H-Bridge Schematic Correct?

Thread Starter

tracecom

Joined Apr 16, 2010
3,940
In your LED test circuit, your transistor saturates fairly well with a 100k base resistor. But a weaker (but passing) 2N2222A transistor needs a 12k base resistor.
I am not disagreeing with you, but am trying to learn. Would you please show the math or the reference on the datasheet that provides that resistor value?

Thanks.

ETA: Is this is it?

Ic = .0089 A
Ib = .0089 / 10
Ib = .00089A

E=IR
12.03 = .00089R
R = 13517
 
Last edited:

Audioguru

Joined Dec 20, 2007
11,251
I am not disagreeing with you, but am trying to learn. Would you please show the math or the reference on the datasheet that provides that resistor value?

Thanks.

ETA: Is this is it?

Ic = .0089 A
Ib = .0089 / 10
Ib = .00089A

E=IR
12.03 = .00089R
R = 13517
Correct.
Use 12k or 15k.
 

SgtWookie

Joined Jul 17, 2007
22,201
Actually, the generic formula for a base resistor when using it as a saturated switch is:
Rbase = (Vin - Vbe) / (Ic / 10)
where:
Vin = the voltage applied to the resistor opposite the base

Vbe = the voltage on the base when measured using the emitter as a reference; use 0.7v for low collector current (up to ~1/10th the rated collector current or so), use 0.8v for moderate collector current (up to ~1/4 rated collector current) and 0.9v for up to 1/2 of the rated collector current. If you need more than 1/2 of the rated collector current, you need a more capable transistor.

Ic = the desired collector current.

So, if you want 20mA collector current, and you have a 12v supply, then:
Rbase = (12v - 0.8v) / (20mA/10) = 11.2 / 0.002A = 5600 Ohms, which is a standard value of resistance.
Here is a decade table of standard resistance values; use the green E24 columns:
http://www.logwell.com/tech/components/resistor_values.html
You use the value of resistance that is closest to the result of your calculation.
 

Thread Starter

tracecom

Joined Apr 16, 2010
3,940
This thread has helped me quite a lot, but I still would like to understand what was happening in the attached circuit. As I previously described, I removed the LED current limiting resistor and varied the base resistance to control the collector current as shown in the attached table.

It still seems to me that I was operating the transistor as an amplifier and that the Ib to Ic ratio in the table is hFE. If it's not hFE, does it have a name?

Thanks for any further help.
 

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Audioguru

Joined Dec 20, 2007
11,251
This thread has helped me quite a lot, but I still would like to understand what was happening in the attached circuit. As I previously described, I removed the LED current limiting resistor and varied the base resistance to control the collector current as shown in the attached table.

It still seems to me that I was operating the transistor as an amplifier and that the Ib to Ic ratio in the table is hFE. If it's not hFE, does it have a name?
Since your transistor has plenty of collector to emitter voltage then its IC/IB is its hFE. Its hFE changes when the temperature changes and every transistor has a different amount of hFE.
 

Thread Starter

tracecom

Joined Apr 16, 2010
3,940
With regard to BJT saturation, I found the attached BJT Operation Mode information in a textbook.

I decided to test my circuit (shown attached) and placed the DMM leads across the collector base junction and began substituting the base resistor R1. At 220k, the CBJ was forward biased, but at 330k, the CBJ was reverse biased, so the transistor had gone into saturation at some intermediate resistance. If I had a suitable pot, I could determine exactly where, but for my present purpose it doesn't matter. What's important is that I know have a definitive way to determine whether or not the transistor is fully turned on.

With R1 = 330k, Vce = 3.23
With R1 = 220k, Vce = .45
 

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Audioguru

Joined Dec 20, 2007
11,251
What's important is that I know have a definitive way to determine whether or not the transistor is fully turned on.
No you don't.
Your latest circuit does not allow the transistor to saturate because the LED holds its collector voltage high. When a transistor is saturated then its collector to emitter voltage becomes low which reduces its hFE.

A transistor is saturated (fully turned on) when its collector to emitter voltage is low so the load gets most of the voltage. That occurs when the LED has a series current-limiting resistor. The resistor becomes part of the load.
 

Thread Starter

tracecom

Joined Apr 16, 2010
3,940
No you don't.
Your latest circuit does not allow the transistor to saturate because the LED holds its collector voltage high. When a transistor is saturated then its collector to emitter voltage becomes low which reduces its hFE.

A transistor is saturated (fully turned on) when its collector to emitter voltage is low so the load gets most of the voltage. That occurs when the LED has a series current-limiting resistor. The resistor becomes part of the load.
I was referring to the circuit with the LED current limiting resistor as shown in the attachment. I have edited post 29 to be more specific.


 

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Audioguru

Joined Dec 20, 2007
11,251
You forgot to show the collector to emitter voltages so we don't know if the transistor is saturated (fully turned on).
Its LED current is fairly low at only about 8.6mA or less so maybe you think the transistor is saturated.
 

Thread Starter

tracecom

Joined Apr 16, 2010
3,940
You forgot to show the collector to emitter voltages so we don't know if the transistor is saturated (fully turned on).
Its LED current is fairly low at only about 8.6mA or less so maybe you think the transistor is saturated.
With R1 = 330k, Vce = 3.23
With R1 = 220k, Vce = .45
 

Audioguru

Joined Dec 20, 2007
11,251
With R1 = 330k, Vce = 3.23
With R1 = 220k, Vce = .45
The datasheet shows that the typical Vce is only 0.3V max when it is saturated with a collector current of 150mA.
With less collector current like you have its saturation voltage loss is much less (if the base current is 1/10th the collector current).

It looks like your base current is not enough.
 

Thread Starter

tracecom

Joined Apr 16, 2010
3,940
The datasheet shows that the typical Vce is only 0.3V max when it is saturated with a collector current of 150mA.
With less collector current like you have its saturation voltage loss is much less (if the base current is 1/10th the collector current).

It looks like your base current is not enough.
Based on your previous advice, I am using a 12k base resistor which results in a Vce of .056. The 220k was based on the "textbook" method that I described in post 29. Although that method seems clear, I need to read more about it to make sure that I am not overlooking something. I also would like to find an explanation for the formula posted by SgtWookie in a textbook or reference book. (Yes, I know that I am seriously pedantic.:))
 

SgtWookie

Joined Jul 17, 2007
22,201
<snip>I also would like to find an explanation for the formula posted by SgtWookie in a textbook or reference book. (Yes, I know that I am seriously pedantic.:))
Look at any datasheet that plots in it, and you'll see a little notes on the plot s that say "Ic/Ib=10" or "Ib=Ic/10" or "Ic = 10 Ib"

Some examples:
http://www.onsemi.com/pub_link/Collateral/PN2222-D.PDF
http://www.onsemi.com/pub_link/Collateral/PN2907A-D.PDF
http://www.onsemi.com/pub_link/Collateral/MPSA05-D.PDF
http://www.fairchildsemi.com/ds/KS/KSH42C.pdf
http://www.fairchildsemi.com/ds/MJ/MJD31C.pdf
http://www.fairchildsemi.com/ds/KS/KSA1156.pdf
There are plenty more out there.

That's where the right-hand part of the equation comes from.

The left side of the equation over the right side is Ohm's Law; R=E/I.
The Vin is the voltage on the left side of the resistor.
Vbe is the voltage on the right side of the resistor.
So, (Vin - Vbe) is the voltage across the resistor for the E part.
Ic / 10 is the I part.

Now, if you start looking in physics textbooks, you will start seeing things like, "Look at the lowest hFE value and use that". The only time you would want to go that way is if the lowest hFE value was less than 10.

If you start skimping on the base current, you risk having your transistor come out of saturation. If that happens, for a given collector current your power dissipation in the transistor greatly increases; if you exceed the power rating of the transistor, you will destroy it.
 

thatoneguy

Joined Feb 19, 2009
6,349
This thread has helped me quite a lot, but I still would like to understand what was happening in the attached circuit. As I previously described, I removed the LED current limiting resistor and varied the base resistance to control the collector current as shown in the attached table.

It still seems to me that I was operating the transistor as an amplifier and that the Ib to Ic ratio in the table is hFE. If it's not hFE, does it have a name?

Thanks for any further help.
An amplifier wasn't created, as there wasn't a signal to amplify. What was made instead was a current sink. It was operating in the linear region, as it would in an amplifier, though the transistor was simply limiting the amount of current via the voltage drop across collector and emitter, since the base current was constant. Semantics, I guess.

When operating non-saturated, the transistor dissipates more heat, the less voltage across the Collector and Emitter, the less power is lost in the transistor as heat. Generally, the less heat a circuit produces, the more efficient it is, especially with power supplies and H-Bridges.
 
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