bjt led driving configuration differences

[seyyah]

What are the differences of the two configurations in the image? What are the drawbacks or advantages?

 

[cameo_2007]

A:
advantage- The supply voltage is entirely coming across the LED
disdv - transistor acts as an inverter.When i/p is 0,LED ON;when i/p is 1,LED OFF
B:
adv:no inverting happens.if i/p is 1,led ON and vice versa
disadv:The complete Vdd is not sourced to the LED as there will be voltage drop across CEjunction

 

[nguyennam]

What are the differences of the two configurations in the image? What are the drawbacks or advantages?

I think there is fault in your circuits: it must be the n-p-n transistor in the first circuit, and the p-n-p in the second.

If that, so these 2 circuits are identical, only different from each other by the polarity of input control voltage.

The first needs the high-level (with respect to the minus terminal of the battery) while the second needs the low-level control voltage to switch the lamp on.

However, the second circuit is more preferrable in practice, for example, if you place it on a car, then the lamp-holder can be installed directly (without isolation) on the chassis which in the most cases connected to the minus terminal of the battery.

You can use the second circuit with the n-p-n transistor, but you have to apply the input control voltage higher than the supply voltage from the battery. That is impractical.

nguyennam

 

[seyyah]

A:
advantage- The supply voltage is entirely coming across the LED
disdv - transistor acts as an inverter.When i/p is 0,LED ON;when i/p is 1,LED OFF
B:
adv:no inverting happens.if i/p is 1,led ON and vice versa
disadv:The complete Vdd is not sourced to the LED as there will be voltage drop across CEjunction

You are wrong i think as simulation approves. Both LEDs are on when transitor is on (or base is triggered).

Added after 1 hours:

What are the differences of the two configurations in the image? What are the drawbacks or advantages?

I think there is fault in your circuits: it must be the n-p-n transistor in the first circuit, and the p-n-p in the second.

If that, so these 2 circuits are identical, only different from each other by the polarity of input control voltage.

The first needs the high-level (with respect to the minus terminal of the battery) while the second needs the low-level control voltage to switch the lamp on.

However, the second circuit is more preferrable in practice, for example, if you place it on a car, then the lamp-holder can be installed directly (without isolation) on the chassis which in the most cases connected to the minus terminal of the battery.

You can use the second circuit with the n-p-n transistor, but you have to apply the input control voltage higher than the supply voltage from the battery. That is impractical.

nguyennam

I know that i must use pnp in second conf. to activate the transitors by using different polarites. But i don't ask that. Both configuration works. Second configuration needs lower base current and has a bit more current gain. But VCE is much more than the first configuration and the transistor is not in the saturation region. This may be a disadvantage since it will dissipate more power.

I don't see any reason that, in the second configuration, a higher voltage is needed to turn it on than the supply voltage as you suggested. Can you explain it?

Somebody who has experince about these two conf. had said me that the second conf. causes led to blink sometimes. What may be the reason of this?

 

[nhoj]

hi there

this is my analysis...
the blinking of the LEDs are dependent to the base voltage or the input. if you have a pusle in the input, then the LED will blink with respect to the time of the pulse.

there's no exact difference in both configuration for that particular application of turning the LED on. why? when a high logic level is applied at the base, the base to emitter diode is on, since its an NPN, current will flow from the collector to the emitter, the emitter is connected to negative supply where we obtain a complete loop. of course the the LED turns on...

ok, when we apply low input at the base, the base to emitter diode is off, then no current will flow from collector to emitter. there's no complete loop, the LED is off

now, the difference on both configuration is where will you place your output: across of which transistor terminal, is it on the collector or on the emitter.

 

[nguyennam]

A:
advantage- The supply voltage is entirely coming across the LED
disdv - transistor acts as an inverter.When i/p is 0,LED ON;when i/p is 1,LED OFF
B:
adv:no inverting happens.if i/p is 1,led ON and vice versa
disadv:The complete Vdd is not sourced to the LED as there will be voltage drop across CEjunction

You are wrong i think as simulation approves. Both LEDs are on when transitor is on (or base is triggered).

Added after 1 hours:

What are the differences of the two configurations in the image? What are the drawbacks or advantages?

I think there is fault in your circuits: it must be the n-p-n transistor in the first circuit, and the p-n-p in the second.

If that, so these 2 circuits are identical, only different from each other by the polarity of input control voltage.

The first needs the high-level (with respect to the minus terminal of the battery) while the second needs the low-level control voltage to switch the lamp on.

However, the second circuit is more preferrable in practice, for example, if you place it on a car, then the lamp-holder can be installed directly (without isolation) on the chassis which in the most cases connected to the minus terminal of the battery.

You can use the second circuit with the n-p-n transistor, but you have to apply the input control voltage higher than the supply voltage from the battery. That is impractical.

nguyennam

I know that i must use pnp in second conf. to activate the transitors by using different polarites. But i don't ask that. Both configuration works. Second configuration needs lower base current and has a bit more current gain. But VCE is much more than the first configuration and the transistor is not in the saturation region. This may be a disadvantage since it will dissipate more power.

I don't see any reason that, in the second configuration, a higher voltage is needed to turn it on than the supply voltage as you suggested. Can you explain it?

Somebody who has experince about these two conf. had said me that the second conf. causes led to blink sometimes. What may be the reason of this?

Well, for the case of circuit 2 with n-p-n transistor:
The best working status for transistors in both figures is like a switch, they will switch off the LED with not any current (of course we do not talk about the collector leakage current) or switch on with maximum current (defined by resistor), at that time transistor is in saturation mode - Vce-sat = 0V.

So, you only can bring the transistor into saturation mode with the input voltage of about 0.7V (Vbe-sat = 0.7V) higher than the power supply voltage.

With the input voltage = or < power supply voltage, the transistor must open in amplifying mode, meanings that Vce is about a few V, that makes transistor heated, LED maybe dimmed, or brightness changed, or even blinking if the circuit is not stable due to low quality transistor, or over-heated transistor.

With transistor working in switch mode, hFE gain value does not effect much on the circuit features, so you cannot see any difference in 2 circuits.

nguyennam