• Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

Emitter follower as current booster

Status
Not open for further replies.

Roff

Well-Known Member
As shown below, the output of the differentiator is only -ve going spikes which are already less than Vcc/3. Hence 4Vpp wouldn't be a problem.
You don't understand.
Pin 2 is normally at 10V, held there by the 10k resistor in your full schematic (with the 555). Pin 2 must go below 3.33V in order to trigger the 555. 10V-3.33V=6.67V. That is why I said you need a pulse that is at least 7V p-p.
 

vinodquilon

Member
You don't understand.
Pin 2 is normally at 10V, held there by the 10k resistor in your full schematic (with the 555). Pin 2 must go below 3.33V in order to trigger the 555. 10V-3.33V=6.67V. That is why I said you need a pulse that is at least 7V p-p.
OK I got it.
 

vinodquilon

Member
If I have exact 5V at top of ladder (by the use of zener to clamp 555 output at exact 5V), then resistors of the ladder should be have 0.01% tolerance levels for equal divisions of this 5V.
My doubt is then the 50K in parallel with them could change the balanced condition of the ladder. Thus no equal voltage divisions takes place. Can I avoid this by placing 100K or 1M ohms (>> 1K ladder) in place of 50K without disturbing both ladder and the op-amp.

If I avoid POT and connect op-amp directly to ladder , then floating condition would be resulted between switching intervals of ladder outs.
 

Roff

Well-Known Member
If I have exact 5V at top of ladder (by the use of zener to clamp 555 output at exact 5V), then resistors of the ladder should be have 0.01% tolerance levels for equal divisions of this 5V.
My doubt is then the 50K in parallel with them could change the balanced condition of the ladder. Thus no equal voltage divisions takes place. Can I avoid this by placing 100K or 1M ohms (>> 1K ladder) in place of 50K without disturbing both ladder and the op-amp.

If I avoid POT and connect op-amp directly to ladder , then floating condition would be resulted between switching intervals of ladder outs.
You can buy rotary switches, and possibly analog multiplexers, that have make-before-break switching. Are you using jumpers to switch the voltage divider? If so, then I think I would do it like this.
See the attachment.
 

Attachments

vinodquilon

Member
You can buy rotary switches, and possibly analog multiplexers, that have make-before-break switching. Are you using jumpers to switch the voltage divider? If so, then I think I would do it like this.
See the attachment.
Thanks Roff.

1. How Q2 can ensure that the top of the ladder truly goes to zero when output of 555 is low ? As it is in emitter follower mode, the top of the ladder (emitter) directly follows the Base input.
Here base input is (out of 555 + 0.7V). Thus at the top of ladder voltage is (out of 555 +0.7V - 0.7V)= out of 555.

Based on the above fact, top of the ladder following even the small changes at the output of 555. That is below 0.5V. (lower limit of required output pulse)

2. I think
Q1=Q2= 2N2222
U2=U3=LM324 or MC3407x
R13= 1K(and load is across 1K)
R12= 22R
 

Roff

Well-Known Member
Thanks Roff.

1. How Q2 can ensure that the top of the ladder truly goes to zero when output of 555 is low ? As it is in emitter follower mode, the top of the ladder (emitter) directly follows the Base input.
Here base input is (out of 555 + 0.7V). Thus at the top of ladder voltage is (out of 555 +0.7V - 0.7V)= out of 555.

Based on the above fact, top of the ladder following even the small changes at the output of 555. That is below 0.5V. (lower limit of required output pulse)

2. I think
Q1=Q2= 2N2222
U2=U3=LM324 or MC3407x
R13= 1K(and load is across 1K)
R12= 22R
Without the emitter follower on U3, the top of the ladder could only go as low as the saturation voltage of U3, which will always be greater than zero volts, especially if you are using an LM324. With the emitter follower, U3 can go low enough to completely cut off the emitter follower, assuring 0V on the top of the ladder. You still have to use an op amp that can swing within ≈0.5V of ground.
 

vinodquilon

Member
You can buy rotary switches, and possibly analog multiplexers, that have make-before-break switching. Are you using jumpers to switch the voltage divider? If so, then I think I would do it like this.
See the attachment.
I think I can simply replace CMOS 555 with 72121 0r 74123 in your scheme without disturbing supply, input, output and even trigger sections.
 

Roff

Well-Known Member
Why you are struck with CMOS ----555 CMOS & CMOS 74123.
Because CMOS logic outputs swing rail-to-rail. If you connect 5.000V and 0.000V to the supply pins, the outputs will swing between those voltages, providing there is no resistive load on them. TTL outputs do not swing rail-to-rail, nor does the output of a bipolar 555.
 

vinodquilon

Member
Without the emitter follower on U3, the top of the ladder could only go as low as the saturation voltage of U3, which will always be greater than zero volts, especially if you are using an LM324. With the emitter follower, U3 can go low enough to completely cut off the emitter follower, assuring 0V on the top of the ladder. You still have to use an op amp that can swing within ≈0.5V of ground.
Output of CMOS 555 is only 4.7V at 5V supply ! But I want 5V at output.

Can you suggest CMOS equivalent of 74121 ? I think CD4047 outputs 5V(typical with min value of 4.95V) at 5V VDD supply.
 
Last edited:

vinodquilon

Member
Can I use any resistor network ICs to simulate the role of good tolerance resistance ladder ?

Can I use any analog MUXs to switch voltage levels from 0 to 5V in 0.5V steps between ladder and output driver transistor ?

Can I use Make-Before-Break contact having NOT latch instead of debouncer using capacitor across the switch as shown below?
 

Attachments

audioguru

Well-Known Member
Most Helpful Member
I got it from the datasheet itself.
The datasheet shows a small voltage drop when the output has a load current! But your Cmos 555 is feeding the extremely low input current of an opamp and the opamp might have Fet inputs which have NO input current.

The datasheets say:
1) The LMC555 has a typical output of 4.7V when its load is 2mA.
2) The TLC555 has a typical output of 4.8V when its load is 1mA.
3) The ICM7555 has a typical output of 4.3V when its load is 0.8mA.
Their outputs are rail-to-rail when there is no load current.
 
Status
Not open for further replies.

Latest threads

EE World Online Articles

Loading
Top