findings, digital ones have displays that need to be driven, thus they need a power source...