AD 149 in place of sk 100 as switch in a circuit

[rajaram04]

Hello sir
please refer the above ciruit , i am replacing sk 100 in second stage with old model AD 149
Is it ok or not , please help in designing . . thanks

 

[mvs sarma]

SK100 apart, i fear 5 white leds in series and 470 ohm, crosses the supply barrier and they may not work. i suppose Vf of white LED is 3.7V.

For initial tests you might better try two chains with 3 leds each. Otherwise plan to step up the supply voltage.

 

[betwixt]

Other than the lack of voltage as mvs sarma pointed out, the AD149 is a germanium device and known for it's high leakage current. Although it may work, the chances are the LEDs will never fully turn off.

Brian.

 

[mvs sarma]

Other than the lack of voltage as mvs sarma pointed out, the AD149 is a germanium device and known for it's high leakage current. Although it may work, the chances are the LEDs will never fully turn off.

Brian.

perhaps AAD149 can deliver current of the order of 3 Amps while SK100 may not. perhaps the OP can try some better device like BD140 that could deliver at least 1Amp odd.

 

[rajaram04]

kk sir thanks fot he ifo i got the above explained . . now i am using 3 battery sources 4 volts 2 amps of each to make a total of 12 volts , is that ok for such circuit arrangments which we modified above ??????????

 

[mvs sarma]

Please indicate the wattage of the LEDs used and the current you want to drive in for each chain. what ever, you need to have only 3 LEDs that too for a very short while before the battery drops to lower voltage. Even three are doubtful as the drop of three LEDs works around 11.1V dropping resistor should have margin depending on current need.
a generic formula can be used
I assume linear operation. forgetting the transistors and their need. I also indicate the formula for one series chain with n LEDS could be
I = [Vsupply-(n*Vf)]/Rs thus
Vsupply > (I*Rs)+(n*Vf)
Rs=[Vsupply-(n*Vf)]/I
Itotal would be i*x where x is number of such chains.

 

[betwixt]

Rajaram, you are missing what we are teling you: If each LED takes say 3.7V, when you wire 5 of them in series it takes (3.7 x 5) = 18.5V to operate them. Unless you can increase the supply to about 20V you have no option but to reduce the number of LEDs. We are suggesting you use no more than 3 LEDs although 2 would be better and increase the number of chains of them, each with it's own series resistor to make up the total number you need.

Brian.

 

[rajaram04]

Rajaram, you are missing what we are teling you: If each LED takes say 3.7V, when you wire 5 of them in series it takes (3.7 x 5) = 18.5V to operate them. Unless you can increase the supply to about 20V you have no option but to reduce the number of LEDs. We are suggesting you use no more than 3 LEDs although 2 would be better and increase the number of chains of them, each with it's own series resistor to make up the total number you need.

Brian.

ok brian sir lets forget the LED chain mentioned in the diagram & think of a simple voltage supply of 12v with 3 batteries of 4v & 1amp each with BD 140 as a swtich in the same circuit to drive some simple light duty load xyz , now is that BD 140 able to handle heavy current ?

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Please indicate the wattage of the LEDs used and the current you want to drive in for each chain. what ever, you need to have only 3 LEDs that too for a very short while before the battery drops to lower voltage. Even three are doubtful as the drop of three LEDs works around 11.1V dropping resistor should have margin depending on current need.
a generic formula can be used
I assume linear operation. forgetting the transistors and their need. I also indicate the formula for one series chain with n LEDS could be
I = [Vsupply-(n*Vf)]/Rs thus
Vsupply > (I*Rs)+(n*Vf)
Rs=[Vsupply-(n*Vf)]/I
Itotal would be i*x where x is number of such chains.

no sarma sir no idea about this LED's wattage etc as i found it on net & in the diagram nothing more than that is mentioned what i ve posted in post #1 , but i am searching the related link & may be we could find some more details there but hopefully here you have mentioned so many things on which i ve to make study . . thanks
But basically i am modifing the unit for diffrent purpose but the base would be same there so i need to know about BD140 current handling nature

 

[betwixt]

The BD140 is rated at 1.5A max collector current and 8W maximum dissipation.

I doubt your LEDs take that much, especially with 470 Ohm resistors in series with them so you should be OK in respect of it's maximum Ic rating. However, it will warm up and could overheat although I would say the risk is low. The power it will dissipate is the voltage across the emitter and collector multiplied by the current through it but without more information on the LEDs it's difficult to make a calculation.

The circuit isn't very well designed, apart from the LED voltage issues, when the first transistor turns off, it leaves the base of the second transistor floating. In practice, this could result in the LEDs still glowing slightly when they should be completely off.

Brian.

 

[Audioguru]

I have an AD149 that is about 48 years old! I will sell it for $100.00 or more. It was a class-A audio amplifier in an AM car radio.

The next car radio was AM/FM and used complementary AD161/AD162 germanium output transistors in a class-AB audio amplifier driving a stepup transformer that drove an 8 ohm speaker to 4W.

 

[mvs sarma]

i have one. can offer free of cost
but Packing and postage to be borne by the taker

 

[rajaram04]

The BD140 is rated at 1.5A max collector current and 8W maximum dissipation.

I doubt your LEDs take that much, especially with 470 Ohm resistors in series with them so you should be OK in respect of it's maximum Ic rating. However, it will warm up and could overheat although I would say the risk is low. The power it will dissipate is the voltage across the emitter and collector multiplied by the current through it but without more information on the LEDs it's difficult to make a calculation.

The circuit isn't very well designed, apart from the LED voltage issues, when the first transistor turns off, it leaves the base of the second transistor floating. In practice, this could result in the LEDs still glowing slightly when they should be completely off.

Brian.

o yeah i was in a doubt that circuit configurations are not very much biased . . speacially second transistor , well i ll have to make practicals with new idea depending on all your explanations & keeping other people's info in mind , let see if we could get something diffrent as i ve to introduce IR devices & related arrangments . . lets start with hope . . thanks

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hehehe we don't have to think on how much old the devices are , Actually in other sence we can compare things here with "old wine & its effect" & whats our motive truly ? Just to make impossible things possible with easy availablity low cost & reasonable speed then we ll have to do by hook or crook , no matter . . cheer up . . we are designing a new idea again & ll proceed from here . . Surely it ll defeat the previous one . . somebody told me about the above design that its from EFY . . so lets see