Emitter follower as current booster
Re: Emitter follower as current booster
LM317 should have a load of at least 10mA to ensure regulation. The 555 doesn't draw much current.
Just add a 1k from the LM317 output to ground. That will give you 5mA through the 1KΩ resistor, and another ≈5mA through the 220Ω, for a total of slightly more than 10mA.
Just add a 1k from the LM317 output to ground. That will give you 5mA through the 1KΩ resistor, and another ≈5mA through the 220Ω, for a total of slightly more than 10mA.
@VinodQuilon
Re: Emitter follower as current booster
10ma is the maximum rating for the LM317L, 5ma for the LM317L-N. Typical current to maintain regulation is 3.5ma for both. The current draw is 5.6ma for an output of 5 volts which includes the current through the adjustment pin and resistor divider of R1,R2 and R3. I would connect the 5 volt supply to the CD40106 as well.LM317 should have a load of at least 10mA to ensure regulation
Steve G
Re: Emitter follower as current booster
Hello,
Just to note, the 7805 type regulators have rather poor temperature characteristic, and LM317's are better, but better yet is to use a voltage reference diode based regulator. This of course depends on how much you expect the temperature to change.
Just to note, the 7805 type regulators have rather poor temperature characteristic, and LM317's are better, but better yet is to use a voltage reference diode based regulator. This of course depends on how much you expect the temperature to change.
LEDs vs Bulbs, LEDs are winning.
Re: Emitter follower as current booster
vinod said:
Nice! I like it.
Steve G
final circuit using 4047
Nice! I like it.
Steve G
Re: Emitter follower as current booster
I think 820R of LM317T should be replaced with 620R.sghioto wrote:You can but I would use a LM317L with the values chosen to give a more finite adjustmentCan I use 7805 with the selected values of resistances to get the adjustable voltage around 5-7V.
Steve G
From the datasheet of LM317T (I'm ignoring Iadj):
Vout=Vref*(R2/R1+1)
R2=R1*(Vout/Vref-1)
R1=220, Vref=1.25, Vout=5
R2=660
Since pot (Rpot) is 100Ω, R2 should be made up of 610Ω +Rpot/2.
The nearest standard 5% value is 620Ω.
@VinodQuilon
component swapping
I have made the following changes to my circuit as the availability of components. Does it affect the circuit performance ?sghioto wrote:
Nice! I like it.
Steve G
1. 20K tuning resistor at 4047
2. HCF 4047 BE as monoshot
3. LM 317 T as regulator
4. HCF 40106 BE as debouncer
5. 620 ohm instead of 820 ohm at regulator (as per design equation)
6. 1K is added at output of LM 317 T as it unlike LM 317L require minimum 10mA current at load to ensure regulation
@VinodQuilon
Re: Emitter follower as current booster
1. The actual value needed for 1ms pulse is close to 4K. A 20K will work but a 5K might be easier to adjust. Or you can connect a 6.2K resistor across the 20K pot. That will set the control near the middle of it's range to get close to 4K, if using a linear pot.1. 20K tuning resistor at 4047
2. HCF 4047 BE as monoshot
3. LM 317 T as regulator
4. HCF 40106 BE as debouncer
5. 620 ohm instead of 820 ohm at regulator (as per design equation)
6. 1K is added at output of LM 317 T as it unlike LM 317L require minimum 10mA current at load to ensure regulation
2. The 4047 is a good choice. I like the fact that the output pulse is not effected by the duration of the trigger pulse.
3. OK. A lot of overkill but if that's what you have on hand might as well use it.
4. OK. Just about any inverter chip will work
5. Yes that is the correct value needed.
6. OK.
Steve G
Re: Emitter follower as current booster
That is great.sghioto wrote: 1. The actual value needed for 1ms pulse is close to 4K. A 20K will work but a 5K might be easier to adjust. Or you can connect a 6.2K resistor across the 20K pot. That will set the control near the middle of it's range to get close to 4K, if using a linear pot.
Steve G
1. I have changed 10V supply to 15V as I am using this monoshot to source one package which is powered by 15V supply so that I can use same 15V supply to power both monoshot and the package.
2. Can I use 1W ratings for elements connected to collector and emitter arms of 2N2222 ?
Which one can I use as replacement for 1K/1W across the load ?
@VinodQuilon
Re: Emitter follower as current booster
vinod said:
The 1K resistor only needs to be .125 (1/8th) watt.
Steve G
In that case I would change the 22 ohm resistor to a 47 ohm 1w, to drop the dissipation across the 2N2222. Otherwise you will exceed the device rating of the transistor when using the 33 ohm load at 5 volts. But since the pulses are only 1ms the resistors won't even know the difference. So you can keep the load resistors at 1w.I have changed 10V supply to 15V
The 1K resistor only needs to be .125 (1/8th) watt.
Steve G
Re: Emitter follower as current booster
Vce sat @ 150mA for 2N2222 is 0.4Vsghioto wrote: In that case I would change the 22 ohm resistor to a 47 ohm 1w, to drop the dissipation across the 2N2222. Otherwise you will exceed the device rating of the transistor when using the 33 ohm load at 5 volts. But since the pulses are only 1ms the resistors won't even know the difference. So you can keep the load resistors at 1w.
The 1K resistor only needs to be .125 (1/8th) watt.
Steve G
Transisor dissipation will be (Vce * 152mA)= 60mW
The dissipation across collector resistor is [(15-(5+Vce))V * 152mA]= 1.459W [2.22W during 0V at load]
Load dissipation is 5V * 152mA=0.760W (let me to consider most worse case)
So I will prefer 2W ratings for 22 Ohm(1W/47 Ohm may be replaced with) and 1W ratings for Loads with 2N2222 as driver. 1/8W for 1K as less portion of 152mA flows through it.
@VinodQuilon
Re: Emitter follower as current booster
[quote]Vce sat @ 150mA for 2N2222 is 0.4V
Transisor dissipation will be (Vce * 152mA)= 60mW
Not correct. The transistor is not in saturation. The current through the load resistor is the same as through the 22 ohm. So 22 ohms X 152ma = 3.34 volts. Therefore the voltage across the 2N2222 is 15 volts minus 5 volts (worst case across the load) and minus the 3.34 volts or 6.66 volts and the dissipation is just over 1 watt but again for only 1ms.
Steve G
Transisor dissipation will be (Vce * 152mA)= 60mW
Not correct. The transistor is not in saturation. The current through the load resistor is the same as through the 22 ohm. So 22 ohms X 152ma = 3.34 volts. Therefore the voltage across the 2N2222 is 15 volts minus 5 volts (worst case across the load) and minus the 3.34 volts or 6.66 volts and the dissipation is just over 1 watt but again for only 1ms.
Steve G
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