Emitter follower as current booster

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vinod
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Emitter follower as current booster

Post by vinod »

Consider the following Common Collector & Darlington emitter follower circuits,

CASE 1:
Let V(in) = 5V
Vcc(supply) = 10V
Emitter(load) resistance = 100 ohm
thus, V(out) should be 5V (following input)
Corresponding Load current will be 5/100 = 50 mA

CASE 2:
Let V(in) = 5V
Vcc(supply) = 10V
Emitter(load) resistance = 33 ohm
thus, V(out) should be 5V (following input)
Corresponding Load current will be 5/33 = 152 mA

Which source will provide this additional current requirement [Vcc or V(in)] ?
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sghioto
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Re: Emitter follower as current booster

Post by sghioto »

Will depend a lot on the gain ( hfe ) of the xistors used. Obviously the darlington pair will present a much better gain but the output will not follow the input because of the voltage drop through the junctions. You'll lose about 1.2 volts through the darlington and about .6 volts through the single xistor stage. This seems to be related to the 555 pulse circuit. The LM324 opamp regulator in that circuit is the best solution.

Steve G
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vinod
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Re: Emitter follower as current booster

Post by vinod »

sghioto wrote:The LM324 opamp regulator in that circuit is the best solution.

Steve G
What would be the output if I apply 0.5V which is less than junction drop 0.7V to Op Amp input ? Can I get same 0.5V at emitter ?
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sghioto
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Re: Emitter follower as current booster

Post by sghioto »

What would be the output if I apply 0.5V which is less than junction drop 0.7V to Op Amp input ? Can I get same 0.5V at emitter ?
Yes you will get .5 volt at the emitter. You can go down below .1 volt at the op amp input and still get the same on the emitter. :smile:
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MrAl
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Re: Emitter follower as current booster

Post by MrAl »

Hi,

A one transistor emitter follower provides a current gain only and no voltage gain. In fact, the voltage at the emitter will typically be less than the base voltage by about 0.7 volts or so, but it will 'follow' the base voltage in that if the base voltage is 2v and the emitter voltage is 1.3v, if the base voltage increases to 3v then the emitter voltage will increase to 2.3 volts approximately, provided the transistor has enough current gain to handle the load at the emitter to ground. If we apply a base voltage of 2 volts and it provides 1ma and the transistor Beta is 50, then we can get up to maybe 50ma output but no more if we expect the emitter voltage to 'follow' the base voltage.

There is a small portion of the emitter current that is due to the base current itself. So if we apply 1ma to the base the emitter current will include this 1ma plus the collector current, and the collector current is equal to the base current times the Beta. So to be perfectly theoretical with the calculation, we would get the 1ma base current plus the 50ma collector current which totals 51ma through the emitter to the load.
But that's only if the load draws that much. If the load is lighter, we may only get 25ma or even less depending on the load resistance at the emitter to ground. If we had a 100 ohm resistor there and 2v at the emitter we'd see 20ma of course, not 51ma. And that would be a better choice for the emitter current because we dont want to design the circuit using the full potential of the transistor (the Beta spec) we want to design at an operating point less than that. 20ma is less than 51ma so we should be ok.

If we want more current sometimes we go to a Darlington set up using two transistors, where the emitter of the first transistor drives the base of the second transistor. This provides more current gain so we can use a smaller resistor on the emitter of the second transistor. The total gain is the product of the two transistor gains, Beta1 times Beta2, so we get a huge increase in current gain. The drawback is that now we have two emitters in series so we loose more voltage from the base to emitter. If one transistor drops 0.7v then two will drop around 1.4v, so went we apply 2v to the base now we only get 0.6v out of the emitter of the second transistor. Everything else works pretty much the same however, although we might connect a base resistor to ground on the second transistor to make sure it can turn off fully if we need it to turn off completely during normal operation.
Usually when we use this kind of set up the current gain is more important so we dont mind loosing a little more voltage. If the voltage is also important, we have to switch to a different topology.
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vinod
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Output is not following the input

Post by vinod »

I have setup the Op Amp-Current booster circuit as shown below using LM324 & 2N2222 for loads 33/68/100 Ohms one at a time.

But the output voltage is not following the input as shown below:

INPUT(40mA max)// EMITTER VOLTAGE// DEVIATION //SUPPLY(10V/200mA max)

0 // 0 // 0 // 10,< 100mA

0.5 // 0.4 // 0.1 // 10,< 100mA

1 // 0.9 // 0.1 // 10,< 100mA

1.5 // 1.3 // 0.2// 10, 100mA

2 // 1.8 // 0.2 // 10, 100mA

2.5 // 2.3 // 0.2// 10, 100mA

3 // 2.7 // 0.3// 10, 100mA

3.5// 3.2 // 0.3// 10, 100mA

4// 3.6 // 0.4 // 10, 200mA

4.5 // 4 // 0.5 // 10, 200mA

5 // 4.3 // 0.7 // 9.8, 200mA

5.5 // 4.3 // 0.7 // 9.8, 200mA

6 // 4.3 // 0.7 // 9.8, 200mA

6.5 // 4.3 // 0.7 // 9.8, 200mA

7 // 4.3 // 0.7 // 9.8, 200mA

7.5 // 4.3 // 0.7 // 9.8, 200mA

8 // 4.3 // 0.7 // 9.8, 200mA

8.5 // 4.3 // 0.7 // 9.8, 200mA

9 // 4.3 // 0.7 // 9.8, 200mA

9.5 // 4.3 // 0.7 // 9.8, 200mA

10 // 4.3 // 0.7 // 9.8, 200mA

Anomalies:
1. Output is not following the input
2. After 5V input, the output is struck at 4.3V
3. Supply voltage dips to 9.8V when input is >= 5V
4. Deviation incraeses as the input voltage and set at 0.7 level
5. The measurement is repeated for inputs set at 10mA max & 100mA max. But the issue continues.
6. Can I use complementary emitter follower for my above application by leaving the Op Amp?
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vinod
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Re: Emitter follower as current booster

Post by vinod »

MrAl wrote:Hi,


Usually when we use this kind of set up the current gain is more important so we dont mind loosing a little more voltage. If the voltage is also important, we have to switch to a different topology.
Thanks for your detailed explanation,

For my application voltage is also important, that is I don't want to drop my voltage by 0.7v at output. Output should be same as input for three loads 33/68/100 ohms (one is connected at a time). What is your topology?
note: my input would be 0 to 5V in 0.5V steps, then I expect 0, 0.5,1.0,1.5,2.0,2.5,3.0,3.5,4.0,4.5, and 5.0 at outputs for all the loads(common supply voltage settings for all the three loads. i.e, 10V/160mA max).
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Re: Emitter follower as current booster

Post by sghioto »

I have tested the LM324 circuit using a 2N3904 for the output and it works perfectly. :???:
You must be doing something wrong. Check your components and connections again.
Are you sure the power supply is OK?

Steve G
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vinod
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Re: Emitter follower as current booster

Post by vinod »

sghioto wrote:I have tested the LM324 circuit using a 2N3904 for the output and it works perfectly. :???:
You must be doing something wrong. Check your components and connections again.
Are you sure the power supply is OK?

Steve G
1. OK, I will check it and report on this forum
2. Can I use the same Op Amp-2N2222 settings for all three loads 33/68/100 ohms only by setting the supply to its max requirement of above three cases i.e, 10V, 160mA current limit for all the cases.
Thus I can simply swap loads without any settings change while getting output which follows input. :grin:
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Re: Emitter follower as current booster

Post by sghioto »

Yes. I would set the current limit higher at 200 ma for all load resistors. Just make sure the power supply is maintaining regulation at that current

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vinod
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Re: Emitter follower as current booster

Post by vinod »

I am using 1ms single pulses from monoshot(with 0.5v stepped out) as the input to the op-amp and monitoring voltage at emitter on CRO.


Can OP-AMP & transistor can track the 1ms pulse change?
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Re: Emitter follower as current booster

Post by sghioto »

Can OP-AMP & transistor can track the 1ms pulse change?
Yes! No problem. :smile:

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I got it !

Post by vinod »

sghioto wrote:I have tested the LM324 circuit using a 2N3904 for the output and it works perfectly. :???:
You must be doing something wrong. Check your components and connections again.

Steve G
As sghioto told, it was my mistake. I got interchanged the collector and emitter.Thanks man. Now I rectified the problem and the output follows input.

See the attachment for my project.

1. First I have implemented LM324-2N2222 portion without R9 and 50K on a bread board

2. Set supply at 10V with 200mA current limit

3. Gave inputs to the 3rd pin of Op-Amp using a dc source and varied the dc volts in 0,0.5,1,1.5,2,2.5,3,3.5,4,4.5,& 5 to simulate variable mono pulses

4. WoW! :grin: output following the input with saturation occurs at 7.77V

5. Connected the 100R resistor network as shown in the figure with one end at ground and other end connected to dc source(set at 5V) mentioned in 3rd point to simulate the mono pulse

6. Connected voltage legs of the ladder directly to 3rd pin of Op-Amp without 50K.

7. WoW! output following the input for 0,0.5,1,1.5,2,2.5,3,3.5,4,4.5,& 5 at different positions of the resistor ladder

8. Tommorow I am going to implement 1ms 555 monoshot or 1ms 74123 monoshot(additional 7805 IC needed to power it) on separate bread board and will cascade all three sections(1ms monoshot + Resistor ladder + Op-Amp_BJT) to form the final project. Hope it will work fine.

ANOMALIES OBSERVED: :x

1. 50K is rejected intentionally and connected the voltage legs of the ladder to the Op-Amp directly. This is to avoid the disturbance to the ladder resistors when 50K comes in parallel to them as I want exact 0.5V divisions of 5V.
But this causes the emitter volatge to goes to saturation(7.77V) when 3rd pin of Op-Amp not connected to any points.(floats)
To avoid this can I connect a high value resistor like 100K between 3rd pin and ground permanently without disturbing both ladder and Op amp.

2. 1ms pulses are almost too short for a lousy old LM324 opamp.
Which Op Amp should I use to transfer 1ms varying amplitude pulses to load fluently with 0 to 5 volt in/out capacity ?
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sghioto
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Re: Emitter follower as current booster

Post by sghioto »

Fantastic I'm glad to see you figured it out.
A 100K or even a 1meg resistor from pin #3 to ground will suffice.
I don't see anything wrong with using a LM324. They are good to almost a megahz, and cheap.

Steve G
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vinod
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Re: Emitter follower as current booster

Post by vinod »

sghioto wrote:Fantastic I'm glad to see you figured it out.

I don't see anything wrong with using a LM324. They are good to almost a megahz, and cheap.

Steve G
But I use 1ms single pulses produced by pressing the trigger switch once and next pushing may or may not be done as per the requirement of output pulse randomly.

That is there is no specific period for the pulses and the differentiator for triggering is designed for choosen period of 625ms >> 1ms.
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